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23373
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\section{Way Above Closures Form Basis}
Tags: Topological Order Theory, Continuous Lattices
\begin{theorem}
Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.
Then $\set {x^\gg: x \in S}$ is an (analytic) basis of $L$.
\end{theorem}
\begin{proof}
Define $B = \set {x^\gg: x \in S}$.
Thus by Way Above Closure is Open:
:$B \subseteq \tau$
We will prove that:
:for all $x \in S$: there exists a local basis $Q$ of $x$: $Q \subseteq B$
Let $x \in S$.
By Way Above Closures that Way Below Form Local Basis:
:$Q := \set {g^\gg: g \in S \land g \ll x}$ is a local basis at $x$.
Thus by definition of subset:
:$Q \subseteq B$
{{qed|lemma}}
Thus by Characterization of Analytic Basis by Local Bases:
:$B$ is an (analytic) basis of $L$.
{{qed}}
\end{proof}
|
23374
|
\section{Way Above Closures that Way Below Form Local Basis}
Tags: Way Below Relation, Topological Order Theory, Continuous Lattices
\begin{theorem}
Let $L = \struct {S, \preceq, \tau}$ be a complete continuous topological lattice with Scott topology.
Let $p \in S$.
Then $\set {q^\gg: q \in S \land q \ll p}$ is a local basis at $p$.
\end{theorem}
\begin{proof}
Define $B := \set {q^\gg: q \in S \land q \ll p}$
By Way Above Closure is Open:
:$B \subseteq \tau$
By definition of way above closure:
:$\forall X \in B: p \in X$
Thus by definition:
:$B$ is set of open neighborhoods.
{{explain|open neighborhoods of what?}}
Let $U$ be an open subset of $S$ such that
:$p \in U$
By Open implies There Exists Way Below Element:
:$\exists u \in U: u \ll p$
Thus by definition of $B$:
:$u^\gg \in B$
By definition of Scott topology:
:$U$ is upper.
We will prove that
:$u^\gg \subseteq U$
Let $z \in u^\gg$
By definition of way above closure:
:$u \ll z$
By Way Below implies Preceding:
:$u \preceq z$
Thus by definition of upper set:
:$z \in U$
{{qed}}
\end{proof}
|
23375
|
\section{Way Below Closure is Directed in Bounded Below Join Semilattice}
Tags: Way Below Relation, Join and Meet Semilattices
\begin{theorem}
Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $x \in S$.
Then
:$x^\ll$ is directed.
\end{theorem}
\begin{proof}
By Bottom is Way Below Any Element:
:$\bot \ll x$
By definition of way below closure:
:$\bot \in x^\ll$
Thus by definition:
:$x^\ll$ is a non-empty set.
Let $y, z \in x^\ll$
By definition of way below closure:
:$y \ll x$ and $z \ll x$
By Join is Way Below if Operands are Way Below
:$y \vee z \ll x$
By definition of way below closure:
:$y \vee z \in x^\ll$
By Join Succeeds Operands:
:$y \preceq y \vee z$ and $z \preceq y \vee z$
Thus by definition
:$x^\ll$ is directed.
{{qed}}
\end{proof}
|
23376
|
\section{Way Below Closure is Ideal in Bounded Below Join Semilattice}
Tags: Way Below Relation, Join and Meet Semilattices
\begin{theorem}
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $x \in S$.
Then
:$x^\ll$ is ideal in $L$.
\end{theorem}
\begin{proof}
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$x^\ll$ is a non-empty directed set.
Let $y \in x^\ll, z \in S$ such that
:$z \preceq y$
By definition of way below closure:
:$y \ll x$
By definition of reflexivity:
:$x \preceq x$
By Preceding and Way Below implies Way Below:
:$z \ll x$
Thus by definition of way below closure:
:$z \in x^\ll$
Thus by definition
:$x^\ll$ is a lower set.
Thus by definition
:$x^\ll$ is an ideal in $L$.
{{qed}}
\end{proof}
|
23377
|
\section{Way Below Closure is Lower Set}
Tags: Way Below Relation, Join and Meet Semilattices
\begin{theorem}
Let $L = \struct {S, \vee, \preceq}$ be an ordered set.
Let $x \in S$.
Then
:$x^\ll$ is a lower set.
\end{theorem}
\begin{proof}
Let $y \in x^\ll, z \in S$ such that:
:$z \preceq y$
By definition of way below closure:
:$y \ll x$
By definition of reflexivity:
:$x \preceq x$
By Preceding and Way Below implies Way Below:
:$z \ll x$
Thus by definition of way below closure:
:$z \in x^\ll$
Thus by definition:
:$x^\ll$ is a lower set.
{{qed}}
\end{proof}
|
23378
|
\section{Way Below Relation is Antisymmetric}
Tags: Way Below Relation
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $x, y \in S$ such that
:$x \ll y$ and $y \ll x$
Then
:$x = y$
\end{theorem}
\begin{proof}
By Way Below implies Preceding:
:$x \preceq y$ and $y \preceq x$
Thus by definition of antisymmetry:
:$x = y$
{{qed}}
\end{proof}
|
23379
|
\section{Way Below Relation is Auxiliary Relation}
Tags: Way Below Relation, Auxiliary Relations
\begin{theorem}
Lrt $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Then
:$\ll$ is auxiliary relation
where $\ll$ denotes the way below relation.
\end{theorem}
\begin{proof}
By Way Below implies Preceding:
:$\forall x, y \in S: x \ll y \implies x \preceq y$
By Preceding and Way Below implies Way Below:
:$\forall x, y, z, u \in S: x \preceq y \ll z \preceq u \implies x \ll u$
By Join is Way Below if Operands are Way Below:
:$\forall x, y, z \in S: x \ll z \land y \ll z \implies x \vee y \ll z$
By Bottom is Way Below Any Element:
:$\forall x: \bot \ll x$
where $\bot$ denotes the smallest element in $L$.
Thus by definition:
:$\ll$ is auxiliary relation.
{{qed}}
\end{proof}
|
23380
|
\section{Way Below Relation is Multiplicative implies Pseudoprime Element is Prime}
Tags: Way Below Relation, Prime Elements
\begin{theorem}
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice such that
:$\ll$ is multiplicative relation
where $\ll$ denotes the way below relation of $L$.
Let $p \in S$.
Then $p$ is a pseudoprime element is a prime element.
\end{theorem}
\begin{proof}
Let $p$ be a pseudoprime element.
{{AimForCont}}
:$p$ is not a prime element.
By definition of prime element:
:$\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$
By definition of continuous:
:$\forall z \in S: z^\ll$ is directed.
and
:$L$ satisfies axiom of approximation.
By Axiom of Approximation in Up-Complete Semilattice:
:$\exists u \in S: u \ll x \land u \npreceq p$
and
:$\exists v \in S: v \ll y \land v \npreceq p$
By Way Below Relation is Auxiliary Relation:
:$\ll$ is auxiliary relation.
By Multiplicative Auxiliary Relation iff Congruent:
:$u \wedge v \ll x \wedge y$
By definition of transitivity:
:$u \wedge v \ll p$
By Characterization of Pseudoprime Element when Way Below Relation is Multiplicative:
:$u \preceq p$ or $v \preceq p$
This contradicts $u \npreceq p$ and $v \npreceq p$
{{qed}}
\end{proof}
|
23381
|
\section{Way Below Relation is Transitive}
Tags: Way Below Relation
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x, y, z \in S$ such that
:$x \ll y \ll z$
Then
:$x \ll z$
\end{theorem}
\begin{proof}
By Way Below implies Preceding:
:$x \preceq y$
By definition of reflexivity:
:$z \preceq z$
Thus by Preceding and Way Below implies Way Below:
:$x \ll z$
{{qed}}
\end{proof}
|
23382
|
\section{Way Below has Interpolation Property}
Tags: Way Below Relation, Continuous Lattices
\begin{theorem}
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.
Let $x, z \in S$ such that
:$x \ll z$
Then
:$\exists y \in S: x \ll y \land y \ll z$
\end{theorem}
\begin{proof}
Case $x \ne z$:
By Way Below has Strong Interpolation Property:
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
Thus
:$\exists y \in S: x \ll y \land y \ll z$
Case $x = z$:
Define $y = x$
Thus
:$x \ll y \land y \ll z$
{{qed}}
\end{proof}
|
23383
|
\section{Way Below has Strong Interpolation Property}
Tags: Way Below Relation, Continuous Lattices, Approximating Relations
\begin{theorem}
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.
Let $x, z \in S$ such that
:$x \ll z \land x \ne z$
Then
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
\end{theorem}
\begin{proof}
By Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:
:$\ll$ is an auxiliary approximating relation on $S$.
Thus by Auxiliary Approximating Relation has Interpolation Property:
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
{{qed}}
\end{proof}
|
23384
|
\section{Way Below if Between is Compact Set in Ordered Set of Topology}
Tags: Way Below Relation, Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $L = \struct {\tau, \preceq}$ be an ordered set
where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
Let $x, y \in \tau$ such that:
:$\exists H \subseteq S: x \subseteq H \subseteq y \land H$ is compact
Then:
:$x \ll y$
\end{theorem}
\begin{proof}
Let $D$ be a directed subset of $\tau$ such that:
:$y \preceq \sup D$
By proof of Topology forms Complete Lattice:
:$\ds y \subseteq \bigcup D$
By Subset Relation is Transitive:
:$\ds H \subseteq \bigcup D$
By definition:
:$D$ is open cover of $H$
By definition of compact:
:$H$ has finite subcover $\GG$ of $D$
By Directed iff Finite Subsets have Upper Bounds:
:$\exists d \in D: d$ is upper bound for $\GG$
By definitions of upper bound and of $\preceq$:
:$\forall z \in \GG: z \subseteq d$
By Union is Smallest Superset/Set of Sets:
:$\ds \bigcup \GG \subseteq d$
By definition of cover:
:$\ds H \subseteq \bigcup \GG$
By Subset Relation is Transitive:
:$x \subseteq d$
Thus by definition of $\preceq$:
:$x \preceq d$
Thus by definition of way below relation:
:$x \ll y$
{{qed}}
\end{proof}
|
23385
|
\section{Way Below iff Preceding Finite Supremum}
Tags: Way Below Relation
\begin{theorem}
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $x, y \in S$.
Then $x \ll y$ {{iff}}
:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
where
:$\ll$ denotes the way below relation,
:${\it Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$.
\end{theorem}
\begin{proof}
=== Sufficient Condition ===
Let $x \ll y$
Let $X \subseteq S$ such that
:$y \preceq \sup X$
Define $F := \left\{ {\sup A: A \in {\it Fin}\left({X}\right)}\right\}$
By definition of union:
:$X = \bigcup {\it Fin}\left({X}\right)$
By Supremum of Suprema:
:$\sup X = \sup F$
We will prove that
:$F$ is directed
Let $a, b \in F$
By definition of $F$:
:$\exists A \in {\it Fin}\left({X}\right): a = \sup A$
and
:$\exists B \in {\it Fin}\left({X}\right): b = \sup B$
By Union of Subsets is Subset:
:$A \cup B \subseteq X$
By Finite Union of Finite Sets is Finite:
:$A \cup B$ is finite
By definition of {\it Fin}:
:$A \cup B \in {\it Fin}\left({X}\right)$
By definition of $F$:
:$\sup\left({A \cup B}\right) \in F$
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
By Supremum of Subset:
:$a \preceq \sup\left({A \cup B}\right)$ and $b \preceq \sup\left({A \cup B}\right)$
Thus by definition
:$F$ is directed
By definition of way below relation:
:$\exists d \in F: x \preceq d$
Thus by definition of $F$:
:$\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
{{qed|lemma}}
\end{proof}
|
23386
|
\section{Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal}
Tags: Way Below Relation
\begin{theorem}
Let $\mathscr S = \struct {S, \preceq}$ be an up-complete ordered set.
Let $x, y \in S$.
Then $x \ll y$ {{iff}}
:$\forall I \in \map {\operatorname {Ids} } {\mathscr S}: y \preceq \sup I \implies x \in I$
where
:$\ll$ denotes the way below relation,
:$\map {\operatorname {Ids} } {\mathscr S}$ denotes the set of all ideals in $\mathscr S$.
\end{theorem}
\begin{proof}
=== Sufficient Condition ===
Let $x \ll y$
Let $I \in \map {\operatorname {Ids} } {\mathscr S}$ such that
:$y \preceq \sup I$
By definition of ideal:
:$I$ is directed and lower.
By definition of up-complete:
:$I$ admits a supremum.
By definition of way below relation:
:$\exists i \in I: x \preceq i$
Thus by definition of lower set:
:$x \in I$
{{qed|lemma}}
\end{proof}
|
23387
|
\section{Way Below implies Preceding}
Tags: Way Below Relation
\begin{theorem}
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x, y \in S$ such that
:$x \ll y$
where $\ll$ denotes element is way below second element.
Then
:$x \preceq y$
\end{theorem}
\begin{proof}
By Singleton is Directed and Filtered Subset:
:$\left\{ {y}\right\}$ is directed.
By Supremum of Singleton:
:$\left\{ {y}\right\}$ admits a supremum and $\sup \left\{ {y}\right\} = y$
By definition of reflexivity:
:$y \preceq \sup \left\{ {y}\right\}$
By definition of way below:
:$\exists d \in \left\{ {y}\right\}: x \preceq d$
Thus by definition of singleton:
:$x \preceq y$
{{qed}}
\end{proof}
|
23388
|
\section{Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure}
Tags: Continuous Lattices
\begin{theorem}
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.
Let $x, y \in S$ such that
:$x \ll y$
where $\ll$ denotes the way below relation.
Then there exists a way below open filter in $L$: $y \in F \land F \subseteq x^\gg$
where $x^\gg$ denotes the way above closure of $x$.
\end{theorem}
\begin{proof}
We will prove that
:$x^\gg$ is way below open.
Let $z \in x^\gg$.
By definition of way above closure:
:$x \ll z$
By Way Below has Interpolation Property:
:$\exists x' \in S: x \ll x' \land x' \ll z$
Thus by definition of way above closure:
:$x' \in x^\gg$
Thus
:$x' \ll z$
{{qed|lemma}}
Then
:$\forall z \in x^\gg: \exists y \in x^\gg: y \ll z$
By Axiom of Choice define a mapping $f: x^\gg \to x^\gg$:
:$\forall z \in x^\gg: f\left({z}\right) \ll z$
By definition of way above closure:
:$y \in x^\gg$
Define $V := \left\{ {z^\succeq: \exists n \in \N: z = f^n\left({y}\right)}\right\}$
We will prove that
:$\forall X, Y \in V: \exists Z \in V: X \cup Y \subseteq Z$
Let $X, Y \in V$.
By definition of $V$:
:$\exists z_1 \in S: X = z_1^\succeq \land \exists n_1 \in \N: z_1 = f^{n_1}\left({y}\right)$
and
:$\exists z_2 \in S: Y = z_2^\succeq \land \exists n_2 \in \N: z_2 = f^{n_2}\left({y}\right)$
We will prove that
:$\forall n, k \in \N: f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$
Let $n \in \N$.
'''Base Case''':
By definition of reflexivity:
:$f^{n+0}\left({y}\right) \preceq f^n\left({y}\right)$
'''Induction Hypothesis''':
:$f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$
'''Induction Step''':
By definition of $f$:
:$f^{n+k+1}\left({y}\right) \ll f^{n+k}\left({y}\right)$
By Way Below implies Preceding:
:$f^{n+k+1}\left({y}\right) \preceq f^{n+k}\left({y}\right)$
By Induction Hypothesis and definition of transitivity:
:$f^{n+k+1}\left({y}\right) \preceq f^{n}\left({y}\right)$
{{qed|lemma}}
WLOG: suppose $n_1 \le n_2$
Then
:$\exists k \in \N: n_2 = n_1+k$
Then
:$z_2 \preceq z_1$
By Preceding iff Meet equals Less Operand:
:$z_1 \wedge z_2 = z_2$
By definition of $V$:
:$Z := \left({z_1 \wedge z_2}\right)^\succeq \in V$
By Meet Precedes Operands:
:$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$
By Upper Closure is Decreasing:
:$z_1^\succeq \subseteq Z$ and $z_2^\succeq \subseteq Z$
Thus by Union of Subsets is Subset:
:$X \cup Y \subseteq Z$
{{qed|lemma}}
Define $F := \bigcup V$.
We will prove that
:$F$ is way below open.
Let $u \in F$.
By definition of union:
:$\exists Y \in V: u \in Y$
By definition of $V$:
:$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$
By definition of $f$:
:$z \in x^\gg$
By definition of $f$:
:$f\left({z}\right) \ll z$
Then
:$z' := f\left({z}\right) = f^{n+1}\left({y}\right)$
By definition of $V$:
:${z'}^\succeq \in V$
By definition of reflexivity:
:$z' \preceq z'$
By definition of upper closure of element:
:$z' \in {z'}^\succeq$
By definition of union:
:$z' \in F$
By definition of upper closure of element:
:$z \preceq u$
By Preceding and Way Below implies Way Below:
:$z' \ll u$
Hence
:$\exists g \in F: g \ll u$
{{qed|lemma}}
By Upper Closure of Element is Filter:
:$\forall X \in V: X$ is a filtered upper set.
By Union of Upper Sets is Upper:
:$F$ is an upper set.
By Union of Filtered Sets is Filtered:
:$F$ is filtered.
By definition of filter:
:$F$ is a way below open filter in $L$.
Then
:$f^0\left({y}\right) = y$
By definition of $V$:
:$y^\succeq \in V$
By definitions of upper closure of element and reflexivity:
:$y \in y^\succeq$
By definition of union:
:$y \in F$
It remains to prove that
:$F \subseteq x^\gg$
Let $u \in F$.
By definition of union:
:$\exists Y \in V: u \in Y$
By definition of $V$:
:$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$
By definition of $f$:
:$z \in x^\gg$
By definition of way above closure:
:$x \ll z$
By definition of upper closure of element:
:$z \preceq u$ and $x \preceq x$
By Preceding and Way Below implies Way Below:
:$x \ll u$
Thus by definition of way above closure:
:$u \in x^\gg$
{{qed}}
\end{proof}
|
23389
|
\section{Way Below in Lattice of Power Set}
Tags: Way Below Relation, Power Set
\begin{theorem}
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \preceq}$ be a lattice of power set of $X$ where $\mathord\preceq = \mathord\subseteq \cap \paren {\powerset X \times \powerset X}$
Let $x, y \in \powerset X$.
Then $x \ll y$ {{iff}}
:for every a set $Y$ of subsets of $X$ such that $\ds y \subseteq \bigcup Y$
::then there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$
where $\ll$ denotes the way below relation.
\end{theorem}
\begin{proof}
=== Sufficient Condition ===
Let $x \ll y$
Let $Y$ be a set of subsets of $X$ such that
:$\ds y \subseteq \bigcup Y$
By definitions of power set and subset:
:$Y \subseteq \powerset X$
By Power Set is Complete Lattice:
:$\ds \bigcup Y = \sup Y$
By definition of $\preceq$:
:$y \preceq \sup Y$
By Way Below in Complete Lattice:
:there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$
By the proof of Power Set is Complete Lattice:
:$\ds \bigcup Z = \sup Z$
Thus by definition of $\preceq$:
:there exists a finite subset $Z$ of $Y$: $\ds x \subseteq \bigcup Z$
{{qed|lemma}}
\end{proof}
|
23390
|
\section{Way Below is Approximating Relation}
Tags: Way Below Relation, Continuous Lattices, Approximating Relations
\begin{theorem}
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Then $\ll$ is an approximating relation on $S$.
\end{theorem}
\begin{proof}
Let $x \in S$.
Define $\RR := \mathord \ll$.
By definitions of way below closure and $\RR$-segment:
:$x^\ll = x^\RR$
where:
:$x^\ll$ denotes the way below closure of $x$
:$x^\RR$ denotes the $\RR$-segment of $x$
By definition of continuous:
:$L$ satisfies axiom of approximation.
Thus by axiom of approximation:
:$x = \map \sup {x^\ll} = \map \sup {x^\RR}$
Hence $\ll$ is an approximating relation on $S$.
{{qed}}
\end{proof}
|
23391
|
\section{Way Below is Congruent for Join}
Tags: Way Below Relation
\begin{theorem}
Let $L = \left({S, \vee, \preceq}\right)$ be a join semilattice.
Then $\ll$ is congruence relation for $\vee$:
:$\forall a, b, x, y \in S: a \ll x \land b \ll y \implies a \vee b \ll x \vee y$
where $\ll$ denotes the way below relation.
\end{theorem}
\begin{proof}
Let $a, b, x, y \in S$ such that
$a \ll x$ and $b \ll y$
By Join Succeeds Operands:
:$x \preceq x \vee y$ and $y \preceq x \vee y$
By Preceding and Way Below implies Way Below and definition of reflexivity:
:$a \ll x \vee y$ and $b \ll x \vee y$
Thus by Join is Way Below if Operands are Way Below:
:$a \vee b \ll x \vee y$
{{qed}}
\end{proof}
|
23392
|
\section{Weak-* Limit in Normed Dual Space is Unique}
Tags: Weak-* Convergence (Normed Vector Spaces)
\begin{theorem}
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space for $\struct {X, \norm \cdot}$.
Let $f, g \in X^\ast$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ such that:
:$f_n \weakstarconv f$
and:
:$f_n \weakstarconv g$
where $\weakstarconv$ denotes weak-$\ast$ convergence.
Then:
:$f = g$
\end{theorem}
\begin{proof}
By the definition of weak-$\ast$ convergence, we have:
:$\map {f_n} x \to \map f x$ for each $x \in X$
and:
:$\map {f_n} x \to \map g x$ for each $x \in X$.
Then, from Convergent Complex Sequence has Unique Limit we have:
:$\map f x = \map g x$ for each $x \in X$.
That is:
:$f = g$
{{qed}}
\end{proof}
|
23393
|
\section{Weak Convergence in Hilbert Space}
Tags: Weak Convergence in Hilbert Space, Hilbert Spaces, Weak Convergence (Normed Vector Spaces)
\begin{theorem}
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$.
Let $x \in X$.
Then:
:$\sequence {x_n}_{n \mathop \in \N}$ converges weakly to $x$
{{iff}}:
:$\innerprod {x_n} y \to \innerprod x y$ for each $y \in \HH$.
\end{theorem}
\begin{proof}
Let $\struct {\HH^\ast, \norm \cdot_{\HH^\ast} }$ be the normed dual space of $\HH$.
\end{proof}
|
23394
|
\section{Weak Countable Compactness is not Preserved under Continuous Maps}
Tags: Weakly Countably Compact Spaces
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ be a topological space which is weakly countably compact.
Let $T_B = \struct {S_B, \tau_B}$ be another topological space.
Let $\phi: T_A \to T_B$ be a continuous mapping.
Then $T_B$ is not necessarily weakly countably compact.
\end{theorem}
\begin{proof}
Let $\Z_{>0}$ be the strictly positive integers:
:$\Z_{>0} = \set {1, 2, 3, \ldots}$
Let $T_A = \struct {\Z_{>0}, \tau_A}$ be the odd-even topology.
Let $T_B = \struct {\Z_{>0}, \tau_B}$ be the discrete topology on $\Z_{>0}$.
Let $\phi: T_A \to T_B$ be the mapping:
:$\map \phi {2 k} = k, \map \phi {2 k - 1} = k$
Then:
:$\map {\phi^{-1} } k = \set {2 k, 2 k - 1} \in \tau_A$
demonstrating that $\phi$ is continuous.
Now we have that the Odd-Even Topology is Weakly Countably Compact.
But we also have that a Countable Discrete Space is not Weakly Countably Compact.
Hence the result.
{{qed}}
\end{proof}
|
23395
|
\section{Weak Existence of Matrix Logarithm}
Tags: Matrix Algebra, Matrix Logarithm, Matrix Logarithms
\begin{theorem}
Let $T$ be a square matrix of order $n$.
Let $\norm {T - I} < 1$ in the norm on bounded linear operators, where $I$ the identity matrix.
Then there is a square matrix $S$ such that:
:$e^S = T$
where $e^S$ is the matrix exponential.
\end{theorem}
\begin{proof}
Define:
:$\ds S = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n$
$S$ converges since $\norm {T - I} < 1$.
We have that $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \norm {T - I}^n$ is the Newton-Mercator Series.
This converges since $\norm {T - I} < 1$.
Hence the series for $S$ converges absolutely, and so $S$ is well defined.
Using the series definition for the matrix exponential:
{{begin-eqn}}
{{eqn | l = e^S
| r = I + S + \frac 1 {2!} S^2 + \frac 1 {3!} S^3 + \cdots
}}
{{eqn | r = I + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n \paren {T - I}^n}^3 + \cdots
}}
{{eqn | ll= \leadsto
| l = e^S
| r = I + \paren {T - I} + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots
| c = grouping terms by powers of $T - I$
}}
{{eqn | r = T + c_2 \paren {T - I}^2 + c_3 \paren {T - I}^3 + \cdots
}}
{{end-eqn}}
If $c_i = 0$ for $i \ge 2$, then $e^S = T$, and the result is shown.
The Newton-Mercator Series is a Taylor expansion for $\map \ln {1 + x}$.
When combined with the Power Series Expansion for Exponential Function, it gives:
{{begin-eqn}}
{{eqn | l = e^{\map \ln {1 + x} }
| r = 1 + \map \ln {1 + x} + \frac 1 {2!} \paren {\map \ln {1 + x} }^2 + \frac 1 {3!} \paren {\map \ln {1 + x} }^3 + \cdots
}}
{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n + \frac 1 {2!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^2 + \frac 1 {3!} \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} } n x^n}^3 + \cdots
}}
{{eqn | r = 1 + x + c_2 x^2 + c_3 x^3 + \cdots
| c = grouping terms by powers of $x$
}}
{{end-eqn}}
But $e^{\map \ln {1 + x} } = 1 + x$.
Thus:
:$1 + x = 1 + x + c_2 x^2 + c_3 x^3 + \cdots \implies c_i = 0$
for $i \ge 2$.
{{qed}}
\end{proof}
|
23396
|
\section{Weak Inequality of Integers iff Strict Inequality with Integer plus One}
Tags: Orderings on Integers, Integers
\begin{theorem}
Let $a, b \in \Z$ be integers.
{{TFAE}}
:$(1): \quad a \le b$
:$(2): \quad a < b + 1$
where:
:$\le$ is the ordering on the integers
:$<$ is the strict ordering on the integers.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Orderings on Integers
\end{proof}
|
23397
|
\section{Weak Law of Large Numbers}
Tags: Probability Theory
\begin{theorem}
Let $P$ be a population.
Let $P$ have mean $\mu$ and finite variance.
Let $\sequence {X_n}_{n \mathop \ge 1}$ be a sequence of random variables forming a random sample from $P$.
Let:
:$\ds {\overline X}_n = \frac 1 n \sum_{i \mathop = 1}^n X_i$
Then:
:${\overline X}_n \xrightarrow p \mu$
where $\xrightarrow p$ denotes convergence in probability.
\end{theorem}
\begin{proof}
Let $\sigma$ be the standard deviation of $P$.
By the definition of convergence in probability, we aim to show that:
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} = 1$
for all real $\varepsilon > 0$.
Let $\varepsilon > 0$ be a real number.
By Variance of Sample Mean:
:$\var {{\overline X}_n} = \dfrac {\sigma^2} n$
By Chebyshev's Inequality, we have for real $k > 0$:
:$\map \Pr {\size {{\overline X}_n - \mu} \ge \dfrac {k \sigma} {\sqrt n}} \le \dfrac 1 {k^2}$
As $\sigma > 0$ and $n > 0$, we can set:
:$k = \dfrac {\sqrt n} {\sigma} \varepsilon$
This gives:
:$\map \Pr {\size {{\overline X}_n - \mu} \ge \varepsilon} \le \dfrac {\sigma^2} {n \varepsilon^2}$
We therefore have:
{{begin-eqn}}
{{eqn | l = \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon}
| r = 1 - \map \Pr {\size { {\overline X}_n - \mu} \ge \varepsilon}
}}
{{eqn | o = \ge
| r = 1 - \frac {\sigma^2} {n \varepsilon^2}
}}
{{end-eqn}}
So:
:$1 - \dfrac {\sigma^2} {n \varepsilon^2} \le \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} \le 1$
We have:
:$\ds \lim_{n \mathop \to \infty} \paren {1 - \dfrac {\sigma^2} {n \varepsilon^2} } = 1$
and:
:$\ds \lim_{n \mathop \to \infty} 1 = 1$
So by the Squeeze Theorem:
:$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} = 1$
for all real $\varepsilon > 0$.
{{qed}}
\end{proof}
|
23398
|
\section{Weak Limit in Normed Vector Space is Unique}
Tags: Weak Convergence (Normed Vector Space), Weak Convergence (Normed Vector Spaces)
\begin{theorem}
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x, y \in X$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that:
:$x_n \weakconv x$
and:
:$x_n \weakconv y$
where $\weakconv$ denotes weak convergence.
Then:
:$x = y$
\end{theorem}
\begin{proof}
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Since:
:$x_n \weakconv x$
we have:
:$\map f {x_n} \to \map f x$ for each $f \in X^\ast$.
Since:
:$x_n \weakconv x$
we have:
:$\map f {x_n} \to \map f y$ for each $f \in X^\ast$.
From Convergent Complex Sequence has Unique Limit, we have:
:$\map f x = \map f y$ for each $f \in X^\ast$.
From Normed Dual Space Separates Points, we therefore have:
:$x = y$
{{qed}}
\end{proof}
|
23399
|
\section{Weak Local Compactness is Preserved under Open Continuous Surjection}
Tags: Continuity, Local Compactnesss, Continuous Mappings, Compact Spaces, Weakly Locally Compact Spaces, Open Sets, Local Compactness, Locally Compact Spaces, Open Mappings
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping which is also an open mapping and a surjection.
If $T_A$ is weakly locally compact, then $T_B$ is also weakly locally compact.
\end{theorem}
\begin{proof}
Let $\phi$ be a mapping which is surjective, continuous and open.
Let $T_A$ be weakly locally compact.
Take $b \in S_B$.
Let $V$ be a neighbourhood of $b$.
Since $\phi$ is surjective:
:$\forall y \in S_B: \exists x \in S_A: x \in \map {\phi^{-1} } y$
From the weak local compactness of $T_A$ and the continuity of $\phi$, there exists a compact neighbourhood $K$ of $x$ such that $\phi \sqbrk K \subseteq V$.
Since $K$ is a neighbourhood of $x$, then $x \in K^\circ$ and $y \in \phi \sqbrk {K^\circ} \subseteq \phi \sqbrk K$, where $K^\circ$ is the interior of $K$.
$\phi$ is an open mapping and $K^\circ$ is an open set, so $\phi \sqbrk {K^\circ}$ is also open.
Finally we get that $y \in \phi \sqbrk K \subseteq V$, where $\phi \sqbrk K$ is a compact neighbourhood.
Thus $T_B$ is weakly locally compact.
{{qed}}
\end{proof}
|
23400
|
\section{Weak Lower Closure in Restricted Ordering}
Tags: Lower Closures, Order Theory
\begin{theorem}
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\preccurlyeq T} = T \cap t^{\preccurlyeq S}$
where:
:$t^{\preccurlyeq T}$ is the weak lower closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$
:$t^{\preccurlyeq S}$ is the weak lower closure of $t$ in $\left({S, \preccurlyeq}\right)$.
\end{theorem}
\begin{proof}
Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$.
By definition of weak lower closure $t^{\preccurlyeq T}$, this is equivalent to:
:$t' \preccurlyeq \restriction_T t$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t' \preccurlyeq t \land t' \in T$
as it is assumed that $t \in T$.
The first conjunct precisely expresses that $t' \in t^{\preccurlyeq S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\preccurlyeq S}$
{{iff}} $t' \in T$ and $t' \in t^{\preccurlyeq S}$.
Thus, it follows that the following are equivalent:
:$t' \in t^{\preccurlyeq T}$
:$t' \in T \cap t^{\preccurlyeq S}$
and hence the result follows, by definition of set equality.
{{qed}}
Category:Lower Closures
\end{proof}
|
23401
|
\section{Weak Nullstellensatz}
Tags: Algebraic Geometry
\begin{theorem}
Let $K$ be an algebraically closed field.
Let $n \ge 0$ be an natural number.
Let $K \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Let $I \subseteq K \sqbrk {x_1,\ldots, x_n}$ be an ideal.
{{TFAE}}
# $I$ is the unit ideal: $I = (1)$.
# Its zero-locus is empty set: $\map V I = \O$.
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Algebraic Geometry
\end{proof}
|
23402
|
\section{Weak Solution to Dx u + 3yu = 0 with Heaviside Step Function Boundary Condition}
Tags: Examples of Weak Solutions
\begin{theorem}
Consider the boundary value problem:
:$\begin{cases}
\dfrac {\partial u} {\partial x} + 3 y u = 0 & : x \in \R_{>0},~ y \in \R \\
& \\
\map u {0, y} = \map H y & : y \in \R \\
\end{cases}$
Then it has a weak solution of the form:
:$u = e^{-3 y x} \map H y$
\end{theorem}
\begin{proof}
Let $u = e^{-3 y x} \map H y$
We have that:
:Heaviside Step Function is Locally Integrable
:Locally Integrable Function defines Distribution
:Multiplication of Distribution induced by Locally Integrable Function by Smooth Function
Hence, we can define a distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$.
Then in the distributional sense we have that:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u}{\partial x}
| r = \dfrac {\partial}{\partial x} \paren {e^{-3 y x} \map H y}
}}
{{eqn | r = e^{-3 y x} \paren {-3y} \map H y + e^{-3 y x} \dfrac {\partial \map H y}{\partial x}
}}
{{end-eqn}}
That is:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial T_u}{\partial x}
| r = \dfrac {\partial}{\partial x} T_{e^{-3 y x} \map H y}
}}
{{eqn | r = \dfrac {\partial}{\partial x} \paren {e^{-3 y x} T_{\map H y} }
| c = Multiplication of Distribution induced by Locally Integrable Function by Smooth Function
}}
{{eqn | r = e^{-3 y x} \paren {-3y} T_{\map H y} + e^{-3 y x} \dfrac {\partial}{\partial x} T_{\map H y}
| c = Product Rule for Distributional Derivatives of Distributions multiplied by Smooth Functions
}}
{{end-eqn}}
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial}{\partial x} \map {T_{\map H y} } \phi
| r = - \map {T_{\map H y} } {\dfrac {\partial \phi}{\partial x} }
}}
{{eqn | r = - \int_{-\infty}^\infty \int_{-\infty}^\infty \map H y \dfrac {\partial \phi}{\partial x} \rd x \rd y
}}
{{eqn | r = - \int_0^\infty \int_{-\infty}^\infty \dfrac {\partial \phi}{\partial x} \rd x \rd y
}}
{{eqn | r = - \int_0^\infty \paren {\bigintlimits {\map \phi {x, y} } {x \mathop = - \infty} {x \mathop = \infty} } \rd y
}}
{{eqn | r = - \int_0^\infty 0 \rd y
| c = {{Defof|Test Function}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Hence, in the distributional sense we have that:
:$\dfrac \partial {\partial x} \map H y = \mathbf 0$
where $\mathbf 0$ is the zero distribution.
Therefore:
{{begin-eqn}}
{{eqn | l = \dfrac {\partial u} {\partial x} + 3 y u
| r = e^{-3xy} \paren {-3y} \map H y + 3y e^{-3yx} \map H y
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
Moreover:
{{begin-eqn}}
{{eqn | l = \map u {0, y}
| r = e^{-0} \map H y
}}
{{eqn | r = \map H y
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23403
|
\section{Weak Solution to Dx u = Heaviside Step Function}
Tags: Examples of Weak Solutions
\begin{theorem}
Let $H: \R \to \closedint 0 1$ be the Heaviside step function.
Let $u : \R \to \R$ be such that:
:$\map u x = \begin{cases}
c & : x < 0 \\
x + c & : x > 0
\end{cases}$
where $c \in \R$.
Let $T_u$ be the distribution associated with $u$.
Then $u$ is a weak solution of:
:$u' = H$
That is, in the distributional sense it holds that:
:$\dfrac \d {\d x} T_u = T_H$
\end{theorem}
\begin{proof}
$u$ is continuous on $\R$ and continously differentiable on $\R \setminus \set 0$.
For $x < 0$ we have $\map {u'} x = 0$.
For $x > 0$ we have $\map {u'} x = 1$.
That is:
:$\map {u'} x = \map H x$
Furthermore:
:$\ds \lim_{x \mathop \to 0^-} = 0$
:$\ds \lim_{x \mathop \to 0^+} = 1$
By the jump rule:
{{begin-eqn}}
{{eqn | l = T_u'
| r = T_H + 0 \cdot \delta
}}
{{eqn | r = T_H
}}
{{end-eqn}}
{{qed}}
{{Stub|no classical solution; is this obvious?}}
\end{proof}
|
23404
|
\section{Weak Solution to Dx u = u}
Tags: Examples of Distributional Solutions, Examples of Weak Solutions
\begin{theorem}
Let $H$ be the Heaviside step function.
Let $\map u {x, t} = \map H t e^x$
Then $u$ is a weak solution of the partial differential equation $\ds \dfrac {\partial u} {\partial x} = u$.
That is, for the distribution $T_u \in \map {\DD'} {\R^2}$ associated with $u$ in the distributional sense it holds that:
:$\ds \dfrac {\partial T_u} {\partial x} = T_u$
\end{theorem}
\begin{proof}
Let $\phi \in \map \DD {\R^2}$ be a test function.
Then in the distributional sense we have that:
{{begin-eqn}}
{{eqn | l = \paren {\dfrac \partial {\partial x} - 1} \map {T_u} \phi
| r = -\map {T_u} {\dfrac {\partial \phi} {\partial x} } - \map {T_u} \phi
| c = {{Defof|Distributional Partial Derivative}}
}}
{{eqn | r = -\iint_{\R^2} \paren {\map H t e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + \map H t e^x \map \phi {x, t} }\rd x \rd t
| c = {{Defof|Distribution}}
}}
{{eqn | r = -\int_0^\infty \int_{-\infty}^\infty \paren {e^x \dfrac {\partial \map \phi {x, t} } {\partial x} + e^x \map \phi {x, t} }\rd x \rd t
| c = {{Defof|Heaviside Step Function}}
}}
{{eqn | r = -\int_0^\infty \int_{-\infty}^\infty \dfrac \partial {\partial x} \paren {e^x \map \phi {x, t} }\rd x \rd t
| c = Product Rule for Derivatives
}}
{{eqn | r = -\int_0^\infty \bigintlimits {e^x \map \phi {x, t} } {x \mathop = -\infty} {x \mathop = \infty} \rd t
| c = Fundamental Theorem of Calculus
}}
{{eqn | r = -\int_0^\infty 0 \rd t
| c = {{Defof|Test Function}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23405
|
\section{Weak Upper Closure in Restricted Ordering}
Tags: Upper Closures, Order Theory
\begin{theorem}
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S}$
where:
: $t^{\succcurlyeq T}$ is the weak upper closure of $t$ in $\left({T, \preccurlyeq \restriction_T}\right)$
: $t^{\succcurlyeq S}$ is the weak upper closure of $t$ in $\left({S, \preccurlyeq}\right)$.
\end{theorem}
\begin{proof}
Let $t \in T$.
Suppose that:
:$t' \in t^{\succcurlyeq T}$
By definition of weak upper closure $t^{\succcurlyeq T}$, this is equivalent to:
:$t \preccurlyeq \restriction_T t'$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t \preccurlyeq t' \land t' \in T$
as it is assumed that $t \in T$.
The first conjunct precisely expresses that $t' \in t^{\succcurlyeq S}$.
By definition of set intersection, it also holds that:
:$t' \in T \cap t^{\succcurlyeq S}$
{{iff}} $t' \in T$ and $t' \in t^{\succcurlyeq S}$.
Thus it follows that the following are equivalent:
:$t' \in t^{\succcurlyeq T}$
:$t' \in T \cap t^{\succcurlyeq S}$
and hence the result follows, by definition of set equality.
{{qed}}
Category:Upper Closures
\end{proof}
|
23406
|
\section{Weak Whitney Immersion Theorem}
Tags: Topology, Manifolds, Named Theorems
\begin{theorem}
Every $k$-dimensional manifold $X$ admits a one-to-one immersion in $\R^{2 k + 1}$.
{{MissingLinks|$k$-dimensional}}
\end{theorem}
\begin{proof}
Let $M > 2 k + 1$ be a natural number such that $f: X \to \R^M$ is an injective immersion.
Define a map $h: X \times X \times \R \to \R^M$ by:
:$\map h {x, y, t} = \map t {\map f x - \map f y}$
Define a map $g: \map T X \to \R^M$ by:
:$\map g {x, v} = \map {\d f_x} v$
where $\map T X$ is the tangent bundle of $X$.
Since $M > 2 k + 1$, the Morse-Sard Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.
Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.
Suppose:
:$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$
for some $x, y$.
Then:
:$\map f x - \map f y = t a$
for some scalar $t$.
{{AimForCont}} $x \ne y$.
Then as $f$ is injective:
:$t \ne 0$
But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$.
By Proof by Contradiction:
:$x = y$
and so $\pi \circ f$ is injective.
Let $v$ be a nonzero vector in $\map {T_x} X$ (the tangent space of $X$ at $x$) for which:
:$\map {\d \paren {\pi \circ f}_x} v = 0$
Since $\pi$ is linear:
:$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$
Thus:
:$\map {\pi \circ \d f_x} v = 0$
so:
:$\map {\d f_x} v = t a$
for some scalar $t$.
Because $f$ is an immersion, $t \ne 0$.
Hence:
:$\map g {x, 1/t} = a$
contradicting the choice of $a$.
Hence $\pi \circ f: X \to H$ is an immersion.
$H$ is obviously isomorphic to $\R^{M-1}$.
Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.
\end{proof}
|
23407
|
\section{Weakly Convergent Sequence in Hilbert Space with Convergent Norm is Convergent}
Tags: Hilbert Spaces, Weak Convergence (Normed Vector Spaces)
\begin{theorem}
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\HH$.
Let $x \in X$ be such that:
:$x_n \weakconv x$
and:
:$\norm {x_n} \to \norm x$
where $\weakconv$ denotes weak convergence.
Then:
:$x_n \to x$
\end{theorem}
\begin{proof}
Let $\norm \cdot$ be the inner product norm for $\struct {\HH, \innerprod \cdot \cdot}$.
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have:
:$x_n \to x$
{{iff}}:
:$\norm {x_n - x} \to 0$
From Complex Sequence is Null iff Positive Integer Powers of Sequence are Null, it suffices to show that:
:$\norm {x_n - x}^2 \to 0$
We have:
{{begin-eqn}}
{{eqn | l = \norm {x_n - x}^2
| r = \innerprod {x_n - x} {x_n - x}
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {x_n} {x_n} - \innerprod x {x_n} - \innerprod {x_n} x + \innerprod x x
| c = Inner Product is Sesquilinear
}}
{{eqn | r = \norm {x_n}^2 - \innerprod x {x_n} - \innerprod {x_n} x + \norm x^2
| c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
From Weak Convergence in Hilbert Space, we have:
:$\innerprod {x_n} x \to \innerprod x x = \norm x^2$
since $x_n$ converges weakly to $x$.
From Weak Convergence in Hilbert Space: Corollary, we also have:
:$\innerprod x {x_n} \to \innerprod x x = \norm x^2$
From hypothesis, we have:
:$\norm {x_n}^2 \to \norm x$
Then, from Sum Rule for Real Sequences, we have:
:$\norm {x_n - x}^2 \to 0$
so:
:$x_n \to x$
{{qed}}
\end{proof}
|
23408
|
\section{Weakly Convergent Sequence in Normed Dual Space is Weakly-* Convergent}
Tags: Weak Convergence (Normed Vector Spaces), Weak-* Convergence (Normed Vector Spaces), Weak Convergence (Normed Vector Space), Weak-* Convergence (Normed Vector Space)
\begin{theorem}
Let $\mathbb F$ be a subfield of $\C$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\mathbb F$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $f \in X^\ast$.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence in $X^\ast$ converging weakly to $f$.
Then $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
\end{theorem}
\begin{proof}
Let $x \in X$.
We aim to show that:
:$\map {f_n} x \to \map f x$
Then, since $x \in X$ was arbitrary, we will obtain that $\sequence {f_n}_{n \mathop \in \N}$ converges weakly-$\ast$ to $f$.
Since $\sequence {f_n}_{n \mathop \in \N}$ converges weakly to $f$, we have:
:$\map F {f_n} \to \map F f$ for each $F \in \paren {X^\ast}^\ast$.
Define $x^\wedge : X^\ast \to \mathbb F$ by:
:$\map {x^\wedge} f = \map f x$
for each $f \in X^\ast$.
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have:
:$x^\wedge \in \paren {X^\ast}^\ast$
So, setting $F = x^\wedge$, we obtain:
:$\map {x^\wedge} {f_n} \to \map {x^\wedge} f$
That is:
:$\map {f_n} x \to \map f x$
{{qed}}
\end{proof}
|
23409
|
\section{Weakly Convergent Sequence in Normed Vector Space is Bounded}
Tags: Weak Convergence (Normed Vector Spaces)
\begin{theorem}
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weakly convergent sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is bounded.
\end{theorem}
\begin{proof}
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $X$.
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $X$.
Let $J : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$.
Let:
:$x^\wedge = \map J x$
for each $x \in X$.
Consider the sequence $\sequence {x_n^\wedge}_{n \mathop \in \N}$ in $X^{\ast \ast}$.
Then, for each $f \in X^\ast$, we have:
:$\map {x_n^\wedge} f = \map f {x_n}$
Since $\sequence {x_n}_{n \mathop \in \N}$ converges weakly we have that:
:$\sequence {\map f {x_n} }_{n \mathop \in \N}$ converges for each $f \in X^\ast$.
So, from Convergent Complex Sequence has Unique Limit, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is bounded.
So, there exists a real number $M > 0$ such that:
:$\cmod {\map f {x_n} } \le M$
for each $n \in \N$ and $f \in X^\ast$.
That is:
:$\cmod {\map {x_n^\wedge} f} \le M$
That is:
:$\ds \sup_{n \mathop \in \N} \cmod {\map {x_n^\wedge} f}$ is finite for each $f \in X^\ast$.
From the Banach-Steinhaus Theorem, we then obtain:
:$\ds \sup_{n \mathop \in \N} \norm {x_n^\wedge}_{X^{\ast \ast} }$ is finite.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have:
:$\norm {x_n^\wedge}_{X^{\ast \ast} } = \norm {x_n}_X$
So:
:$\ds \sup_{n \mathop \in \N} \norm {x_n}_X$ is finite.
So we may conclude that:
:$\sequence {x_n}_{n \mathop \in \N}$ is bounded.
{{qed}}
\end{proof}
|
23410
|
\section{Weakly Locally Compact Hausdorff Space is Strongly Locally Compact}
Tags: Strongly Locally Compact Spaces, Hausdorff Spaces, Compact Spaces, Weakly Locally Compact Spaces, Local Compactness, Locally Compact Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $T$ be weakly locally compact.
Then $T$ is strongly locally compact.
\end{theorem}
\begin{proof}
Let $x \in S$.
As $T$ is weakly locally compact, $x$ is contained in a compact neighborhood $N_x$.
As $T$ is a $T_2$ (Hausdorff) space, we can use the result Compact Subspace of Hausdorff Space is Closed.
Thus the interior of $N_x$ has a closure which is compact.
Hence the result, from definition of strongly locally compact space.
{{qed}}
\end{proof}
|
23411
|
\section{Wedderburn's Theorem}
Tags: Division Rings, Fields, Named Theorems, Finite Fields, Galois Fields
\begin{theorem}
Every finite division ring $D$ is a field.
\end{theorem}
\begin{proof}
Let $D$ be a finite division ring.
If $D$ is shown commutative then, by definition, $D$ is a field.
Let $\map Z D$ be the center of $D$, that is:
:$\map Z D := \set {z \in D: \forall d \in D: z d = d z}$
From Center of Division Ring is Subfield it follows that $\map Z D$ is a Galois field.
Thus from Characteristic of Galois Field is Prime the characteristic of $\map Z D$ is a prime number $p$.
Let $\Z / \ideal p$ denote the quotient ring over the principal ideal $\ideal p$ of $\Z$.
From Field of Prime Characteristic has Unique Prime Subfield, the prime subfield of $\map Z D$ is isomorphic to $\Z / \ideal p$.
From Division Ring is Vector Space over Prime Subfield, $\map Z D$ is thus a vector space over $\Z / \ideal p$.
From Vector Space over Division Subring is Vector Space, $D$ is a vector space over $\map Z D$.
Since $\map Z D$ and $D$ are finite, both vector spaces are of finite dimension.
Let $n$ and $m$ be the dimension of the two vector spaces respectively.
It now follows from Cardinality of Finite Vector Space that $\map Z D$ has $p^n$ elements and $D$ has $\paren {p^n}^m$ elements.
Now the idea behind the rest of the proof is as follows.
We want to show $D$ is commutative.
By definition, $\map Z D$ is commutative.
Hence it is to be shown that $D = \map Z D$.
It is shown that:
:$\order D = \order {\map Z D}$
Hence $D = \map Z D$, and the proof is complete.
$\map Z D$ and $D$ are considered as modules.
We have that if $m = 1$ then:
:$\order D = \order {\map Z D}$
and the result then follows.
Thus it remains to show that $m = 1$.
In a finite group, let $x_j$ be a representative of the conjugacy class $\tuple {x_j}$ (the representative does not matter).
{{finish|Invoke Normalizer of Conjugate is Conjugate of Normalizer to formalise the independence of representative choice}}
Let there be $l$ (distinct) non-singleton conjugacy classes.
Let $\map {N_D} x$ denote the normalizer of $x$ with respect to $D$.
Then we know by the Conjugacy Class Equation that:
:$\ds \order D = \order {\map Z D} + \sum_{j \mathop = 0}^{l - 1} \index D {\map {N_D} {x_j} }$
which by Lagrange's theorem is:
:$\ds \order D + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {x_j} } }$
Consider the group of units $\map U D$ in $D$.
Consider what the above equation tells if we start with $\map U D$ instead of $D$.
{{explain|We cannot take $D$ in the first place, since $D$ is not a group under multiplication. Doesn't it make sense to start with $\map U D$ directly? --Wandynsky (talk) 16:51, 30 July 2021 (UTC)}}
If we centralize a multiplicative unit that is in the center, from Conjugacy Class of Element of Center is Singleton we get a singleton conjugacy class.
Bear in mind that the above sum only considers non-singleton classes.
Thus choose some element $u$ not in the center, so $\map {N_D} u$ is not $D$.
However, $\map Z D \subset \map {N_D} u$ because any element in the center commutes with everything in $D$ including $u$.
Then:
:$\order {\map {N_D} u} = \paren {p^n}^m$
for $r < m$.
Suppose there are $l$ such $u$.
Then:
{{begin-eqn}}
{{eqn | l = \order {\map U D}
| r = \order {\map Z {\map U D} } - 1 + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {u_j} } }
| c =
}}
{{eqn | r = p^n - 1 + \sum_{\alpha_i} \frac {\paren {p^n}^m - 1} {\paren {p^n}^{\alpha_i} - 1}
| c =
}}
{{end-eqn}}
{{stub|Clean the following up. This bit is due to {{AuthorRef|Ernst Witt}}.}}
We need two results to finish.
:$(1):\quad$ If $p^k - 1 \divides p^j - 1$, then $k \divides j$
where $\divides$ denotes divisibility.
:$(2)\quad$ If $j \divides k$ then $\Phi_n \divides \dfrac{x^j - 1} {x^k - 1}$
where $\Phi_n$ denotes the $n$th cyclotomic polynomial.
{{refactor|The above two results need to be proved, on their own pages.}}
{{AimForCont}} $m > 1$.
Let $\gamma_i$ be an $m$th primitive root of unity.
Then the above used conjugacy class theorem tells us how to compute size of $\map U D$ using non-central elements $u_j$.
However, in doing so, we have that:
:$\paren {q^n}^{\alpha_i} - 1 \divides \paren {q^n}^m - 1$
Thus by the first result:
:$\alpha_i \divides m$
Thus:
:$\Phi_m \divides \dfrac {x^m - 1} {x^{\alpha_i} - 1}$
However:
:$\size {p^n - \gamma_i} > p^n - 1$
Thus the division is impossible.
This contradicts our assumption that $m > 1$.
Hence $m = 1$ and the result follows, as determined above.
{{qed}}
\end{proof}
|
23412
|
\section{Weierstrass's Theorem}
Tags: Named Theorems: Weierstrass, Named Theorems, Analysis
\begin{theorem}
There exists a real function $f: \closedint 0 1 \to \closedint 0 1$ such that:
:$(1): \quad f$ is continuous
:$(2): \quad f$ is nowhere differentiable.
\end{theorem}
\begin{proof}
Let $C \closedint 0 1$ denote the set of all real functions $f: \closedint 0 1 \to \R$ which are continuous on $\closedint 0 1$.
By Continuous Function on Closed Interval is Complete, $C \closedint 0 1$ is a complete metric space under the supremum norm $\norm {\,\cdot \,}_\infty$.
Let $X$ consist of the $f \in C \closedint 0 1$ such that:
:$\map f 0 = 0$
:$\map f 1 = 1$
:$\forall x \in \closedint 0 1: 0 \le \map f x \le 1$
Then we have the following lemma:
\end{proof}
|
23413
|
\section{Weierstrass-Casorati Theorem}
Tags: Complex Analysis, Named Theorems
\begin{theorem}
Let $f$ be a holomorphic function defined on the open ball $\map B {a, r} \setminus \set a$.
Let $f$ have an essential singularity at $a$.
Then:
:$\forall s < r: f \sqbrk {\map B {a, s} \setminus \set a}$ is a dense subset of $\C$.
{{Disambiguate|Definition:Dense}}
\end{theorem}
\begin{proof}
{{WLOG}}, suppose $a = 0$ and $r = 1$.
{{AimForCont}} $\exists s < 1$ such that:
:$f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$.
Then, by definition of dense subset:
:$\exists z_0 \in \C: \exists r_0 > 0: \map B {z_0, r_0} \cap f \sqbrk {\map B {0, s} \setminus \set 0} = \O$
{{explain|This definition of dense still needs to be added to {{ProofWiki}}. Whichever definition of denseness (either everywhere dense or dense-in-itself) will need to be expanded for the Complex case so as to make it clear that the above definition follows.}}
Hence, the function $\varphi$ defined on $\map B {z_0, r_0}$ by:
:$\map \varphi z = \dfrac 1 {\map f z - z_0}$
is analytic on $\map B {0, s} \setminus \set 0$ and bounded near to $0$, because:
:$\forall z \in \map B {0, s} \setminus \set 0: \cmod {\map f z - z_0} > r_0 \implies \cmod {\map \varphi z} < \dfrac 1 {r_0}$
Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).
Let $\map \varphi 0 \ne 0$.
Then:
:$\map f 0 = z_0 + \dfrac 1 {\map \varphi 0}$
and the singularity of $f$ was removable.
Otherwise, let the power series of $\varphi$ be written:
:$\ds \map \varphi z = \sum_{n \mathop = 1}^{+\infty} a_n z^n$
Then as $\varphi \ne 0$:
:$E = \set {k \in \N: a_k \ne 0} \ne \O$
Let $p = \min E$.
Then $0$ is a pole of order $p$ of $f$.
In each case, the assumption that:
:$\exists s < 1: f \sqbrk {\map B {0, s} \setminus \set 0}$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$.
Hence the result, by Proof by Contradiction.
{{qed}}
\end{proof}
|
23414
|
\section{Weierstrass Approximation Theorem}
Tags: Weierstrass Approximation Theorem, Real Analysis
\begin{theorem}
Let $f$ be a real function which is continuous on the closed interval $\Bbb I$.
Then $f$ can be uniformly approximated on $\Bbb I$ by a polynomial function to any given degree of accuracy.
\end{theorem}
\begin{proof}
Let $\map f t: \Bbb I = \closedint a b \to \R$ be a continuous function.
Introduce $\map x t$ with a rescaled domain:
:$\map f t \mapsto \map x {a + t \paren {b - a} } : \closedint a b \to \closedint 0 1$
From now on we will work with $x: \closedint 0 1 \to \R$, which is also continuous.
Let $n \in \N$.
For $t \in \closedint 0 1$ consider the Bernstein polynomial:
:$\ds \map {B_n x} t = \sum_{k \mathop = 0}^n \map x {\frac k n} \binom n k t^k \paren {1 - t}^{n - k}$
For $t \in \closedint 0 1$, $0 \le k \le n$, let:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
By the binomial theorem:
:$\ds \sum_{k \mathop = 0}^n \map {p_{n, k} } t = 1$
\end{proof}
|
23415
|
\section{Weierstrass Approximation Theorem/Lemma 1}
Tags: Weierstrass Approximation Theorem
\begin{theorem}
Let $\map {p_{n, k} } t : \N^2 \times \closedint 0 1 \to \R$ be a real valued function defined by:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
where:
:$n, k \in \N$
:$t \in \closedint 0 1$
:$\dbinom n k$ denotes the binomial coefficient.
Then:
:$\ds \sum_{k \mathop = 0}^n k \map {p_{n, k} } t = n t$
\end{theorem}
\begin{proof}
From the binomial theorem:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^n
| r = \sum_{k \mathop = 0}^n \binom n k y^k x^{n - k}
}}
{{eqn | ll= \leadsto
| l = 1
| r = \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}
| c = $y = t, ~x = 1 - t$
}}
{{eqn | l = 0
| r = \sum_{k \mathop = 0}^n \binom n k \paren {k t^{k - 1} \paren {1 - t}^{n - k} - t^k \paren{n - k} \paren {1 - t}^{n - k - 1} }
| c = Derivative {{WRT|Differentiation}} $t$
}}
{{eqn | r = \sum_{k \mathop = 0}^n k p_{n,k} \paren {\frac 1 t + \frac 1 {1 - t} } - \frac n {1 - t}
}}
{{eqn | ll= \leadsto
| l = n t
| r = \sum_{k \mathop = 0}^n k p_{n,k}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23416
|
\section{Weierstrass Approximation Theorem/Lemma 2}
Tags: Weierstrass Approximation Theorem
\begin{theorem}
Let $\map {p_{n, k} } t : \N^2 \times \closedint 0 1 \to \R$ be a real valued function defined as:
:$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$
where:
:$n, k \in \N$
:$t \in \closedint 0 1$
:$\dbinom n k$ denotes a binomial coefficient.
Then:
:$\ds \sum_{k \mathop = 0}^n \paren {k - n t}^2 \map {p_{n, k} } t = n t \paren {1 - t}$
\end{theorem}
\begin{proof}
From the binomial theorem:
:$\ds 1 = \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}$
From Lemma 1:
{{begin-eqn}}
{{eqn | l = n t
| r = \sum_{k \mathop = 0}^n \binom n k k t^k \paren {1 - t}^{n - k}
}}
{{eqn | ll= \leadsto
| l = n
| r = \sum_{k \mathop = 0}^n \binom n k k \paren {k t^{k - 1} \paren {1 - t}^{n - k} - t^k \paren {n - k} \paren {1 - t}^{n - k - 1} }
| c = {{Defof|Derivative of Real Function}} {{WRT|Differentiation}} $t$
}}
{{eqn | r = \frac 1 t \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t - \frac n {1 - t} \sum_{k \mathop = 0}^n k \map {p_{n, k} } t + \frac 1 {1 - t} \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t
}}
{{eqn | r = \frac 1 {t \paren {1 - t} } \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t - \frac n {1 - t} n t
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^n k^2 \map {p_{n, k} } t
| r = n t \paren {1 - t} + n^2 t^2
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \paren {k - n t}^2 \map {p_{n, k} } t
| r = \sum_{k \mathop = 0}^n \paren {k^2 - 2 n t k + n^2 t^2} \map {p_{n, k} } t
}}
{{eqn | r = n t \paren {1 - t} + n^2 t^2 - 2 n t n t + n^2 t^2
}}
{{eqn | r = n t \paren {1 - t}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23417
|
\section{Weierstrass Factorization Theorem}
Tags: Complex Analysis, Analysis, Entire Functions, Named Theorems: Weierstrass, Infinite Products
\begin{theorem}
Let $f$ be an entire function.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let the sequence $\sequence {a_n}$ consist of the nonzero zeroes of $f$, repeated according to multiplicity.
\end{theorem}
\begin{proof}
From Weierstrass Product Theorem, the function:
:$\ds \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
defines an entire function that has the same zeroes as $f$ counting multiplicity.
Thus $f / h$ is both an entire function and non-vanishing.
As $f / h$ is both holomorphic and nowhere zero there exists a holomorphic function $g$ such that:
:$e^g = f / h$
Therefore:
:$f = e^g h$
as desired.
{{qed}}
\end{proof}
|
23418
|
\section{Weierstrass Function is Continuous}
Tags: Weierstrass Function
\begin{theorem}
Let $a \in \openint 0 1$.
Let $b$ be a strictly positive odd integer such that:
:$\ds a b > 1 + \frac 3 2 \pi$
Let $f: \R \to \R$ be a real function defined by:
:$\ds \map f x = \sum_{n \mathop = 0}^\infty a^n \map \cos {b^n \pi x}$
for each $x \in \R$.
Then $f$ is well-defined and continuous.
\end{theorem}
\begin{proof}
Note that:
:$\ds \sup_{x \mathop \in \R} \size {a^n \map \cos {b^n \pi x} } = a^n$
Since $a \in \openint 0 1$:
:$\ds \sum_{n \mathop = 0}^\infty a^n$ converges.
So, by the Weierstrass M-Test:
:$\ds \sum_{n \mathop = 0}^\infty a^n \map \cos {b^n \pi x}$ converges uniformly on $\R$.
That is, $f$ is well-defined.
Further, from Uniformly Convergent Series of Continuous Functions Converges to Continuous Function: Corollary:
:$f$ is continuous.
{{qed}}
Category:Weierstrass Function
\end{proof}
|
23419
|
\section{Weierstrass M-Test}
Tags: Named Theorems, Real Analysis, Convergence, Analysis
\begin{theorem}
Let $f_n$ be a sequence of real functions defined on a domain $D \subseteq \R$.
Let $\ds \sup_{x \mathop \in D} \size {\map {f_n} x} \le M_n$ for each integer $n$ and some constants $M_n$
Let $\ds \sum_{i \mathop = 1}^\infty M_i < \infty$.
Then $\ds \sum_{i \mathop = 1}^\infty f_i$ converges uniformly on $D$.
\end{theorem}
\begin{proof}
Let:
:$\ds S_n = \sum_{i \mathop = 1}^n f_i$
Let:
:$\ds f = \lim_{n \mathop \to \infty} S_n$
To show the sequence of partial sums converge uniformly to $f$, we must show that:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in D} \size {f - S_n} = 0$
But:
{{begin-eqn}}
{{eqn | l = \sup_{x \mathop \in D} \size {f - S_n}
| r = \sup_{x \mathop \in D} \size {\paren {f_1 + f_2 + \dotsb} - \paren {f_1 + f_2 + \dotsb + f_n} }
| c =
}}
{{eqn | r = \sup_{x \mathop \in D} \size {f_{n + 1} + f_{n + 2} + \dotsc}
| c =
}}
{{end-eqn}}
By the Triangle Inequality, this value is less than or equal to:
:$\ds \sum_{i \mathop = n + 1}^\infty \sup_{x \mathop \in D} \size {\map {f_i} x} \le \sum_{i \mathop = n + 1}^\infty M_i$
We have that:
:$\ds 0 \le \sum_{i \mathop = 1}^\infty M_n < \infty$
It follows from Tail of Convergent Series tends to Zero:
:$\ds 0 \le \lim_{n \mathop \to \infty} \sum_{i \mathop = n + 1}^\infty \sup_{x \mathop \in D} \size {\map {f_i} x} \le \lim_{n \mathop \to \infty} \sum_{i \mathop = n + 1}^\infty M_i = 0$
So:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in D} \size {f - S_n} = 0$
Hence the series converges uniformly on the domain.
{{qed}}
{{expand|Establish how broadly this can be applied - it's defined for real functions but should also apply to complex ones and possibly a general metric space.}}
\end{proof}
|
23420
|
\section{Weierstrass Product Inequality}
Tags: Inequalities, Analysis, Infinite Products
\begin{theorem}
For $n \ge 1$:
:$\ds \prod_{i \mathop = 1}^n \paren {1 - a_i} \ge 1 - \sum_{i \mathop = 1}^n a_i$
where all of $a_i$ are in the closed interval $\closedint 0 1$.
\end{theorem}
\begin{proof}
For $n = 1$ we have:
:$1 - a_1 \ge 1 - a_1$
which is clearly true.
Suppose the proposition is true for $n = k$, that is:
:$\ds \prod_{i \mathop = 1}^k \paren {1 - a_i} \ge 1 - \sum_{i \mathop = 1}^k a_i$
Then:
{{begin-eqn}}
{{eqn | l = \prod_{i \mathop = 1}^{k + 1} \paren {1 - a_i}
| r = \paren {1 - a_{k + 1} } \prod_{i \mathop = 1}^k \paren {1 - a_i}
| c =
}}
{{eqn | o = \ge
| r = \paren {1 - a_{k + 1} } \paren {1 - \sum_{i \mathop = 1}^k a_i}
| c = as $1 - a_{k + 1} \ge 0$
}}
{{eqn | r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i + a_{k + 1} \sum_{i \mathop = 1}^k a_i
| c =
}}
{{eqn | o = \ge
| r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i
| c = as $a_i \ge 0$
}}
{{eqn | r = 1 - \sum_{i \mathop = 1}^{k + 1} a_i
| c =
}}
{{end-eqn}}
Thus, by Principle of Mathematical Induction, the proof is complete.
{{qed}}
\end{proof}
|
23421
|
\section{Weierstrass Product Theorem}
Tags: Complex Analysis, Infinite Products
\begin{theorem}
Let $\sequence {a_k}$ be a sequence of non-zero complex numbers such that:
:$\cmod {a_n} \to \infty$ as $n \to \infty$
Let $\sequence {p_n}$ be a sequence of non-negative integers for which the series:
:$\ds \sum_{n \mathop = 1}^\infty \size {\dfrac r {a_n} }^{1 + p_n}$
converges for every $r \in \R_{> 0}$.
Let:
:$\ds \map f z = \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
where $E_{p_n}$ are Weierstrass elementary factors.
Then $f$ is entire and its zeroes are the points $a_n$, counted with multiplicity.
\end{theorem}
\begin{proof}
By:
:Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent
:Infinite Product of Analytic Functions is Analytic
:Zeroes of Infinite Product of Analytic Functions
it suffices to show that the product $\ds \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ converges locally uniformly absolutely.
By Bounds for Weierstrass Elementary Factors and Weierstrass M-Test, this is the case.
{{stub}}
{{qed}}
\end{proof}
|
23422
|
\section{Weight of Body at Earth's Surface}
Tags: Weight (Physics), Weight
\begin{theorem}
Let $B$ be a body of mass $m$ situated at (or near) the surface of Earth.
Then the weight of $B$ is given by:
:$W = m g$
where $g$ is the value of the acceleration due to gravity at the surface of Earth.
\end{theorem}
\begin{proof}
The weight of $B$ is the magnitude of the force exerted on it by the influence of the gravitational field it is in.
By Newton's Second Law of Motion, that force is given by:
:$\mathbf W = -m g \mathbf k$
where:
:$g$ is the value of the acceleration due to gravity at the surface of Earth
:$\mathbf k$ is a unit vector directed vertically upwards.
Hence the magnitude of $\mathbf W$ is given by:
:$W = \size {-m g \mathbf k} = m g$
{{qed}}
\end{proof}
|
23423
|
\section{Weight of Discrete Topology equals Cardinality of Space}
Tags: Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then:
:$\map w T = \size S$
where:
:$\map w T$ denotes the weight of $T$
:$\card S$ denotes the cardinality of $S$.
\end{theorem}
\begin{proof}
By Basis for Discrete Topology the set $\BB = \set {\set x: x \in S}$ is a basis of $T$.
By Set of Singletons is Smallest Basis of Discrete Space $\BB$ is smallest basis of $T$:
:for every basis $\CC$ of $T$, $\BB \subseteq \CC$.
Then by Subset implies Cardinal Inequality:
:for every basis $\CC$ of $T$, $\card \BB \le \card \CC$.
Hence $\card \BB$ is minimal cardinalty of basis of $T$:
:$\map w T = \card \BB$ by definition of weight.
Thus by Cardinality of Set of Singletons:
:$\map w T = \card S$
{{qed}}
\end{proof}
|
23424
|
\section{Weight of Sorgenfrey Line is Continuum}
Tags: Sorgenfrey Line
\begin{theorem}
Let $T = \struct {\R, \tau}$ be the Sorgenfrey line.
Then $\map w T = \mathfrak c$
where
:$\map w T$ denotes the weight of $T$
:$\mathfrak c$ denotes continuum, the cardinality of real numbers.
\end{theorem}
\begin{proof}
By definition of Sorgenfrey line, the set:
:$\BB = \set {\hointr x y: x, y \in \R \land x < y}$
is a basis of $T$.
By definition of weight:
:$\map w T \le \card \BB$
where $\card \BB$ denotes the cardinality of $\BB$.
By Cardinality of Basis of Sorgenfrey Line not greater than Continuum:
:$\card \BB \le \mathfrak c$
Thus
:$\map w T \le \mathfrak c$
It remains to show that:
:$\mathfrak c \le \map w T$
{{AimForCont}}
:$\mathfrak c \not \le \map w T$
Then:
:$\map w T < \mathfrak c$
By definition of weight, there exists a basis $\BB_0$ of $T$:
:$\map w T = \card {\BB_0}$
Then by Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure:
:$\exists x, y \in \R: x < y \land \hointr x y \notin \set {\bigcup A: A \subseteq \BB_0} = \tau$
By definition of $\BB$:
:$\hointr x y \in \BB \subseteq \tau$
Thus this contradicts by definition of subset with:
:$\hointr x y \notin \tau$
{{qed}}
\end{proof}
|
23425
|
\section{Well-Founded Induction}
Tags: Set Theory, Axiom of Foundation
\begin{theorem}
Let $\struct {A, \RR}$ be a strictly well-founded relation.
Let $\RR^{-1} \sqbrk x$ denote the preimage of $x$ for each $x \in A$.
Let $B$ be a class such that $B \subseteq A$.
Suppose that:
:$(1): \quad \forall x \in A: \paren {\RR^{-1} \sqbrk x \subseteq B \implies x \in B}$
Then:
:$A = B$
That is, if a property passes from the preimage of $x$ to $x$, then this property is true for all $x \in A$.
\end{theorem}
\begin{proof}
{{AimForCont}} $A \nsubseteq B$.
Then $A \setminus B \ne 0$.
By Strictly Well-Founded Relation determines Strictly Minimal Elements, $A \setminus B$ must have some strictly minimal element under $\RR$.
Then:
:$\exists x \in A \setminus B: \paren {A \setminus B} \cap \RR^{-1} \sqbrk x = \O$
so:
:$A \cap \RR^{-1} \sqbrk x \subseteq B$
Since $\RR^{-1} \sqbrk x \subseteq A$:
:$\RR^{-1} \sqbrk x \subseteq B$
Thus, by hypothesis $(1)$:
:$x \in B$
But this contradicts the fact that:
:$x \in A \setminus B$
By Proof by Contradiction it follows that:
:$A \setminus B = \O$
and so:
:$A \subseteq B$
Therefore:
:$A = B$
{{qed}}
\end{proof}
|
23426
|
\section{Well-Founded Proper Relational Structure Determines Minimal Elements}
Tags: Relational Closures, Relational Closure
\begin{theorem}
Let $A$ and $B$ be classes.
Let $\struct {A, \prec}$ be a proper relational structure.
Let $\prec$ be a strictly well-founded relation.
Suppose $B \subset A$ and $B \ne \O$.
Then $B$ has a strictly minimal element under $\prec$.
\end{theorem}
\begin{proof}
$B$ is not empty.
So $B$ has at least one element $x$.
By Singleton of Element is Subset:
:$\set x \subseteq B$
By Relational Closure Exists for Set-Like Relation:
:$\set x$ has a $\prec$-relational closure.
{{explain|Set-like relation: it needs to be shown that $\prec$ is one of those.}}
This $\prec$-relational closure shall be denoted $y$.
By Intersection is Subset:
:$y \cap B \subseteq A$
As $x \in y$ and $x \in B$:
:$y \cap B \ne \O$
By the definition of strictly well-founded relation:
:$(1): \quad \exists x \in \paren {y \cap B}: \forall w \in \paren {y \cap B}: w \nprec x$
{{AimForCont}} that $w \in B$ such that $w \prec x$.
Since $x \in y$, it follows that $w \in y$, so $w \in \paren {B \cap y}$ by the definition of intersection.
This contradicts $(1)$.
Therefore:
:$w \nprec x$
and so $B$ has a strictly minimal element under $\prec$.
{{qed}}
\end{proof}
|
23427
|
\section{Well-Founded Relation has no Relational Loops}
Tags: Well-Founded Relations
\begin{theorem}
Let $\RR$ be a well-founded relation on $S$.
Let $x_1, x_2, \ldots, x_n \in S$.
Then:
:$\neg \paren {\paren {x_1 \mathrel \RR x_2} \land \paren {x_3 \mathrel \RR x_4} \land \cdots \land \paren {x_n \mathrel \RR x_1} }$
That is, there are no relational loops within $S$.
\end{theorem}
\begin{proof}
Since $x_1, x_2, \ldots, x_n \in S$, there exists a non-empty subset $T$ of $S$ such that:
:$T = \set {x_1, x_2, \ldots, x_n}$
By the definition of a well-founded relation:
:$(1): \quad \exists z \in T: \forall y \in T \setminus z: \neg y \mathrel \RR z$
{{AimForCont}} $\paren {x_1 \mathrel \RR x_2} \land \paren {x_2 \mathrel \RR x_3} \land \cdots \land \paren {x_n \mathrel \RR x_1}$.
We have that the elements of $T$ are $x_1, x_2, \ldots, x_n$.
Hence:
:$\forall y \in T: \exists z \in T \setminus z: y \mathrel \RR z$
This then contradicts $(1)$.
Hence:
:$\neg \paren {\paren {x_1 \mathrel \RR x_2} \land \paren {x_3 \mathrel \RR x_4} \land \cdots \land \paren {x_n \mathrel \RR x_1} }$
and so it follows that a well-founded relation has no relational loops.
{{qed}}
\end{proof}
|
23428
|
\section{Well-Founded Relation is not necessarily Ordering}
Tags: Well-Founded Relations, Orderings, Order Theory
\begin{theorem}
Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be a well-founded relation on $S$.
Then it is not necessarily the case that $\RR$ is also either an ordering or a strict ordering.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $P$ be the set of all polynomials over $\R$ in one variable with real coefficients.
Let $\DD$ be a relation on $P$ defined as:
:$\forall p_0, p_1 \in P: \tuple {p_0, p_1} \in \DD$ {{iff}} $p_0$ is the derivative of $p_1$.
From Differentiation of Polynomials induces Well-Founded Relation, we have that $\DD$ is a well-founded relation on $P$.
Let $P_a$ be the polynomial defined as:
:$P_a = x^3$
Then the derivative of $P_a$ {{WRT|Differentiation}} $x$ is:
:$P_a' = \dfrac \d {\d x} x^3 = 3 x^2$
Then the derivative of $P_a$ {{WRT|Differentiation}} $x$ is:
:$P_a'' = \dfrac \d {\d x} 3 x^2 = 6 x$
So we have that:
:$\tuple {P_a', P_a} \in \DD$
and:
:$\tuple {P_a'', P_a'} \in \DD$
but it is not the case that $\tuple {P_a'', P_a} \in \DD$.
That is, $\DD$ is not transitive.
It follows that $\DD$ is neither an ordering nor a strict ordering.
{{qed}}
\end{proof}
|
23429
|
\section{Well-Ordered Induction}
Tags: Order Theory, Principle of Mathematical Induction, Well-Orderings, Mathematical Induction, Class Theory
\begin{theorem}
Let $\struct {A, \prec}$ be a strict well-ordering.
For all $x \in A$, let the $\prec$-initial segment of $x$ be a small class.
Let $B$ be a class such that $B \subseteq A$.
Let:
:$(1): \quad \forall x \in A: \paren {\paren {A \mathop \cap \map {\prec^{-1} } x} \subseteq B \implies x \in B}$
Then:
:$A = B$
That is, if a property passes from the initial segment of $x$ to $x$, then this property is true for all $x \in A$.
\end{theorem}
\begin{proof}
{{AimForCont}} that $A \nsubseteq B$.
Then:
:$A \setminus B \ne 0$.
By Proper Well-Ordering Determines Smallest Elements, $A \setminus B$ must have some $\prec$-minimal element.
Thus:
:$\ds \exists x \in \paren {A \setminus B}: \paren {A \setminus B} \cap \map {\prec^{-1} } x = \O$
implies that:
:$A \cap \map {\prec^{-1} } x \subseteq B$
Hence this fulfils the hypothesis for $(1)$.
We have that $x \in A$.
so by $(1)$:
:$x \in B$
But this contradicts the fact that $x \in \paren {A \setminus B}$.
Thus by Proof by Contradiction:
:$A \subseteq B$
It follows by definition of set equality that:
:$A = B$
{{qed}}
\end{proof}
|
23430
|
\section{Well-Ordered Transitive Subset is Equal or Equal to Initial Segment}
Tags: Well-Orderings, Order Theory
\begin{theorem}
Let $\struct {\prec, A}$ be a well-ordered set.
For every $x \in A$, let every $\prec$-initial segment $A_x$ be a set.
Let $B$ be a subclass of $A$ such that
:$\forall x \in A: \forall y \in B: \paren {x \prec y \implies x \in B}$.
That is, $B$ must be $\prec$-transitive.
Then:
:$A = B$
or:
:$\exists x \in A: B = A_x$
\end{theorem}
\begin{proof}
Let $A \ne B$.
Then $B \subsetneq A$.
Therefore, by Set Difference with Proper Subset:
:$A \setminus B \ne \O$
Then:
{{begin-eqn}}
{{eqn | l = A \setminus B
| o = \ne
| r = \O
}}
{{eqn | ll= \leadsto
| q = \exists x \in A \setminus B
| l = \paren {A \setminus B} \cap A_x
| r = \O
| c = Proper Well-Ordering Determines Smallest Elements
}}
{{eqn | ll= \leadsto
| q = \exists x \in A \setminus B
| l = A \cap A_x
| o = \subseteq
| r = B
| c = Set Difference with Superset is Empty Set
}}
{{end-eqn}}
One direction of inclusion is proven.
By the hypothesis:
{{begin-eqn}}
{{eqn | l = x \in A \land x \prec y \land y \in B
| o = \implies
| r = x \in B
}}
{{end-eqn}}
But $x \in A \land x \notin B$, so:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = B
}}
{{eqn | ll= \leadsto
| l = \neg x
| o = \prec
| r = y
| c = Modus Tollendo Tollens and other propositional manipulations
}}
{{eqn | ll= \leadsto
| l = y
| o = \prec
| r = x
| c = $\prec$ is totally ordered and $y \ne x$
}}
{{end-eqn}}
Therefore:
:$B \subseteq A_x$
and so
:$B \subseteq A \cap A_x$
{{qed}}
\end{proof}
|
23431
|
\section{Well-Ordering Minimal Elements are Unique}
Tags: Well-Orderings
\begin{theorem}
Let $\struct {S,\preceq}$ be a well-ordered set.
Then every non-empty subset of $S$ has a unique minimal element.
\end{theorem}
\begin{proof}
The proof consists of a uniqueness and an existence part.
Let $S'$ be a non-empty subset of $S$.
\end{proof}
|
23432
|
\section{Well-Ordering Principle}
Tags: Number Theory, Well-Ordering Principle, Ordering on Natural Numbers, Natural Numbers, Named Theorems, Well-Orderings
\begin{theorem}
Every non-empty subset of $\N$ has a smallest (or '''first''') element.
That is, the relational structure $\struct {\N, \le}$ on the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.
This is called the '''well-ordering principle'''.
\end{theorem}
\begin{proof}
Consider the natural numbers $\N$ defined as the naturally ordered semigroup $\struct {S, \circ, \preceq}$.
From its definition, $\struct {S, \circ, \preceq}$ is well-ordered by $\preceq$.
The result follows.
As $\N_{\ne 0} = \N \setminus \set 0$, by Set Difference is Subset $\N_{\ne 0} \subseteq \N$.
As $\N$ is well-ordered, by definition, every subset of $\N$ has a smallest element.
{{qed}}
\end{proof}
|
23433
|
\section{Well-Ordering Theorem}
Tags: Set Theory, Named Theorems, Ordinals, Axiom of Choice
\begin{theorem}
Every set is well-orderable.
\end{theorem}
\begin{proof}
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
By the Axiom of Choice, there is a choice function $c$ defined on $\powerset S \setminus \set \O$.
We will use $c$ and the Principle of Transfinite Induction to define a bijection between $S$ and some ordinal.
Intuitively, we start by pairing $\map c S$ with $0$, and then keep extending the bijection by pairing $\map c {S \setminus X}$ with $\alpha$, where $X$ is the set of elements already dealt with.
\end{proof}
|
23434
|
\section{Well-Ordering Theorem implies Hausdorff Maximal Principle}
Tags: Well-Orderings
\begin{theorem}
Let the Well-Ordering Theorem hold.
Then the Hausdorff Maximal Principle holds.
\end{theorem}
\begin{proof}
Let $X$ be a non-empty set.
Let $X$ contain at least two elements; otherwise, any non-empty ordering on $X$ is trivially a maximal chain.
By the Well-Ordering Theorem, $X$ can be well-ordered.
Fix such a well-ordering.
Let $\le$ be any ordering on $X$.
Let $\map P {a, Y}$ be the predicate:
:$a$ is $\le$-comparable with every $y \in Y$.
Here, $a$ and $Y$ are bound variables.
That is, $\map P {a, Y}$ holds if, for every $y \in Y$, $a \le y$ or $y \le a$.
Define the mapping:
:$\map \rho {f: S_x \to \powerset X} = \begin {cases}
f \sqbrk {S_x} \cup \set x & : \map P {x, S_x} \\
f \sqbrk {S_x} & : \text {otherwise}
\end{cases}$
where $S_x$ is the initial segment in $X$ defined by $x$, and $\map f {\,\cdot\,}$ denotes an image set.
Using the Principle of Recursive Definition for Well-Ordered Sets, we can use $\rho$ to uniquely define:
:$h: X \to \powerset X$:
:$\map h \alpha = \map \rho {h {\restriction_{S_\alpha} } } = \begin {cases}
h \sqbrk {S_\alpha} \cup \set \alpha & : \map P {\alpha, S_\alpha} \\ h \sqbrk {S_\alpha} & : \text {otherwise} \end{cases}$
Then $\map h \alpha$ is a $\le$-totally ordered set, by virtue of the construction of $h$ in terms of the predicate $P$.
Similarly, $h \sqbrk {S_\alpha}$ is totally ordered, for the same reason.
Then the union:
:$Z = \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$
admits an ordered sum in terms of $\le$ imposed on each $\map h \alpha$.
By Ordered Sum of Tosets is Totally Ordered Set, this is a totally ordered set.
In particular, it is a chain of $\struct {X, \le}$.
We claim this chain is maximal.
To see this, consider $x_0 \notin \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$.
Then $x_0$ is not $\le$-comparable with the elements of $Z$, because $\neg \map P {x, Z}$.
That is, there are no proper supersets of $Z$ that are totally ordered.
Therefore the ordered sum on $Z$ is a maximal chain in $\struct {X, \le}$.
{{qed}}
\end{proof}
|
23435
|
\section{Westwood's Puzzle}
Tags: Euclidean Geometry, Named Theorems
\begin{theorem}
:500px
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.
\end{theorem}
\begin{proof}
Construct the perpendicular from $E$ to $AC$, and call its foot $G$.
Call the intersection of $IE$ and $AC$ $K$, and the intersection of $EH$ and $AC$ $L$.
:500px
{{begin-eqn}}
{{eqn|l=\angle CKI|r=\angle EKG|c=By Two Straight Lines make Equal Opposite Angles}}
{{eqn|l=\angle EGK|r=\mbox{Right Angle}|c=By Tangent to Circle is Perpendicular to Radius}}
{{eqn|l=\angle KIC|r=\mbox{Right Angle}|c=Because $IF \perp CD$}}
{{eqn|l=\angle EGK|r=\angle KIC|c=By Euclid's Fourth Postulate}}
{{eqn|l=IC|r=EJ|c=By Parallel Lines are Everywhere Equidistant}}
{{eqn|l=EJ|r=EG|c=Because both are radii of the same circle}}
{{eqn|l=IC|r=EG|c=By Euclid's First Common Notion}}
{{eqn|l=\mbox{Area}\triangle IKC|r=\mbox{Area}\triangle GKE|c=By Triangle Angle-Angle-Side Equality}}
{{eqn|l=\angle HLA|r=\angle GLE|c=By Two Straight Lines make Equal Opposite Angles}}
{{eqn|l=\angle EGL|r=\mbox{Right Angle}|c=By Tangent to Circle is Perpendicular to Radius}}
{{eqn|l=\angle AHL|r=\mbox{Right Angle}|c=Because $HJ \perp AD$}}
{{eqn|l=\angle EGL|r=\angle AHL|c=By Euclid's Fourth Postulate}}
{{eqn|l=HA|r=EF|c=By Parallel Lines are Everywhere Equidistant}}
{{eqn|l=EF|r=EG|c=Because both are radii of the same circle}}
{{eqn|l=HA|r=EG|c=By Euclid's First Common Notion}}
{{eqn|l=\mbox{Area}\triangle HAL|r=\mbox{Area}\triangle GEL|c=By Triangle Angle-Angle-Side Equality}}
{{eqn|l=\mbox{Area}\triangle ADC|r=\frac{AD\cdot CD} 2|c=By Area of a Triangle in Terms of Side and Altitude}}
{{eqn|l=\frac{\mbox{Area}\Box ABCD} 2|r=\frac{AD\cdot CD} 2|c=By Area of a Parallelogram}}
{{eqn|l=\frac{\mbox{Area}\Box ABCD} 2|r=\mbox{Area}\triangle ADC|c=By Euclid's First Common Notion}}
{{eqn|r=\mbox{Area}\triangle HAL + \mbox{Area}\triangle IKC + \mbox{Area}\Box DHLKI}}
{{eqn|r=\mbox{Area}\triangle GEL + \mbox{Area}\triangle GKE+ \mbox{Area}\Box DHLKI}}
{{eqn|r=\mbox{Area}\Box DHEI}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23436
|
\section{Westwood's Puzzle/Proof 1}
Tags: Euclidean Geometry, Named Theorems
\begin{theorem}
:500px
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.
\end{theorem}
\begin{proof}
Construct the perpendicular from $E$ to $AC$, and call its foot $G$.
Let $K$ be the intersection of $IE$ and $AC$.
Let $L$ be the intersection of $EH$ and $AC$.
:500px
First we have:
{{begin-eqn}}
{{eqn | n = 1
| l = \angle CKI
| r = \angle EKG
| c = Two Straight Lines make Equal Opposite Angles
}}
{{eqn | l = \angle EGK
| r = \text {Right Angle}
| c = Tangent to Circle is Perpendicular to Radius
}}
{{eqn | l = \angle KIC
| r = \text {Right Angle}
| c = as $IF \perp CD$
}}
{{eqn | n = 2
| ll= \therefore
| l = \angle EGK
| r = \angle KIC
| c = Euclid's Fourth Postulate
}}
{{eqn | l = IC
| r = EJ
| c = Opposite Sides and Angles of Parallelogram are Equal
}}
{{eqn | l = EJ
| r = EG
| c = as both are radii of the same circle
}}
{{eqn | n = 3
| ll= \therefore
| l = IC
| r = EG
| c = Euclid's First Common Notion
}}
{{eqn | ll= \therefore
| l = \Area \triangle IKC
| r = \Area \triangle GKE
| c = Triangle Angle-Angle-Side Equality: $(1)$, $(2)$ and $(3)$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | n = 4
| l = \angle HLA
| r = \angle GLE
| c = Two Straight Lines make Equal Opposite Angles
}}
{{eqn | l = \angle EGL
| r = \text {Right Angle}
| c = Tangent to Circle is Perpendicular to Radius
}}
{{eqn | l = \angle AHL
| r = \text {Right Angle}
| c = as $HJ \perp AD$
}}
{{eqn | n = 5
| ll= \therefore
| l = \angle EGL
| r = \angle AHL
| c = Euclid's Fourth Postulate
}}
{{eqn | l = HA
| r = EF
| c = Opposite Sides and Angles of Parallelogram are Equal
}}
{{eqn | l = EF
| r = EG
| c = as both are radii of the same circle
}}
{{eqn | n = 6
| ll= \therefore
| l = HA
| r = EG
| c = Euclid's First Common Notion
}}
{{eqn | ll= \therefore
| l = \Area \triangle HAL
| r = \Area \triangle GEL
| c = Triangle Angle-Angle-Side Equality: $(4)$, $(5)$ and $(6)$
}}
{{end-eqn}}
Finally:
{{begin-eqn}}
{{eqn | l = \frac {\Area \Box ABCD} 2
| r = \frac {AD \cdot CD} 2
| c = Area of Parallelogram
}}
{{eqn | r = \Area \triangle ADC
| c = Area of Triangle in Terms of Side and Altitude
}}
{{eqn | r = \Area \triangle HAL + \Area \triangle IKC + \Area \Box DHLKI
}}
{{eqn | r = \Area \triangle GEL + \Area \triangle GKE + \Area \Box DHLKI
}}
{{eqn | r = \Area \Box DHEI
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Matt Westwood}}
Category:Euclidean Geometry
\end{proof}
|
23437
|
\section{Westwood's Puzzle/Proof 2}
Tags: Euclidean Geometry, Named Theorems
\begin{theorem}
:500px
Take any rectangle $ABCD$ and draw the diagonal $AC$.
Inscribe a circle $GFJ$ in one of the resulting triangles $\triangle ABC$.
Drop perpendiculars $IEF$ and $HEJ$ from the center of this incircle $E$ to the sides of the rectangle.
Then the area of the rectangle $DHEI$ equals half the area of the rectangle $ABCD$.
\end{theorem}
\begin{proof}
The crucial geometric truth to note is that:
:$CJ = CG, AG = AF, BF = BJ$
This follows from the fact that:
:$\triangle CEJ \cong \triangle CEG$, $\triangle AEF \cong \triangle AEG$ and $\triangle BEF \cong \triangle BEJ$
This is a direct consequence of the point $E$ being the center of the incircle of $\triangle ABC$.
Then it is just a matter of algebra.
Let $AF = a, FB = b, CJ = c$.
{{begin-eqn}}
{{eqn | l = \paren {a + b}^2 + \paren {b + c}^2
| r = \paren {a + c}^2
| c = Pythagoras's Theorem
}}
{{eqn | ll= \leadsto
| l = a^2 + 2 a b + b^2 + b^2 + 2 b c + c^2
| r = a^2 + 2 a c + c^2
| c =
}}
{{eqn | ll= \leadsto
| l = a b + b^2 + b c
| r = a c
| c =
}}
{{eqn | ll= \leadsto
| l = a b + b^2 + b c + a c
| r = 2 a c
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {a + b} \paren {b + c}
| r = 2 a c
| c =
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Matt Westwood}}
Category:Euclidean Geometry
\end{proof}
|
23438
|
\section{Whitney Embedding Theorem}
Tags: Topology, Manifolds
\begin{theorem}
Every smooth $m$-dimensional manifold admits a smooth embedding into Euclidean space $\R^{2m+1}$.
\end{theorem}
\begin{proof}
{{ProofWanted}}
{{Namedfor|Hassler Whitney|cat = Whitney}}
Category:Manifolds
\end{proof}
|
23439
|
\section{Whitney Immersion Theorem}
Tags: Differential Topology
\begin{theorem}
Let $m > 1$ be a natural number.
Every smooth $m$-dimensional manifold can be immersed in Euclidean $\left({2m-1}\right)$-space.
\end{theorem}
\begin{proof}
{{ProofWanted}}
{{Namedfor|Hassler Whitney|cat = Whitney}}
Category:Differential Topology
\end{proof}
|
23440
|
\section{Whole Space is Open in Neighborhood Space}
Tags: Neighborhood Spaces
\begin{theorem}
Let $\struct {S, \NN}$ be a neighborhood space.
Then $S$ itself is an open set of $\struct {S, \NN}$.
\end{theorem}
\begin{proof}
Let $x \in S$.
Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$.
As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$.
As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$.
{{qed}}
\end{proof}
|
23441
|
\section{Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals}
Tags: Complex Analysis
\begin{theorem}
Let $z_1$ be a non-zero wholly real number.
Let $z_2$ be a non-zero wholly imaginary number.
Then, $z_1$ and $z_2$ are linearly independent over the rational numbers $\Q$, where the group is the complex numbers $\C$.
\end{theorem}
\begin{proof}
From Rational Numbers form Subfield of Complex Numbers, the unitary module $\struct {\C, +, \times}_\Q$ over $\Q$ satisfies the unitary module axioms:
* $\Q$-Action: $\C$ is closed under multiplication, so $\Q \times \C \subset \C$.
* Distributive: $\times$ distributes over $+$.
* Associativity: $\times$ is associative.
* Multiplicative Identity: $1$ is the multiplicative identity in $\C$.
Let $a, b \in \Q$ such that:
:$a z_1 + b z_2 = 0$
By the definition of wholly imaginary, there is a real number $c$ such that:
:$c i = z_2$
where $i$ is the imaginary unit.
Therefore:
:$a z_1 + b c i = 0$
Equating real parts and imaginary parts:
:$a z_1 = 0$
:$b c = 0$
Since $z_1$ and $c$ are both non-zero, $a$ and $b$ must be zero.
The result follows from the definition of linear independence.
{{qed}}
Category:Complex Analysis
\end{proof}
|
23442
|
\section{Wilson's Theorem}
Tags: Factorials, Number Theory, Wilson's Theorem, Named Theorems, Prime Numbers, Modulo Arithmetic
\begin{theorem}
A (strictly) positive integer $p$ is a prime {{iff}}:
:$\paren {p - 1}! \equiv -1 \pmod p$
\end{theorem}
\begin{proof}
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
\end{proof}
|
23443
|
\section{Wilson's Theorem/Corollary 1}
Tags: Wilson's Theorem
\begin{theorem}
Let $p$ be a prime number.
Then $p$ is the smallest prime number which divides $\paren {p - 1}! + 1$.
\end{theorem}
\begin{proof}
From Wilson's Theorem, $p$ divides $\paren {p - 1}! + 1$.
Let $q$ be a prime number less than $p$.
Then $q$ is a divisor of $\paren {p - 1}!$ and so does not divide $\paren {p - 1}! + 1$.
{{qed}}
\end{proof}
|
23444
|
\section{Wilson's Theorem/Corollary 2}
Tags: Prime Numbers, Factorials, Modulo Arithmetic, Wilson's Theorem
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime factor of $n!$ with multiplicity $\mu$.
Let $n$ be expressed in a base $p$ representation as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m a_j p^j
| c = where $0 \le a_j < p$
}}
{{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + a_m p^m
| c = for some $m > 0$
}}
{{end-eqn}}
Then:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsb a_m! \pmod p$
\end{theorem}
\begin{proof}
Proof by induction:
Let $\map P n$ be the proposition:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$
where $p, a_0, \dots, a_k, \mu$ are as defined above.
\end{proof}
|
23445
|
\section{Wilson's Theorem/Necessary Condition}
Tags: Wilson's Theorem
\begin{theorem}
Let $p$ be a prime number.
Then:
:$\paren {p - 1}! \equiv -1 \pmod p$
\end{theorem}
\begin{proof}
If $p = 2$ the result is obvious.
Therefore we assume that $p$ is an odd prime.
Let $p$ be prime.
Consider $n \in \Z, 1 \le n < p$.
As $p$ is prime, $n \perp p$.
From Law of Inverses (Modulo Arithmetic), we have:
:$\exists n' \in \Z, 1 \le n' < p: n n' \equiv 1 \pmod p$
By Solution of Linear Congruence, for each $n$ there is '''exactly one''' such $n'$, and $\paren {n'}' = n$.
So, provided $n \ne n'$, we can pair any given $n$ from $1$ to $p$ with another $n'$ from $1$ to $p$.
We are then left with the numbers such that $n = n'$.
Then we have $n^2 \equiv 1 \pmod p$.
Consider $n^2 - 1 = \paren {n + 1} \paren {n - 1}$ from Difference of Two Squares.
So either $n + 1 \divides p$ or $n - 1 \divides p$.
Observe that these cases do not occur simultaneously, as their difference is $2$, and $p$ is an odd prime.
From Congruence Modulo Negative Number, we have that $p - 1 \equiv -1 \pmod p$.
Hence $n = 1$ or $n = p - 1$.
So, we have that $\paren {p - 1}!$ consists of numbers multiplied together as follows:
:in pairs whose product is congruent to $1 \pmod p$
:the numbers $1$ and $p - 1$.
The product of all these numbers is therefore congruent to $1 \times 1 \times \cdots \times 1 \times p - 1 \pmod p$ by modulo multiplication.
From Congruence Modulo Negative Number we therefore have that $\paren {p - 1}! \equiv -1 \pmod p$.
{{Namedfor|John Wilson}}
\end{proof}
|
23446
|
\section{Wilson's Theorem/Sufficient Condition}
Tags: Wilson's Theorem
\begin{theorem}
Let $p$ be a (strictly) positive integer such that:
:$\paren {p - 1}! \equiv -1 \pmod p$
Then $p$ is a prime number.
\end{theorem}
\begin{proof}
Assume $p$ is composite, and $q$ is a prime such that $q \divides p$.
Then both $p$ and $\paren {p - 1}!$ are divisible by $q$.
If the congruence $\paren {p - 1}! \equiv -1 \pmod p$ were satisfied, we would have $\paren {p - 1}! \equiv -1 \pmod q$.
However, this amounts to $0 \equiv -1 \pmod q$, a contradiction.
Hence for $p$ composite, the congruence $\paren {p - 1}! \equiv -1 \pmod p$ cannot hold.
{{qed}}
\end{proof}
|
23447
|
\section{Word Metric is Metric}
Tags: Group Theory, Word Metric, Examples of Metric Space, Examples of Metric Spaces
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $S$ be a generating set for $G$ which is closed under inverses (that is, $x^{-1} \in S \iff x \in S$).
Let $d_S$ be the associated word metric.
Then $d_S$ is a metric on $G$.
\end{theorem}
\begin{proof}
Let $g, h \in G$.
It is given that $S$ is a generating set for $G$.
It follows that there exist $s_1, \ldots, s_n \in S$ such that $g^{-1} \circ h = s_1 \circ \cdots \circ s_n$.
Therefore $\map {d_S} {g, h} \le n$, establishing that $\R$ is a valid codomain for the mapping $d_S$ with domain $G \times G$.
This is the form a mapping must have to be able to be a metric.
Now checking the other defining properties for a metric in turn:
\end{proof}
|
23448
|
\section{Woset is Isomorphic to Set of its Initial Segments}
Tags: Order Isomorphisms, Well-Orderings, Orderings, Order Morphisms
\begin{theorem}
Let $\struct {S, \preceq}$ be a well-ordered set.
Let:
:$A = \set {a^\prec: a \in S}$
where $a^\prec$ is the strict lower closure of $S$ determined by $a$.
Then:
:$\struct {S, \preceq} \cong \struct {A, \subseteq}$
where $\cong$ denotes order isomorphism.
\end{theorem}
\begin{proof}
Define $f: S \to A$ as:
:$\forall a \in S: \map f a = a^\prec$
where $a^\prec$ is the initial segment determined by $a$.
\end{proof}
|
23449
|
\section{Wosets are Isomorphic to Each Other or Initial Segments}
Tags: Well-Orderings, Wosets are Isomorphic to Each Other or Initial Segments
\begin{theorem}
Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be well-ordered sets.
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is order isomorphic to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is order isomorphic to an initial segment in $\struct {T, \preceq_T}$
or:
:$\struct {T, \preceq_T}$ is order isomorphic to an initial segment in $\struct {S, \preceq_S}$
\end{theorem}
\begin{proof}
We assume $S \ne \varnothing \ne T$; otherwise the theorem holds vacuously.
Define:
:$S' = S \cup \text{ initial segments in } S$
:$T' = T \cup \text{ initial segments in } T$
:$\mathcal F = \left\{ { f:S' \to T' \ \vert \ f \text{ is an order isomorphism} } \right\}$
We note that $\mathcal F$ is not empty, because it at least contains a trivial order isomorphism between singletons:
:$f_0: \left\{ { \text{smallest element in } S} \right\} \to \left\{ { \text{smallest element in } T} \right\}$
Such smallest elements are guaranteed to exist by virtue of $S$ and $T$ being well-ordered.
By the definition of initial segment, the initial segments of $S$ are subsets of $S$.
For any initial segment $I_{\alpha}$ of $S$, such a segment has an upper bound by definition, namely, $\alpha$.
Also, no initial segment of $S'$ is the entirety of $S$.
Thus every initial segment of $S'$ has an upper bound, $S$ itself.
Every chain in $S'$ has an upper bound, because it defines an initial segment.
The previous reasoning also applies to $T'$.
By Ordered Product of Tosets is Totally Ordered Set, $S' \times T'$ is itself a totally ordered set, with upper bound $S \times T$.
Thus the hypotheses of Zorn's Lemma are satisfied for $\mathcal F$.
Let $f_1$ be a maximal element of $\mathcal F$. Call its domain $A$ and its codomain $B$.
Suppose $A$ is an initial segment $I_a$ in $S$ and $B$ is an initial segment $I_b$ in $T$.
Then $f_1$ can be extended by defining $f_1\left({a}\right) = b$.
This would contradict $f_1$ being maximal, so it cannot be the case that both $A$ and $B$ are initial segments.
Then precisely one of the following hold:
:$A = S$, with $S$ order isomorphic to an initial segment in $T$
or:
:$B = T$, with $T$ order isomorphic to an initial segment in $S$
or:
*both $A = S$ and $B = T$, with $S$ order isomorphic to $T$.
{{qed}}{{improve|I suspect this proof can be adjusted to not use choice}}
{{proofread}}
{{AoC|Zorn's Lemma}}
\end{proof}
|
23450
|
\section{X + y + z equals 1 implies xy + yz + zx less than Half}
Tags: Inequalities, X + y + z equals 1 implies xy + yz + zx less than Half
\begin{theorem}
Let $x$, $y$ and $z$ be real numbers such that:
:$x + y + z = 1$
Then:
:$x y + y z + z x < \dfrac 1 2$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 1
| r = \paren {x + y + z}^2
}}
{{eqn | r = \paren {\paren {x + y} + z}^2
}}
{{eqn | r = \paren {x + y}^2 + 2 z \paren {x + y} + z^2
| c = Square of Sum
}}
{{eqn | r = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2
| c = Square of Sum
}}
{{eqn | r = 2 \paren {x y + y z + z x} + \paren {x^2 + y^2 + z^2}
}}
{{end-eqn}}
So:
:$2 \paren {x y + y z + z x} = 1 - \paren {x^2 + y^2 + z^2}$
We have:
:$x^2 + y^2 + z^2 \ge 0$
for all real numbers $x$, $y$, $z$ with equality only if:
:$x = y = z = 0$
This cannot be the case since $x + y + z = 1$, so we have:
:$x^2 + y^2 + z^2 > 0$
so:
:$2 \paren {x y + y z + z x} < 1$
giving:
:$x y + y z + z x < \dfrac 1 2$
{{qed}}
\end{proof}
|
23451
|
\section{X Choose n leq y Choose n + z Choose n-1 where n leq y leq x leq y+1 and n-1 leq z leq y}
Tags: Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Let $z$ be the unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
Its uniqueness is proved at Uniqueness of Real $z$ such that $\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$.
Then:
:$\dbinom x n \le \dbinom y n + \dbinom z {n - 1}$
\end{theorem}
\begin{proof}
If $z \ge n$, then from Ordering of Binomial Coefficients:
:$\dbinom z {n + 1} \le \dbinom y {n + 1}$
Otherwise $n - 1 \le z \le n$, and:
:$\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$
In either case:
:$(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$
Therefore:
{{begin-eqn}}
{{eqn | l = \dbinom {z + 1} {n + 1}
| r = \dbinom z {n + 1} + \dbinom z n
| c = Pascal's Rule
}}
{{eqn | o = \le
| r = \dbinom y {n + 1} + \dbinom z n
| c =
}}
{{eqn | r = \dbinom x {n + 1}
| c = by hypothesis
}}
{{end-eqn}}
and so $x \ge z + 1$.
Now we are to show that every term of the summation:
:$\ds \binom x {n + 1} - \binom y {n + 1} = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - k} t_k$
where:
:$t_k = \dbinom {x - z - 1 + k} {k + 1} - \dbinom {y - z - 1 + k} {k + 1}$
is negative.
Because $z \ge n - 1$, the binomial coefficient $\dbinom {z - k} {n - k}$ is non-negative.
Because $x \ge z + 1$, the binomial coefficient $\dbinom {x - z - 1 + k} {k + 1}$ is also non-negative.
Therefore:
:$z \le y \le x$
implies that:
:$\dbinom {y - z - 1 + k} {k + 1} \le \dbinom {x - z - 1 + k} {k + 1}$
When $x = y$ and $z = n - 1$ the result becomes:
:$\dbinom x n \le \dbinom x n + \dbinom {n - 1} {n - 1}$
which reduces to:
:$\dbinom x n \le \dbinom x n + 1$
which is true.
Otherwise:
{{begin-eqn}}
{{eqn | l = \dbinom x n - \dbinom y n - \dbinom z {n - 1}
| r = \sum_{k \mathop \ge 0} \dbinom {z - k} {n - 1 - k} \left({t_k - \delta_{k 0} }\right)
| c = where $\delta_{k 0}$ is the Kronecker delta
}}
{{eqn | r = \sum_{k \mathop \ge 0} \dfrac {n - k} {z - n + 1} \dbinom {z - k} {n - k} \left({t_k - \delta_{k 0} }\right)
| c = Factors of Binomial Coefficient
}}
{{end-eqn}}
This is less than or equal to:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \dfrac {n - 1} {z - n + 1} \dbinom {z - k} {n - k} \left({t_k - \delta_{k 0} }\right)
| r = \dfrac {n - 1} {z - n + 1} \left({\dbinom x {n + 1} - \dbinom y {n + 1} - \dbinom z n}\right)
| c =
}}
{{eqn | r = 0
| c = because $t_0 - 1 = x - y - 1 \le 0$
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
23452
|
\section{Yff's Conjecture}
Tags: Triangles
\begin{theorem}
Let $\triangle ABC$ be a triangle.
Let $\omega$ be the Brocard angle of $\triangle ABC$.
Then:
:$8 \omega^3 < ABC$
where $A, B, C$ are measured in radians.
\end{theorem}
\begin{proof}
The Abi-Khuzam Inequality states that
:$\sin A \cdot \sin B \cdot \sin C \le \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 A \cdot B \cdot C$
The maximum value of $A B C - 8 \omega^3$ occurs when two of the angles are equal.
So taking $A = B$, and using $A + B + C = \pi$, the maximum occurs at the maximum of:
:$\map f A = A^2 \paren {\pi - 2 A} - 8 \paren {\map \arccot {2 \cot A - \cot 2 A} }^3$
which occurs when:
:$2 A \paren {\pi - 3 A} - \dfrac {48 \paren {\map \arccot {\frac 1 2 \paren {3 \cot A + \tan A} } }^2 \paren {1 + 2 \cos 2 A} } {5 + 4 \cos 2 A} = 0$
{{finish|Needs expanding and completing}}
\end{proof}
|
23453
|
\section{Yoneda Embedding Theorem}
Tags: Category Theory
\begin{theorem}
Let $C$ be a locally small category.
Let $\mathbf {Set}$ be the category of sets.
Let $\sqbrk {C^{\operatorname {op} }, \mathbf {Set} }$ be the contravariant functor category.
Then the Yoneda embedding $h_- : C \to \sqbrk {C^{\operatorname {op} }, \mathbf {Set} }$ is a fully faithful embedding.
\end{theorem}
\begin{proof}
{{proof wanted|use Yoneda Lemma for Contravariant Functors}}
Category:Category Theory
\end{proof}
|
23454
|
\section{Yoneda Lemma for Covariant Functors}
Tags: Category theory, Category Theory
\begin{theorem}
Let $C$ be a locally small category.
Let $\mathbf{Set}$ be the category of sets.
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Category Theory
333203
333199
2017-12-28T16:38:51Z
Barto
3079
333203
wikitext
text/x-wiki
\end{proof}
|
23455
|
\section{Young's Inequality for Convolutions}
Tags: Measure Theory, Inequalities, Young's Inequality for Convolutions
\begin{theorem}
Let $p, q, r \in \R_{\ge 1}$ satisfy:
:$1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$
Let $\map {L^p} {\R^n}$, $\map {L^q} {\R^n}$, and $\map {L^r} {\R^n}$ be Lebesgue spaces with seminorms $\norm {\, \cdot \,}_p$, $\norm {\, \cdot \,}_q$, and $\norm {\, \cdot \,}_r$ respectively.
Let $f \in \map {L^p} {\R^n}$ and $g \in \map {L^q} {\R^n}$.
Then the convolution $f * g$ is in $\map {L^r} {\R^n}$ and the following inequality is satisfied:
:$\norm {f * g}_r \le \norm f_p \cdot \norm g_q$
\end{theorem}
\begin{proof}
{{Proofread}}
We begin by seeking to bound $\size {\map {\paren {f * g} } x}$:
{{begin-eqn}}
{{eqn | l = \map {\paren {f * g} } x
| r = \int \map f {x - y} \map g y \rd y
}}
{{eqn | l = \size {\map {\paren {f * g} } x}
| o = \le
| r = \int \size {\map f {x - y} } \cdot \size {\map g y} \rd y
}}
{{eqn | r = \int \size {\map f {x - y} }^{1 + p/r - p/r} \cdot \size {\map g y}^{1 + q/r - q/r} \rd y
}}
{{eqn | r = \int \size {\map f {x - y} }^{p/r} \cdot \size {\map g y}^{q/r} \cdot \size {\map f {x - y} }^{1 - p / r} \cdot \size {\map g y}^{1-q/r} \rd y
}}
{{eqn | r = \int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} \cdot \size {\map f {x - y} }^{\paren {r - p} / r} \cdot \size {\map g y}^{\paren {r - q} / r} \rd y
}}
{{eqn | o = \le
| r = \underset {(1)} {\underbrace {\norm {\paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} }_r} } \cdot \underset {(2)} {\underbrace {\norm {\size {\map f {x - y} }^{\paren {r - p} / r} }_{\textstyle \frac {p r} {r - p} } } } \cdot \underset {(3)} {\underbrace {\norm {\size {\map g y}^{\paren {r - q} / r} }_{\textstyle \frac {q r} {r - q} } } }
}}
{{end-eqn}}
where the last inequality is via the Generalized Hölder Inequality applied to three functions.
Note that the relation of conjugate exponents in the Generalized Hölder Inequality is satisfied:
{{begin-eqn}}
{{eqn | l = \frac 1 r + \frac {r - p} {p r} + \frac {r - q} {q r}
| r = \frac 1 r + \frac 1 p - \frac 1 r + \frac 1 q - \frac 1 r
}}
{{eqn | r = \frac 1 p + \frac 1 q - \frac 1 r
}}
{{eqn | r = 1
| c = {{hypothesis}} on $p$, $q$, $r$
}}
{{end-eqn}}
We now analyze terms $(1)$, $(2)$ and $(3)$ in turn:
{{begin-eqn}}
{{eqn | n = 1
| l = \norm {\paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} }_r
| r = \paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{\textstyle \frac 1 r \times r} \rd y}^{1/r}
}}
{{eqn | r = \paren {\int \size {\map f {x - y} }^p \cdot \size {\map g y}^q \rd y}^{1/r}
}}
{{eqn | n = 2
| l = \norm {\size {\map f {x - y} }^{\paren {r - p} /r } }_{\textstyle \frac {p r} {r - p} }
| r = \paren {\int \size {\map f {x - y} }^{\textstyle \frac {r - p} r \times \frac {p r} {r - p} } \rd y}^{\textstyle\frac {r - p} {p r} }
}}
{{eqn | r = \paren {\int \size {\map f {x - y} }^p \rd y}^{\textstyle\frac 1 p \times \frac {r - p} r}
}}
{{eqn | r = {\norm f_p}^{\paren {r - p} / r}
}}
{{eqn | n = 3
| l = \norm {\size {\map g y}^{\paren {r - q} / r} }_{\textstyle\frac {q r} {r - q} }
| r = \paren {\int \size {\map g y}^{\textstyle \frac {r - q} r \times \frac {q r} {r - q} } \rd y}^{\textstyle \frac {r - q} {q r} }
}}
{{eqn | r = \paren {\int \size {\map g y}^{\textstyle \frac {r - q} r \times \frac {q r} {r - q} } \rd y}^{\textstyle \frac {r - q} {q r} }
}}
{{eqn | r = \paren {\int \size {\map g y}^q \rd y}^{\textstyle \frac 1 q \times \frac {r - q} r}
}}
{{eqn | r = {\norm g_q}^{\paren {r - q} / r}
}}
{{end-eqn}}
With these preliminary calculations out of the way, we turn to the main proof:
{{begin-eqn}}
{{eqn | l = {\norm {f * g}_r}^r
| r = \int \size {\map {\paren {f * g} } x}^r \rd x
}}
{{eqn | o = \le
| r = \int \paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q}^{1/r} \cdot \size {\map f {x - y} }^{\paren {r - p} / r} \cdot \size {\map g y}^{\paren {r - q} / r} \rd y}^r \rd x
}}
{{eqn | r = \int \paren {\paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q} \rd y}^{1/r} \cdot {\norm f_p}^{\paren {r - p} / r} \cdot {\norm g_q}^{\paren {r - q} / r} }^r \rd x
| c = from $(1)$, $(2)$, $(3)$
}}
{{eqn | r = \int \paren {\int \paren {\size {\map f {x - y} }^p \cdot \size {\map g y}^q} \rd y} \cdot {\norm f_p}^{r - p} \cdot {\norm g_q}^{r - q} \rd x
}}
{{eqn | r = {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \iint \size {\map g y}^q \size {\map f {x - y} }^p \rd y \rd x
}}
{{eqn | r = {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \int \size {\map g y}^q \paren {\int \size {\map f {x - y} }^p \rd x} \rd y
| c = Fubini's Theorem
}}
{{eqn | r = {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \int \size {\map g y}^q \rd y \int \size {\map f x}^p \rd x
}}
{{eqn | r = {\norm f_p}^{r - p} \ {\norm g_q}^{r - q} \ {\norm g_q}^q \ {\norm f_p}^p
}}
{{eqn | r = {\norm f_p}^r \ {\norm g_q}^r
}}
{{eqn | ll= \leadsto
| l = \norm {f * g}_r
| o = \le
| r = \norm f_p \norm g_q
}}
{{end-eqn}}
{{qed}}
{{Namedfor|William Henry Young|cat = Young}}
Category:Measure Theory
Category:Inequalities
Category:Young's Inequality for Convolutions
\end{proof}
|
23456
|
\section{Young's Inequality for Increasing Functions}
Tags: Integral Calculus
\begin{theorem}
Let $a_0$ and $b_0$ be strictly positive real numbers.
Let $f: \closedint 0 {a_0} \to \closedint 0 {b_0}$ be a strictly increasing bijection.
Let $a$ and $b$ be real numbers such that $0 \le a \le a_0$ and $0 \le b \le b_0$.
Then:
:$\ds ab \le \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$
where $\ds \int$ denotes the definite integral.
\end{theorem}
\begin{proof}
200pxthumbrightThe blue colored region corresponds to $\ds \int_0^a \map f u \rd u$ and the red colored region to $\ds \int_0^b \map {f^{-1} } v \rd v$.
{{ProofWanted}}
{{Namedfor|William Henry Young|cat = Young}}
\end{proof}
|
23457
|
\section{Young's Inequality for Products}
Tags: Real Analysis, Inequalities, Analysis, Young's Inequality for Products, Named Theorems
\begin{theorem}
Let $p, q \in \R_{> 0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Then, for any $a, b \in \R_{\ge 0}$:
:$a b \le \dfrac {a^p} p + \dfrac {b^q} q$
Equality occurs {{iff}}:
:$b = a^{p - 1}$
\end{theorem}
\begin{proof}
The result is obvious if $a=0$ or $b=0$, so assume WLOG that $a > 0$ and $b > 0$.
Then:
{{begin-eqn}}
{{eqn | l = ab
| r = \exp \left({\ln\left({ab}\right)}\right)
| c = exponential and logarithm functions are inverses
}}
{{eqn | l =
| r = \exp \left({ \ln a + \ln b }\right)
| c = "sum property" of logarithms
}}
{{eqn | l =
| r = \exp \left({ \frac 1 p p \ln a + \frac 1 q q \ln b }\right)
| c = definitions of the multiplicative identity and inverse
}}
{{eqn | l =
| r = \exp \left({ \frac 1 p \ln \left( {a^p} \right) + \frac 1 q \ln \left( {b^q} \right) }\right)
| c = "power property" of logarithms
}}
{{eqn | l =
| r = \frac 1 p \exp \left( {\ln \left( {a^p} \right)} \right) + \frac 1 q \exp \left( {\ln \left( {b^q} \right)} \right)
| o = \le
| c = exponential function is convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$
}}
{{eqn | l =
| r = \frac{a^p} p + \frac{b^q} q
| c = exponential and logarithm functions are inverses
}}
{{end-eqn}}
{{qed}}
{{namedfor|William Henry Young}}
Category:Analysis
83328
83326
2012-03-12T22:08:16Z
Prime.mover
59
83328
wikitext
text/x-wiki
\end{proof}
|
23458
|
\section{Z-Module Associated with Abelian Group is Unitary Z-Module}
Tags: Abelian Groups, Z-Module Associated with Abelian Group, Examples of Unitary Modules, Group Theory, Unitary Modules, Modules
\begin{theorem}
Let $\struct {G, *}$ be an abelian group with identity $e$.
Let $\struct {G, *, \circ}_\Z$ be the $Z$-module associated with $G$.
Then $\struct {G, *, \circ}_\Z$ is a unitary $Z$-module.
\end{theorem}
\begin{proof}
The notation $*^n x$ can be written as $x^n$.
Let us verify that $\struct {G, *, \circ}_\Z$ is a unitary $\Z$-module by verifying the axioms in turn.
\end{proof}
|
23459
|
\section{Z/(m)-Module Associated with Ring of Characteristic m}
Tags: Unitary Modules, Group Theory
\begin{theorem}
Let $\left({R,+,*}\right)$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $m$.
Let $\left({\Z_m, +_m, \times_m}\right)$ be the ring of integers modulo $m$.
Let $\circ$ be the mapping from $\Z_m \times R$ to $R$ defined as:
:$\forall \left[\!\left[a\right]\!\right]_m \in \Z_m: \forall x \in R: \left[\!\left[a\right]\!\right]_m \circ x = a \cdot x$
where $\left[\!\left[a\right]\!\right]_m$ is the residue class of $a$ modulo $m$ and $a \cdot x$ is the $a$th power of $x$.
Then $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module.
\end{theorem}
\begin{proof}
Let us verify that the definition of $\circ$ is well-defined.
Let $\left[\!\left[a\right]\!\right]_m=\left[\!\left[b\right]\!\right]_m$.
Then we need to show that:
:$\forall x \in R : \left[\!\left[a\right]\!\right]_m \circ x = \left[\!\left[b\right]\!\right]_m \circ x$
By the definition of congruence:
:$\left[\!\left[a\right]\!\right]_m = \left[\!\left[b\right]\!\right]_m \iff \exists k \in \Z : a = b + k m$
Then:
{{begin-eqn}}
{{eqn | l = \left[\!\left[a\right]\!\right]_m \circ x
| r = a \cdot x
| c = Definition of $\circ$
}}
{{eqn | r = \left(b+km\right) \cdot x
}}
{{eqn | r = b \cdot x + km \cdot x
| c = Powers of Group Elements: Sum of Indices
}}
{{eqn | r = b \cdot x + k \cdot \left( m \cdot x \right)
| c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = b \cdot x + k \cdot 0_R
| c = Characteristic times Ring Element is Ring Zero
}}
{{eqn | r = b \cdot x + 0_R
| c = Power of Identity is Identity
}}
{{eqn | r = b \cdot x
}}
{{eqn | r = \left[\!\left[b\right]\!\right]_m \circ x
| c = Definition of $\circ$
}}
{{end-eqn}}
Thus, the definition of $\circ$ is well-defined.
{{qed|lemma}}
Let us verify that $\left({R, +, \circ}\right)_{\Z_m}$ is a unitary $\Z_m$-module by verifying the axioms in turn.
\end{proof}
|
23460
|
\section{Zariski's Lemma}
Tags: Commutative Algebra, Field Extensions
\begin{theorem}
Let $L / k$ be a field extension.
Let $L$ be finitely generated as an algebra over $k$.
Then $L / k$ is a finite field extension.
\end{theorem}
\begin{proof}
{{MissingLinks}}
By Noether Normalization Lemma, we find a finite and injective morphism:
:$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to L$
If we can prove that $n = 0$, the proof is complete.
{{AimForCont}} $n > 0$.
Then:
:$x_1 \in k \sqbrk {x_1, \dotsc, x_n}$
and:
:$\map \alpha {x_1} \ne 0$
We have that $\map \alpha {x_1}^{-1}$ is integral over $k \sqbrk {x_1, \dotsc, x_n}$.
Thus there exists a $m \in \N$ and $a_0, \dotsc, a_{m - 1} \in k \sqbrk {x_1, \dotsc, x_n}$ such that:
:$\ds \map \alpha {x_1}^{-m} + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{-i} = 0$
Multiplying by $\map \alpha {x_1}^m$:
{{begin-eqn}}
{{eqn | l = 0
| r = 1 + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{m - i}
}}
{{eqn | r = \map \alpha {1 + x_1 \paren {\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} } }
}}
{{end-eqn}}
and thus, since $\alpha$ is injective, we find that:
:$\ds 1 = x_1 \paren {-\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }$
which means that $x_1$ is invertible.
This contradiction shows that $n = 0$.
{{qed}}
{{Namedfor|Oscar Zariski|cat = Zariski}}
Category:Commutative Algebra
Category:Field Extensions
\end{proof}
|
23461
|
\section{Zeckendorf's Theorem}
Tags: Zeckendorf Representation, Named Theorems, Fibonacci Numbers
\begin{theorem}
Every positive integer has a unique Zeckendorf representation.
That is:
Let $n$ be a positive integer.
Then there exists a unique increasing sequence of integers $\left\langle{c_i}\right\rangle$ such that:
:$\forall i \in \N: c_i \ge 2$
:$c_{i + 1} > c_i + 1$
:$\ds n = \sum_{i \mathop = 0}^k F_{c_i}$
where $F_m$ is the $m$th Fibonacci number.
For any given $n$, such a $\left\langle{c_i}\right\rangle$ is unique.
\end{theorem}
\begin{proof}
First note that every Fibonacci number $F_n$ is itself a '''Zeckendorf representation''' of itself, where the sequence $\left\langle{c_i}\right\rangle$ contains $1$ term.
\end{proof}
|
23462
|
\section{Zeckendorf Representation of Integer shifted Left}
Tags: Zeckendorf Representations, Zeckendorf Representation
\begin{theorem}
Let $\map f x$ be the real function defined as:
:$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$
where:
:$\floor {\, \cdot \,}$ denotes the floor function
:$\phi$ denotes the golden mean.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $n$ be expressed in Zeckendorf representation:
:$n = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$
with the appropriate restrictions on $k_1, k_2, \ldots, k_r$.
Then:
:$F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1} = \map f {\phi n}$
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = F_{k_j + 1} \hat \phi + F_{k_j}
| r = \hat \phi^{k_j + 1}
| c = Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less
}}
{{eqn | ll= \leadsto
| l = F_{k_j + 1}
| r = \hat \phi^{k_j} - \frac {F_{k_j} } {\hat \phi}
| c =
}}
{{eqn | r = \hat \phi^{k_j} + \phi F_{k_j}
| c = Reciprocal Form of One Minus Golden Mean
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = F_{k_1 + 1} + F_{k_2 + 1} + \cdots + F_{k_r + 1}
| r = \phi \paren {F_{k_1} + F_{k_2} + \cdots + F_{k_r} } + \paren {\hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} }
| c =
}}
{{eqn | r = \phi n + \paren {\hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} }
| c =
}}
{{end-eqn}}
We have that:
:$\hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots \le \hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} \le \hat \phi^2 + \hat \phi^4 + \hat \phi^6 + \cdots$
Then:
{{begin-eqn}}
{{eqn | l = \hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots
| r = \hat \phi^3 \paren {1 + \hat \phi^2 + \hat \phi^4 + \cdots}
| c =
}}
{{eqn | r = \dfrac {\hat \phi^3} {1 - \hat \phi^2}
| c = Sum of Infinite Geometric Sequence
}}
{{eqn | r = \dfrac {\phi^2 \hat \phi^3} {\phi^2 - \phi^2 \hat \phi^2}
| c =
}}
{{eqn | r = \dfrac {\paren {-1}^2 \hat \phi} {\phi^2 - \paren {-1}^2}
| c = Golden Mean by One Minus Golden Mean equals Minus 1
}}
{{eqn | r = \dfrac {\hat \phi} {\phi^2 - 1}
| c =
}}
{{eqn | r = \dfrac {1 - \phi} {\phi^2 - 1}
| c = {{Defof|One Minus Golden Mean}}
}}
{{eqn | r = \dfrac {1 - \phi} {1 + \phi - 1}
| c = Square of Golden Mean equals One plus Golden Mean
}}
{{eqn | r = \phi^{-1} - 1
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \hat \phi^2 + \hat \phi^4 + \hat \phi^6 + \cdots
| r = \frac 1 {\hat \phi} \paren {\hat \phi^3 + \hat \phi^5 + \hat \phi^7 + \cdots}
| c =
}}
{{eqn | r = \frac 1 {\hat \phi} \paren {\phi^{-1} - 1}
| c = from above
}}
{{eqn | r = -\phi \paren {\frac 1 \phi - 1}
| c = Reciprocal Form of One Minus Golden Mean
}}
{{eqn | r = -1 + \phi
| c =
}}
{{eqn | r = \frac 1 {\phi}
| c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | r = \phi^{-1}
| c =
}}
{{end-eqn}}
Thus:
:$\phi^{-1} - 1 \le \hat \phi^{k_1} + \hat \phi^{k_2} + \cdots + \hat \phi^{k_r} \le \phi^{-1}$
and the result follows.
{{qed}}
\end{proof}
|
23463
|
\section{Zeckendorf Representation of Integer shifted Right}
Tags: Zeckendorf Representation
\begin{theorem}
Let $f \left({x}\right)$ be the real function defined as:
:$\forall x \in \R: f \left({x}\right) = \left\lfloor{x + \phi^{-1} }\right\rfloor$
where:
:$\left\lfloor{\, \cdot \,}\right\rfloor$ denotes the floor function
:$\phi$ denotes the golden mean.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $n$ be expressed in Zeckendorf representation:
:$n = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$
with the appropriate restrictions on $k_1, k_2, \ldots, k_r$.
Then:
:$F_{k_1 - 1} + F_{k_2 - 1} + \cdots + F_{k_r - 1} = f \left({\phi^{-1} n}\right)$
\end{theorem}
\begin{proof}
Follows directly from Zeckendorf Representation of Integer shifted Left, substituting $F_{k_j - 1}$ for $F_{k_j}$ throughout.
{{qed}}
\end{proof}
|
23464
|
\section{Zenith Distance is Complement of Celestial Altitude}
Tags: Spherical Astronomy
\begin{theorem}
Let $X$ be the position of a star (or other celestial body) on the celestial sphere.
The zenith distance $z$ of $X$ is the complement of the altitude $a$ of $X$:
:$z = 90 \degrees - a$
\end{theorem}
\begin{proof}
The vertical circle through $X$ is defined as the great circle that passes through $Z$.
By definition, the angle of the arc from $Z$ to the horizon is a right angle.
Hence $z + a = 90 \degrees$.
The result follows.
{{qed}}
\end{proof}
|
23465
|
\section{Zenith Distance of North Celestial Pole equals Colatitude of Observer}
Tags: Spherical Astronomy
\begin{theorem}
Let $O$ be an observer of the celestial sphere.
Let $P$ be the position of the north celestial pole with respect to $O$.
Let $z$ denote the zenith distance of $P$.
Let $\psi$ denote the (terrestrial) colatitude of $O$.
Then:
:$z = \psi$
\end{theorem}
\begin{proof}
Let $Z$ denote the zenith.
The zenith distance $z$ of $P$ is by definition the length of the arc $PZ$ of the prime vertical.
This in turn is defined as the angle $\angle POZ$ subtended by $PZ$ at $O$.
This is equivalent to the angle between the radius of Earth through $O$ and Earth's axis.
This is by definition the (terrestrial) colatitude of $O$.
Hence the result.
{{qed}}
\end{proof}
|
23466
|
\section{Zermelo's Theorem (Set Theory)}
Tags: Set Theory, Named Theorems, Axiom of Choice
\begin{theorem}
Every set of cardinals is well-ordered with respect to $\le$.
\end{theorem}
\begin{proof}
Let $S_1$ and $S_2$ be sets which are not empty.
Suppose there exists an injection $f: S_1 \to S_2$ and another injection $g: S_2 \to S_1$.
Then by the Cantor-Bernstein-Schröder Theorem there exists a bijection between $S_1$ and $S_2$ and by definition $S_1$ is equivalent to $S_2$.
Let $\AA$ be the set of invertible mappings $\phi: A \to B$ where $A \subseteq S_1$ and $B \subseteq S_2$.
Since $S_1$ and $S_2$ are not empty, $\exists s_1 \in S_1$ and $\exists s_2 \in S_2$.
Thus we can construct the mapping $\alpha: \set {s_1} \to \set {s_2}$ such that $\map \alpha {s_1} = s_2$.
This is trivially an invertible mapping, so $\AA$ is not empty.
We can impose an ordering $\le$ on $\AA$ by letting $\phi_1 \le \phi_2 \iff \phi_1 \subseteq \phi_2$, that is, if $\phi_2$ is an extension of $\phi_1$.
Let $\CC$ be a chain in $\struct {\AA, \le}$.
Then $\ds \bigcup \set {\phi \in \CC}$ is an upper bound of every $\phi \in \CC$, and it lies in $\AA$.
The conditions of Zorn's Lemma are satisfied, so we can find a maximal element $M$ in $\AA$.
Let:
:$M_1 = \set {s_1: \tuple {s_1, s_2} \in M}$
:$M_2 = \set {s_2: \tuple {s_1, s_2} \in M}$
We have that $M_1 \subsetneq S_1$ and $M_2 \subsetneq S_2$ both together contradict the fact that $M$ is maximal element.
Thus either $M_1 = S_1$ or $M_2 = S_2$, and possibly both.
Thus either:
:$M$ is an injection of $S_1$ into $S_2$
or:
:$M^{-1}$ is an injection of $S_2$ into $S_1$.
Thus either $S_1 \le S_2$ or $S_2 \le S_1$, and the result follows.
{{qed}}
{{AoC||Zorn's Lemma}}
\end{proof}
|
23467
|
\section{Zero (Category) is Initial Object}
Tags: Category of Categories
\begin{theorem}
Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf 0$ be the zero category.
Then $\mathbf 0$ is an initial object of $\mathbf{Cat}$.
\end{theorem}
\begin{proof}
Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category.
By Empty Mapping is Unique, there are unique mappings:
:$F_0: \mathbf 0_0 = \O \to \mathbf C_0$
:$F_1: \mathbf 0_1 = \O \to \mathbf C_1$
making $F: \mathbf 0 \to \mathbf C$ a functor by vacuous truth.
That $F_0$ and $F_1$ are actually mappings follows from $\mathbf C$ being a small category.
Hence the result, by definition of initial object.
{{qed}}
\end{proof}
|
23468
|
\section{Zero Choose n}
Tags: Binomial Coefficients, Examples of Binomial Coefficients
\begin{theorem}
:$\dbinom 0 n = \delta_{0 n}$
where:
:$\dbinom 0 n$ denotes a binomial coefficient
:$\delta_{0 n}$ denotes the Kronecker delta.
\end{theorem}
\begin{proof}
By definition of binomial coefficient:
:$\dbinom m n = \begin{cases}
\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\
& \\
0 & : \text { otherwise } \end{cases}$
Thus when $n > 0$:
:$\dbinom 0 n = 0$
and when $n = 0$:
:$\dbinom 0 0 = \dfrac {0!} {0! \paren {0 - 0}!} = 1$
by definition of factorial.
Hence the result by definition of the Kronecker delta.
{{qed}}
\end{proof}
|
23469
|
\section{Zero Complement is Not Empty}
Tags: Naturally Ordered Semigroup
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $S^*$ be the zero complement of $S$.
Then $S^*$ is not empty.
\end{theorem}
\begin{proof}
From {{NOSAxiom|4}}, we have:
:$\exists m, n \in S: m \ne n$
That is, there are at least two distinct elements in $S$.
Therefore, there must be at least one element in $S^* = S \setminus \set 0$.
So:
:$S^* = S \setminus \set 0 \ne \O$
{{qed}}
\end{proof}
|
23470
|
\section{Zero Definite Integral of Nowhere Negative Function implies Zero Function}
Tags: Integral Calculus, Definite Integrals
\begin{theorem}
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $h: \closedint a b \to \R$ be a continuous real function such that:
:$\forall x \in \closedint a b: \map h x \ge 0$
Let:
:$\ds \int_a^b \map h x \rd x = 0$
Then:
:$\forall x \in \closedint a b: \map h x = 0$
\end{theorem}
\begin{proof}
{{AimForCont}} that:
:$\exists c \in \closedint a b: \map h c > 0$
As $h$ is continuous, there exists some closed real interval $\closedint r s \subseteq \closedint a b$ where $r < s$ such that:
:$\exists \epsilon \in \R_{>0}: \forall x \in \closedint r s: \map h x > \dfrac {\map h c} 2$
From Sign of Function Matches Sign of Definite Integral:
:$\ds \int_r^s \map h x \rd x > 0$
This makes a strictly positive contribution to the integral.
$h$ is still continuous on $\closedint a r$ and $\closedint s b$.
So, again from Sign of Function Matches Sign of Definite Integral:
:$\forall x \in \closedint a r: \map f x \ge 0 \implies \ds \int_a^r \map f x \rd x \ge 0$
:$\forall x \in \closedint s b: \map f x \ge 0 \implies \ds \int_s^b \map f x \rd x \ge 0$
Thus as:
:$\ds \int_a^b \map h x \rd x = \int_a^r \map f x \rd x + \int_r^s \map f x \rd x + \int_s^b \map f x \rd x$
it follows that:
:$\ds \int_a^b \map h x \rd x > 0$
But by hypothesis:
:$\ds \int_a^b \map h x \rd x = 0$
Thus by contradiction, there can be no $c \in \closedint a b$ such that $\map h c > 0$.
Hence the result.
{{qed}}
\end{proof}
|
23471
|
\section{Zero Derivative implies Constant Complex Function}
Tags: Complex Analysis
\begin{theorem}
Let $D \subseteq \C$ be a connected domain of $\C$.
Let $f: D \to \C$ be a complex-differentiable function.
For all $z \in D$, let $\map {f'} z = 0$.
Then $f$ is constant on $D$.
\end{theorem}
\begin{proof}
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations:
:$\map u {x, y} = \map \Re {\map f z}$
:$\map v {x, y} = \map \Im {\map f z}$
By the Cauchy-Riemann Equations:
:$f' = \dfrac {\partial f} {\partial x} = \dfrac {\partial u} {\partial x} + i \dfrac {\partial v} {\partial x}$
:$f' = -i \dfrac {\partial f} {\partial y} = \dfrac {\partial v} {\partial y} - i \dfrac {\partial u} {\partial y}$
As $f' = 0$ by assumption, this implies:
:$0 = \dfrac {\partial u} {\partial x} = \dfrac {\partial u} {\partial y} = \dfrac {\partial v} {\partial x} = \dfrac {\partial v} {\partial y}$
Then Zero Derivative implies Constant Function shows that $\map u {x + t, y} = \map u {x, y}$ for all $t \in \R$.
Similar results apply for the other three partial derivatives.
Let $z, w \in D$.
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C$ in $D$ with endpoints $z$ and $w$.
The contour $C$ is a concatenation of directed smooth curves that can be parameterized as line segments on one of these two forms:
:$(1): \quad \map \gamma t = z_0 + t r$
:$(2): \quad \map \gamma t = z_0 + i t r$
for some $z_0 \in D$ and $r \in \R$ for all $t \in \closedint 0 1$.
If $z_1 \in D$ lies on the same line segment as $z_0$, it follows that for parameterizations of type $(1)$:
{{begin-eqn}}
{{eqn | l = \map f {z_1}
| r = \map u {z_0 + t r} + \map v {z_0 + t r}
| c = for some $t \in \closedint 0 1$
}}
{{eqn | r = \map u {z_0} + \map v {z_0}
| c = Zero Derivative implies Constant Function
}}
{{eqn | r = \map f {z_0}
}}
{{end-eqn}}
Similarly, for parameterizations of type $(2)$:
{{begin-eqn}}
{{eqn | l = \map f {z_1}
| r = \map u {z_0 + i t r} + \map v {z_0 + i t r}
| c = for some $t \in \closedint 0 1$
}}
{{eqn | r = \map f {z_0}
| c = Zero Derivative implies Constant Function
}}
{{end-eqn}}
As the image of $C$ is connected by these line segments, it follows that for all $z_0$ and $z_1$ in the image of $C$:
:$\map f {z_0} = \map f {z_1}$
In particular:
:$\map f z = \map f w$
so $f$ is constant on $D$.
{{qed}}
\end{proof}
|
23472
|
\section{Zero Derivative implies Constant Function}
Tags: Differential Calculus, Constant Mappings
\begin{theorem}
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Suppose that:
:$\forall x \in \openint a b: \map {f'} x = 0$
Then $f$ is constant on $\closedint a b$.
\end{theorem}
\begin{proof}
When $x = a$ then $\map f x = \map f a$ by definition of mapping.
Otherwise, let $x \in \hointl a b$.
We have that:
:$f$ is continuous on the closed interval $\closedint a b$
:$f$ is differentiable on the open interval $\openint a b$
Hence it satisfies the conditions of the Mean Value Theorem on $\closedint a b$.
Hence:
:$\exists \xi \in \openint a x: \map {f'} \xi = \dfrac {\map f x - \map f a} {x - a}$
But by our supposition:
:$\forall x \in \openint a b: \map {f'} x = 0$
which means:
:$\forall x \in \openint a b: \map f x - \map f a = 0$
and hence:
:$\forall x \in \openint a b: \map f x = \map f a$
{{qed}}
\end{proof}
|
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