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22673
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\section{Topological Completeness is not Hereditary}
Tags: Topologically Complete Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is topologically complete.
Let $H \subseteq S$ be a subset of $S$.
Let $\struct {H, \tau_H}$ be the topological subspace of $T$ induced by $H$.
Then it is not necessarily the case that $\struct {H, \tau_H}$ is also topologically complete.
That is, topological completeness is not hereditary.
\end{theorem}
\begin{proof}
Let $\struct {\R, d}$ denote the real number line under the Euclidean metric.
Let $\struct {\Q, d}$ denote the rational number space under the Euclidean metric.
We have that $\Q \subset \R$ by definition.
Let $T = \struct {\R, \tau_d}$ be the topological space induced on $\R$ by $d$.
By Real Number Line is Complete Metric Space, $\struct {\R, d}$ is a complete metric space.
Thus, by definition, $T$ is topologically complete.
Let $T' = \struct {\Q, \tau_d}$ be the topological space induced on $\Q$ by $d$.
By Rational Number Space is not Complete Metric Space, $\struct {\Q, d}$ is not a complete metric space.
Thus, by definition, $T'$ is not topologically complete.
Hence the result.
{{qed}}
\end{proof}
|
22674
|
\section{Topological Equivalence is Equivalence Relation}
Tags: Examples of Equivalence Relations, Equivalence Relations, Topologically Equivalent Metrics, Topological Equivalence, Metric Spaces
\begin{theorem}
Let $A$ be a set.
Let $\DD$ be the set of all metrics on $A$.
Let $\sim$ be the relation on $\DD$ defined as:
:$\forall d_1, d_2 \in \DD: d_1 \sim d_2 \iff d_1$ is topologically equivalent to $d_2$
Then $\sim$ is an equivalence relation.
\end{theorem}
\begin{proof}
Let $A$ be a set and let $\DD$ be the set of all metrics on $A$.
In the following, let $d_1, d_2, d_3 \subseteq \DD$ be arbitrary.
Checking in turn each of the criteria for equivalence:
\end{proof}
|
22675
|
\section{Topological Product of Compact Spaces/Finite Product}
Tags: Topology
\begin{theorem}
Let $T_1, T_2, \ldots, T_n$ be topological spaces.
Let $\ds \prod_{i \mathop = 1}^n T_i$ be the product space of $T_1, T_2, \ldots, T_n$.
Then $\ds \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact.
\end{theorem}
\begin{proof}
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact
\end{proof}
|
22676
|
\section{Topological Properties of Non-Archimedean Division Rings/Centers of Closed Balls}
Tags: Normed Division Rings, Topological Properties of Non-Archimedean Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r \in \R_{\gt 0}$.
Then:
:$y \in \map { {B_r}^-} x \implies \map { {B_r}^-} y = \map { {B_r}^-} x$
\end{theorem}
\begin{proof}
Let $y \in \map { {B_r}^-} x$.
Let $a \in \map { {B_r}^-} y$.
By the definition of an closed ball, then:
:$\norm {a - y} \le r$
:$\norm {y - x} \le r$
Hence:
{{begin-eqn}}
{{eqn | l = \norm {a - x}
| r = \norm {a - y + y - x}
}}
{{eqn | r = \max \set {\norm {a - y}, \norm {y - x} }
| o = \le
| c = {{Defof|Non-Archimedean Division Ring Norm}}
}}
{{eqn | r = r
| o = \le
| c =
}}
{{end-eqn}}
By the definition of a closed ball, then:
:$a \in \map { {B_r}^-} x$.
Hence:
:$\map { {B_r}^-} y \subseteq \map { {B_r}^-} x$
By Norm of Negative then:
:$\norm {x - y} \le r$
By the definition of a closed ball, then:
:$x \in \map { {B_r}^-} y$
Similarly it follows that:
:$\map { {B_r}^-} x \subseteq \map { {B_r}^-} y$
By set equality:
:$\map { {B_r}^-} x = \map { {B_r}^-} y$
{{qed}}
\end{proof}
|
22677
|
\section{Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls}
Tags: Normed Division Rings, Topological Properties of Non-Archimedean Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
For $a \in R$ and $\epsilon \in \R_{>0}$ let:
:$\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$
Let $x, y \in R$.
Let $r \in \R_{>0}$.
Then:
:$y \in \map {B_r} x \implies \map {B_r} y = \map {B_r} x$
\end{theorem}
\begin{proof}
Let $y \in \map {B_r} x$.
Let $a \in \map {B_r} y$.
By the definition of an open ball, then:
:$\norm {a - y} < r$
:$\norm {y - x} < r$
Hence:
{{begin-eqn}}
{{eqn | l = \norm {a - x}
| r = \norm {a - y + y - x}
}}
{{eqn | o = \le
| r = \max \set {\norm {a - y}, \norm{y - x} }
| c = {{Defof|Non-Archimedean Division Ring Norm}}
}}
{{eqn | o = <
| r = r
| c =
}}
{{end-eqn}}
By the definition of an open ball:
:$a \in \map {B_r} x$
Hence:
:$\map {B_r} y \subseteq \map {B_r} x$
By Norm of Negative:
:$\norm {x - y} < r$
By the definition of an open ball:
:$x \in \map {B_r} y$
Similarly it follows that:
:$\map {B_r} x \subseteq \map {B_r} y$
By set equality:
:$\map {B_r} y = \map {B_r} x$
{{qed}}
\end{proof}
|
22678
|
\section{Topological Properties of Non-Archimedean Division Rings/Closed Balls are Clopen}
Tags: Normed Division Rings, Topological Properties of Non-Archimedean Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{\gt 0}$.
Let $\map { {B_r}^-} x$ be the closed $r$-ball of $x$ in $\struct {R,\norm {\,\cdot\,} }$
Then:
:The closed $r$-ball of $x$, $\map { {B_r}^-} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
\end{theorem}
\begin{proof}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By the definition of a closed ball in $\norm {\,\cdot\,}$ then:
:$\map { {B_r}^-} x$ is a closed ball in the metric space $\struct {R, d}$.
By Closed Ball is Closed in Metric Space then $\map { {B_r}^-} c$ is closed in $d$.
So it remains to show that $\map { {B_r}^-} x$ is open in $d$.
Let $y \in \map { {B_r}^-} x$.
By Centers of Closed Balls then:
:$\map { {B_r}^-} y = \map { {B_r}^-} x$
By the definition of an open ball then:
:$y \in \map {B_r} y \subseteq \map { {B_r}^-} y = \map { {B_r}^-} x$
By the definition of an open set in a metric space, $\map { {B_r}^-} x$ is open.
{{qed}}
\end{proof}
|
22679
|
\section{Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen}
Tags: Normed Division Rings, Topological Properties of Non-Archimedean Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm {\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{>0}$.
Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $\struct {R, \norm {\,\cdot\,} }$
Then:
:The open $r$-ball of $x$, $\map {B_r} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
\end{theorem}
\begin{proof}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By the definition of an open ball in $\norm {\,\cdot\,}$:
:$\map {B_r} x$ is an open ball in the metric space $\struct {R, d}$.
By Open Ball of Metric Space is Open Set then $\map {B_r} x$ is open in $\struct {R, d}$.
So it remains to show that $\map {B_r} x$ is closed in $\struct {R, d}$.
Let $\map \cl {\map {B_r} x}$ denote the closure of $\map {B_r} x$.
Let $y \in \map \cl {\map {B_r} x}$.
By the definition of the closure of $\map {B_r} x$ then:
:$\forall s > 0: \map {B_s} y \cap \map {B_r} x \ne \O$
In particular:
:$\map {B_r} y \cap \map {B_r} x \ne \O$
Let $z \in \map {B_r} y \cap \map {B_r} x$.
By Centers of Open Balls:
:$\map {B_r} y = B_r \paren{z} = \map {B_r} x$
By the definition of an open ball:
:$y \in \map {B_r} y = \map {B_r} x$.
Hence:
:$\map \cl {\map {B_r} x} \subseteq \map {B_r} x$
By Subset of Metric Space is Subset of its Closure then:
:$\map {B_r} x \subseteq \map \cl {\map {B_r} x}$
So by definition of set equality:
:$\map \cl {\map {B_r} x} = \map {B_r} x$
By Set is Closed iff Equals Topological Closure then $\map {B_r} x$ is closed.
{{qed}}
\end{proof}
|
22680
|
\section{Topological Properties of Non-Archimedean Division Rings/Spheres are Clopen}
Tags: Normed Division Rings, Topological Properties of Non-Archimedean Division Rings
\begin{theorem}
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,
Let $x \in R$.
Let $r \in \R_{>0}$.
Let $\map {S_r} x$ denote the $r$-sphere of $x$ in $\struct {R, \norm {\,\cdot\,}}$
Then:
:The $r$-sphere of $x$, $\map {S_r} x$, is both open and closed in the metric induced by $\norm {\,\cdot\,}$.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \map {S_r} x
| r = \set {y \in R : \norm {y - x} = r}
| c = {{Defof|Sphere in Normed Division Ring}}
}}
{{eqn | r = \set {y \in R : \norm {y - x} \le r} \cap \set {y \in R : \norm{y - x} \ge r}
}}
{{eqn | r = \map { {B_r}^-} x \cap \paren {R \setminus \map {B_r} x}
| c = {{Defof|Open Ball of Normed Division Ring}} and {{Defof|Closed Ball of Normed Division Ring}}
}}
{{end-eqn}}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
By Open Balls Are Clopen then $\map {B_r} x$ is both open and closed in $d$.
By Metric Induces Topology then $R \setminus \map {B_r} x$ is is both open and closed in $d$.
By Closed Balls Are Clopen then $\map { {B_r}^-} x$ is both open and closed in $d$.
By Metric Induces Topology then the intersection of a finite number of open sets is open.
Hence $\map {S_r} x$ is open in the metric space $\struct {R, d}$.
By Intersection of Closed Sets is Closed in Topological Space then $\map {S_r} x$ is closed in $\struct {R, d}$.
The result follows.
{{qed}}
\end{proof}
|
22681
|
\section{Topological Space Induced by Neighborhood Space is Topological Space}
Tags: Neighborhood Spaces
\begin{theorem}
Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\struct {S, \NN}$.
Then $\struct {S, \tau}$ is a topological space.
\end{theorem}
\begin{proof}
From Neighborhood Space is Topological Space, $\struct {S, \NN}$ is a topological space.
Consequently, the open sets of $\struct {S, \tau}$ are exactly the open sets of $\struct {S, \NN}$.
{{qed}}
Category:Neighborhood Spaces
\end{proof}
|
22682
|
\section{Topological Space induced by Neighborhood Space induced by Topological Space}
Tags: Neighborhood Spaces, Topology
\begin{theorem}
Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\struct {S, \tau'}$ be the topological space induced by $\NN$ on $S$.
Then $\tau = \tau'$.
\end{theorem}
\begin{proof}
Let $U \in \tau$ be an open set of $\struct {S, \tau}$.
By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.
By definition, $U$ is an open set of $\struct {S, \NN}$.
Thus by definition of neighborhood space induced by $\tau$ on $S$:
:$U \in \NN$
Then, by definition of the topological space induced by $\NN$ on $S$:
:$U \in \tau'$.
Thus $\tau \subseteq \tau'$.
Now let $U \in \tau'$ be an open set of $\struct {S, \tau'}$.
Then, by definition, in the neighborhood space $\struct {S, \NN}$, $U$ is an open set of $\struct {S, \NN}$.
That is, $U$ is a neighborhood in $\struct {S, \NN}$ of each of its points.
But the neighborhoods of $\struct {S, \NN}$ are the neighborhoods of $\struct {S, \tau}$.
Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\struct {S, \tau}$.
That is:
:$U \in \tau$.
It follows by definition of set equality that $\tau = \tau'$.
{{qed}}
\end{proof}
|
22683
|
\section{Topological Space is Open Neighborhood of Subset}
Tags: Topology, Neighborhoods
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Then $S$ is an open neighborhood of $H$.
\end{theorem}
\begin{proof}
From Underlying Set of Topological Space is Clopen, $S$ is open in $T$.
By hypothesis, $H \subseteq S$.
The result follows from Open Superset is Open Neighborhood.
{{qed}}
Category:Topology
Category:Neighborhoods
\end{proof}
|
22684
|
\section{Topological Space is Quasiuniformizable}
Tags: Uniformities
\begin{theorem}
Every topological space is quasiuniformizable.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be defined as:
:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
Then $\BB$ is a filter sub-basis for a quasiuniformity on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a quasiuniform space.
{{finish|Prove that above assertion.}}
\end{proof}
|
22685
|
\section{Topological Space with Generic Point is Path-Connected}
Tags: Irreducible Spaces, Path-Connected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have a generic point $g \in S$.
Then $T$ is path-connected.
\end{theorem}
\begin{proof}
By Path-Connectedness is Equivalence Relation, it suffices to prove that every point is path-connected with $g$.
Let $x \in S$.
Define a path $\gamma: \closedint 0 1 \to S$ by:
:$\map \gamma t = \begin{cases}
x & : t \le \dfrac 1 2 \\
g & : t > \dfrac 1 2
\end{cases}$
We show that $\gamma$ is indeed continuous.
Let $U \subseteq S$ be open and non-empty.
Because $g$ is a generic point, $g \in U$.
If $x \in U$, then its preimage $\gamma^{-1} \sqbrk U = \closedint 0 1$ is open.
If $x \notin U$, then its preimage $\gamma^{-1} \sqbrk U = \hointl {\dfrac 1 2} 1$ is also open.
Thus $x$ and $g$ are path-connected.
{{qed}}
Category:Path-Connected Spaces
\end{proof}
|
22686
|
\section{Topological Space with One Quasicomponent is Connected}
Tags: Connectedness Between Two Points, Connected Spaces, Connectedness, Definitions: Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which has one quasicomponent.
Then $T$ is connected.
\end{theorem}
\begin{proof}
Let $x \in S$.
By hypothesis, the quasicomponent of $x$ is $S$ itself.
Thus by definition of quasicomponent:
:$\forall y \in S: y \sim x$
where $\sim$ is the relation defined on $T$ as:
:$x \sim y \iff T$ is connected between the two points $x$ and $y$
Let $K = \ds \bigcap_{x \mathop \in U} U: U$ is clopen in $T$.
By Quasicomponent is Intersection of Clopen Sets:
:$\ds \bigcap K = S$
Thus there is no non-empty clopen set of $T$ apart from $S$.
The result follows by definition of connected space.
{{qed}}
\end{proof}
|
22687
|
\section{Topological Subspace of Real Number Line is Lindelöf}
Tags: Real Number Line with Euclidean Topology, Lindelöf Spaces, Real Number Space
\begin{theorem}
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $W$ be a non-empty subset of $\R$.
Then $R_W$ is Lindelöf
where $R_W$ denotes the topological subspace of $R$ on $W$.
\end{theorem}
\begin{proof}
Let $\CC$ be a open cover for $W$.
Define $Q := \set {\openint a b: a, b \in \Q}$
Define a mapping $h: Q \to \Q \times \Q$:
:$\forall \openint a b \in Q: \map h {\openint a b} = \tuple {a, b}$
It is easy to see by definition that
:$h$ is an injection.
By Injection iff Cardinal Inequality:
:$\card Q \le \card {\Q \times \Q}$
where $\card Q$ deontes the cardinality of $Q$.
By Rational Numbers are Countably Infinite:
:$\Q$ is countably infinite.
By definition of countably infinite:
:there exists a bijection $\Q \to \N$
By definitions of set equality and cardinality:
:$\card \Q = \card \N$
By Aleph Zero equals Cardinality of Naturals:
:$\card \Q = \aleph_0$
By Cardinal Product Equal to Maximum:
:$\card {\Q \times \Q} = \max \set {\aleph_0, \aleph_0} = \aleph_0$
By Countable iff Cardinality not greater than Aleph Zero:
:$Q$ is countable.
By definition of cover:
:$W \subseteq \bigcup \CC$
By definition of imion:
:$\forall x \in W: \exists U \in \CC: x \in U$
By Axiom of Choice define a mapping $f: W \to \CC$:
:$\forall x \in W: x \in \map f x$
We will prove that
:$\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq \map f x$
Let $x \in W$.
By definition of open cover:
:$\map f x$ is open in $R_W$.
By definition of topological subspace:
:there exists U a subset of $\R$ such that
::$U$ is open in $R$ and $U \cap W = \map f x$
By definition of $f$:
:$x \in \map f x$
By definition of open set in metric space:
:$\exists r > 0: \map {B_r} x \subseteq U$
:$\openint {x - r} {x + r} \subseteq U$
By Between two Real Numbers exists Rational Number:
:$\exists q \in \Q: x - r < q < x$ and $\exists p \in \Q: x < p < x + r$
By definition of $Q$:
:$\openint q p \in Q$
Thus by definition of open real interval:
:$x \in \openint q p \subseteq \openint {x - r} {x + r}$
By Subset Relation is Transitive:
:$\openint q p \subseteq U$
Thus by Set Intersection Preserves Subsets/Corollary:
:$\openint q p \cap W \subseteq \map f x$
By Axiom of Choice define a mapping $f_1: W \to Q$:
:$\forall x \in W: x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$
By definitions of image of set and image of mapping:
:$\forall A \in \Img {f_1}: \exists x \in W: x \in f_1^{-1} \sqbrk {\set A}$
By Axiom of Choice define a mapping $c: \Img {f_1} \to W$:
:$\forall A \in \Img {f_1}: \map c A \in f_1^{-1} \sqbrk {\set A}$
Define a mapping $g: \Img {f_1} \to \CC$:
:$g := f \circ c$
Define $\GG = \Img g$.
Thus $\GG \subseteq \CC$
By definition of image of mapping:
:$\Img {f_1} \subseteq Q$
By Subset of Countable Set is Countable;
:$\Img {f_1}$ is countable.
By Surjection iff Cardinal Inequality:
:$\size {\Img g} \le \size {\Img {f_1} }$
Thus by Countable iff Cardinality not greater than Aleph Zero:
:$\GG$ is countable.
It remains to prove that
:$\GG$ is a cover for $W$.
Let $x \in W$.
By definition of $f_1$:
:$ x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$
By definition of image of mapping:
:$\map {f_1} x \in \Img {f_1}$
Then by definition of $c$:
:$y := \map c {\map {f_1} x} \in f_1^{-1} \sqbrk {\set {\map {f_1} x} }$
By definition of $f_1$:
:$y \in \map {f_1} y \land \map {f_1} y \cap W \subseteq \map f y$
By definition of image of set:
:$\map {f_1} y = \map {f_1} x$
Then by definitions of subset and intersection:
:$x \in \map f y$
By definition of composition of mappings:
:$\map f y = \map g {\map {f_1} x}$
By definition of image of mapping:
:$\map g {\map {f_1} x} \in \GG$
Thus by definition of union:
:$x \in \bigcup \GG$
Thus the result.
{{qed}}
\end{proof}
|
22688
|
\section{Topologically Distinguishable Points are Distinct}
Tags: Separation Axioms
\begin{theorem}
Let $T = \left({X, \tau}\right)$ be a topological space.
Let $x, y \in X$ be topologically distinguishable.
Then the singleton sets $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are disjoint and so:
: $x \ne y$
\end{theorem}
\begin{proof}
Let $x$ and $y$ be '''topologically distinguishable'''.
Then either:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$
or:
:$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$
Suppose $x = y$.
Then:
:$\neg \left({\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}\right)$
and:
:$\neg \left({\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}\right)$
Hence the result by Proof by Contradiction:
:$x \ne y$
and so by Singleton Equality:
:$\left\{{x}\right\} \ne \left\{{y}\right\}$
{{qed}}
Category:Separation Axioms
\end{proof}
|
22689
|
\section{Topologically Equivalent Metrics induce Equal Topologies}
Tags: Topologically Equivalent Metrics
\begin{theorem}
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be topologically equivalent.
Let $\tau_1$ and $\tau_2$ denote the topologies on $A$ induced by $d_1$ and $d_2$ respectively.
Then $\tau_1$ and $\tau_2$ are equal.
\end{theorem}
\begin{proof}
Let $d_1$ and $d_2$ be topologically equivalent by hypothesis.
By definition of topological equivalence:
$d_1$ and $d_2$ are '''topologically equivalent''' {{iff}}:
:$U \subseteq A$ is $d_1$-open {{iff}} $U \subseteq A$ is $d_2$-open.
By definition of the induced topology:
:The '''topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$''' is defined as the set $\tau$ of all open sets of $M$.
Hence it follows that:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_1}$
{{iff}}:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_2}$
That is:
:$U \in \tau_1 \iff U \in \tau_2$
and the result follows.
{{qed}}
Category:Topologically Equivalent Metrics
\end{proof}
|
22690
|
\section{Topologies are not necessarily Comparable by Coarseness}
Tags: Topology
\begin{theorem}
Let $S$ be a set with at least $2$ elements.
Let $\mathbb T$ be the set of all topologies on $S$.
For two topologies $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is coarser than $\tau_b$.
Then there exist $\tau_1, \tau_2 \in \mathbb T$ such that neither:
:$\tau_1 \le \tau_2$
nor:
:$\tau_2 \le \tau_1$
That is, there are always topologies on $S$ which are non-comparable.
\end{theorem}
\begin{proof}
Let $a, b \in S$.
Let:
:$\tau_a$ be the particular point topology with respect to $a$ on $S$
:$\tau_b$ be the particular point topology with respect to $b$ on $S$
From Homeomorphic Non-Comparable Particular Point Topologies:
:neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$.
Hence the result.
{{qed}}
\end{proof}
|
22691
|
\section{Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic}
Tags: Homeomorphisms, Discrete Topology
\begin{theorem}
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Let $\tau_d$ denote the metric topology for $d$.
Let $\tau_\delta$ denote the metric topology for $\delta$.
Then $\struct {\Z_{>0}, \tau_d}$ and $\struct {\Z_{>0}, \tau_\delta}$ are homeomorphic.
\end{theorem}
\begin{proof}
From Topology induced by Usual Metric on Positive Integers is Discrete:
:$\struct {\Z_{>0}, \tau_d}$ is a discrete space.
From Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete:
:$\struct {\Z_{>0}, \tau_\delta}$ is a discrete space.
Let $I_{\Z_{>0} }$ be the identity mapping from $\Z_{>0}$ to itself.
From Mapping from Discrete Space is Continuous:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_d} \to \struct {\Z_{>0}, \tau_\delta}$ is continuous
and:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_\delta} \to \struct {\Z_{>0}, \tau_d}$ is continuous.
Hence the result by definition of homeomorphic.
{{qed}}
\end{proof}
|
22692
|
\section{Topologies on Doubleton}
Tags: Doubletons, Examples of Topologies
\begin{theorem}
Let $S = \set {a, b}$ be a doubleton.
Then there exist $4$ possible different topologies on $S$:
{{begin-eqn}}
{{eqn | l = \tau_a
| r = \set {\O, \set {a, b} }
| c = Indiscrete Topology
}}
{{eqn | l = \tau_b
| r = \set {\O, \set a, \set {a, b} }
| c = Sierpiński Topology
}}
{{eqn | l = \tau_c
| r = \set {\O, \set b, \set {a, b} }
| c = Sierpiński Topology
}}
{{eqn | l = \tau_d
| r = \set {\O, \set a, \set b, \set {a, b} }
| c = Discrete Topology
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
The power set of $S$ is the set:
:$\powerset S = \set {\O, \set a, \set b, \set {a, b} }$
Because all topologies on $S$ are subsets of $\powerset S$, one of the following must hold:
{{begin-eqn}}
{{eqn | l = \tau_1
| r = \O
| c =
}}
{{eqn | l = \tau_2
| r = \set \O
| c =
}}
{{eqn | l = \tau_3
| r = \set {\set a}
| c =
}}
{{eqn | l = \tau_4
| r = \set {\set b}
| c =
}}
{{eqn | l = \tau_5
| r = \set {\O, \set a}
| c =
}}
{{eqn | l = \tau_6
| r = \set {\O, \set b}
| c =
}}
{{eqn | l = \tau_7
| r = \set {\set a, \set b}
| c =
}}
{{eqn | l = \tau_8
| r = \set {\O, \set a, \set b}
| c =
}}
{{eqn | l = \tau_9
| r = \set {\set a, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{10}
| r = \set {\set b, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{11}
| r = \set {\set a, \set b, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{12}
| r = \set {\set {a, b} }
| c =
}}
{{eqn | l = \tau_{13}
| r = \set {\O, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{14}
| r = \set {\O, \set a, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{15}
| r = \set {\O, \set b, \set {a, b} }
| c =
}}
{{eqn | l = \tau_{16}
| r = \set {\O, \set a, \set b, \set {a, b} }
| c =
}}
{{end-eqn}}
By definition of a topology, $S$ itself must be an element of the topology.
Thus $\tau_1$ up to $\tau_8$ are not topologies on $S$.
By Empty Set is Element of Topology, for $\tau$ to be a topology for $S$, it is necessary that $\O \in \tau$.
Therefore $\tau_9$ up to $\tau_{12}$ are also not topologies on $S$.
By Indiscrete Topology is Topology, $\tau_{13}$ is a topology on $S$.
By Discrete Topology is Topology, $\tau_{16}$ is a topology on $S$.
It is then seen by inspection that $\tau_{14}$ and $\tau_{15}$ are particular point topologies
Indeed, they are Sierpiński topologies.
By Particular Point Topology is Topology, both $\tau_{14}$ and $\tau_{15}$ are topologies.
Hence the result.
The topologies can be assigned arbitrary labels.
{{qed}}
\end{proof}
|
22693
|
\section{Topologies on Set form Complete Lattice}
Tags: Topology, Complete Lattices
\begin{theorem}
Let $X$ be a non-empty set.
Let $\LL$ be the set of topologies on $X$.
Then $\struct {\LL, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
Let $\KK \subseteq \LL$.
Then by Intersection of Topologies is Topology:
:$\bigcap \KK \in \LL$
By Intersection is Largest Subset, $\bigcap \LL$ is the infimum of $\KK$.
{{explain}}
Let $\tau$ be the topology generated by the sub-basis $\bigcup \KK$.
Then $\tau \in \LL$ and $\tau$ is the supremum of $\KK$.
We have that each subset of $\LL$ has a supremum and an infimum in $\LL$.
Thus it follows that $\struct {\LL, \subseteq}$ is a complete lattice.
{{qed}}
Category:Topology
Category:Complete Lattices
\end{proof}
|
22694
|
\section{Topologies on Set with 3 Elements}
Tags: Examples of Topologies
\begin{theorem}
Let $S = \set {a, b, c}$ be a set with $3$ elements.
Then there exist $29$ possible different topologies on $S$:
{{begin-eqn}}
{{eqn | l = \tau_1
| r = \set {\O, \set {a, b, c} }
| c = Indiscrete Topology
}}
{{eqn | l = \tau_2
| r = \set {\O, \set a, \set b, \set c, \set {a, b}, \set {a, c}, \set {b, c}, \set {a, b, c} }
| c = Discrete Topology
}}
{{eqn | l = \tau_3
| r = \set {\O, \set {a, b}, S}
| c = Included Set Topology, where the included set is $\set {a, b}$
}}
{{eqn | l = \tau_4
| r = \set {\O, \set {a, c}, S}
| c = Included Set Topology, where the included set is $\set {a, c}$
}}
{{eqn | l = \tau_5
| r = \set {\O, \set {b, c}, S}
| c = Included Set Topology, where the included set is $\set {b, c}$
}}
{{eqn | l = \tau_6
| r = \set {\O, \set a, S}
| c = Excluded Set Topology, where the excluded set is $\set {b, c}$
}}
{{eqn | l = \tau_7
| r = \set {\O, \set b, S}
| c = Excluded Set Topology, where the excluded set is $\set {a, c}$
}}
{{eqn | l = \tau_8
| r = \set {\O, \set c, S}
| c = Excluded Set Topology, where the excluded set is $\set {a, b}$
}}
{{eqn | l = \tau_9
| r = \set {\O, \set a, \set {b, c}, S}
| c = Partition Topology, where the partition is $\set {a \mid b, c}$
}}
{{eqn | l = \tau_{10}
| r = \set {\O, \set b, \set {a, c}, S}
| c = Partition Topology, where the partition is $\set {b \mid a, c}$
}}
{{eqn | l = \tau_{11}
| r = \set {\O, \set c, \set {a, b}, S}
| c = Partition Topology, where the partition is $\set {c \mid a, b}$
}}
{{eqn | l = \tau_{12}
| r = \set {\O, \set a, \set {a, b}, S}
| c = Order Topology, where the total ordering is $a \preccurlyeq b \preccurlyeq c$
}}
{{eqn | l = \tau_{13}
| r = \set {\O, \set b, \set {a, b}, S}
| c = Order Topology, where the total ordering is $b \preccurlyeq a \preccurlyeq c$
}}
{{eqn | l = \tau_{14}
| r = \set {\O, \set a, \set {a, c}, S}
| c = Order Topology, where the total ordering is $a \preccurlyeq c \preccurlyeq b$
}}
{{eqn | l = \tau_{15}
| r = \set {\O, \set c, \set {a, c}, S}
| c = Order Topology, where the total ordering is $c \preccurlyeq a \preccurlyeq b$
}}
{{eqn | l = \tau_{16}
| r = \set {\O, \set b, \set {b, c}, S}
| c = Order Topology, where the total ordering is $b \preccurlyeq c \preccurlyeq a$
}}
{{eqn | l = \tau_{17}
| r = \set {\O, \set c, \set {b, c}, S}
| c = Order Topology, where the total ordering is $c \preccurlyeq b \preccurlyeq a$
}}
{{eqn | l = \tau_{18}
| r = \set {\O, \set a, \set {a, b}, \set {a, c}, S}
| c = Particular Point Topology, where the particular point is $a$
}}
{{eqn | l = \tau_{19}
| r = \set {\O, \set b, \set {a, b}, \set {b, c}, S}
| c = Particular Point Topology, where the particular point is $b$
}}
{{eqn | l = \tau_{20}
| r = \set {\O, \set c, \set {a, c}, \set {b, c}, S}
| c = Particular Point Topology, where the particular point is $c$
}}
{{eqn | l = \tau_{21}
| r = \set {\O, \set a, \set b, \set {a, b}, S}
| c = Excluded Point Topology, where the excluded point is $c$
}}
{{eqn | l = \tau_{22}
| r = \set {\O, \set a, \set c, \set {a, c}, S}
| c = Excluded Point Topology, where the excluded point is $b$
}}
{{eqn | l = \tau_{23}
| r = \set {\O, \set b, \set c, \set {b, c}, S}
| c = Excluded Point Topology, where the excluded point is $a$
}}
{{eqn | l = \tau_{24}
| r = \set {\O, \set a, \set b, \set {a, b}, \set {a, c}, S}
| c =
}}
{{eqn | l = \tau_{25}
| r = \set {\O, \set a, \set b, \set {a, b}, \set {b, c}, S}
| c =
}}
{{eqn | l = \tau_{26}
| r = \set {\O, \set a, \set c, \set {a, b}, \set {a, c}, S}
| c =
}}
{{eqn | l = \tau_{27}
| r = \set {\O, \set a, \set c, \set {a, b}, \set {b, c}, S}
| c =
}}
{{eqn | l = \tau_{28}
| r = \set {\O, \set b, \set c, \set {a, b}, \set {b, c}, S}
| c =
}}
{{eqn | l = \tau_{29}
| r = \set {\O, \set b, \set c, \set {a, c}, \set {b, c}, S}
| c =
}}
{{end-eqn}}
The numbering is arbitrary.
\end{theorem}
\begin{proof}
The power set of $S$ is the set:
:$\powerset S = \set {\O, \set a, \set b, \set c, \set {a, b}, \set {a, c}, \set {b, c}, \set {a, b, c} }$
A topology on $S$ is a subset of $\powerset S$.
Thus the set of all topologies on $S$ is a subset of the power set of $\powerset S$.
From Cardinality of Power Set of Finite Set:
:$\card {\powerset {\powerset S} } = 256$
Half of the subsets of $\powerset {\powerset S}$ do not contain $\O$ and so are not topologies on $S$.
Half of the remaining subsets of $\powerset {\powerset S}$ do not contain $S$ itself, and so are also not topologies on $S$.
We are left with $64$ subsets of $\powerset {\powerset S}$ to investigate.
In the following, they will be numbered arbitrarily.
First we have the discrete topology and indiscrete topology on $S$:
{{begin-eqn}}
{{eqn | l = \tau_1
| r = \set {\O, S}
}}
{{eqn | l = \tau_2
| r = \set {\O, \set a, \set b, \set c, \set {a, b}, \set {a, c}, \set {b, c}, S}
}}
{{end-eqn}}
{{qed|lemma}}
From here, we investigate topologies according to the number of its singletons.
Before doing that, note that a topology on $S$ with either $3$ singletons or $3$ doubletons is the discrete topology.
Hence in the below we need only investigate topologies with $0$, $1$ and $2$ singletons and doubletons.
\end{proof}
|
22695
|
\section{Topologies on Set with More than One Element may not be Homeomorphic}
Tags: Topologies on Set with More than One Element may not be Homeomorphic, Homeomorphisms
\begin{theorem}
Let $S$ be a set which contains at least $2$ elements.
Let $\tau_1$ and $\tau_2$ be topologies on $S$.
Then it is not necessarily the case that $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ are homeomorphic.
\end{theorem}
\begin{proof}
Let $\tau_1$ be the indiscrete topology on $S$.
Let $\tau_2$ be the discrete topology on $S$.
Then $\struct {S, \tau_1}$ has $2$ elements: $S$ and $\O$.
Let $a, b \in S$ such that $a \ne b$.
Then $\set a \in \tau_2$ and $\set b \in \tau 2$, as well as $S$ and $\O$.
So there cannot be a bijection between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$.
{{qed}}
\end{proof}
|
22696
|
\section{Topology Defined by Basis}
Tags: Topology, Topological Bases
\begin{theorem}
Let $S$ be a set.
Let $\BB$ be a set of subsets of $S$.
Suppose that
:$(\text B1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$
:$(\text B2): \quad \forall x \in X: \exists A \in \BB: x \in A$
::$\tau = \set {\bigcup \GG: \GG \subseteq \BB}$
Then:
:$T = \struct {S, \tau}$ is a topological space
:$\BB$ is a basis of $T$.
\end{theorem}
\begin{proof}
We have to prove Open Set Axioms:
\end{proof}
|
22697
|
\section{Topology Defined by Closed Sets}
Tags: Topology, Closed Sets
\begin{theorem}
Let $S$ be a set.
Let $\tau$ be a set of subsets of $S$.
Then $\tau$ is a topology on $S$ {{iff}}:
:$(1): \quad$ Any intersection of arbitrarily many closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
:$(2): \quad$ The union of any finite number of closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
:$(3): \quad S$ and $\O$ are both closed sets of $S$ under $\tau$
where a closed set $V$ of $S$ under $\tau$ is defined as a subset of $S$ such that $S \setminus V \in \tau$.
\end{theorem}
\begin{proof}
From the definition, if $V$ is a closed set of $S$, then $S \setminus V$ is an open set of $S$.
Let $\mathbb V$ be any arbitrary set of closed sets of $S$.
Then by De Morgan's Laws: Difference with Intersection, we have:
:$\ds S \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {S \setminus V}$
First, let $\tau$ be a topology on $S$.
We have that:
:Intersection of Closed Sets is Closed in Topological Space
:Finite Union of Closed Sets is Closed in Topological Space
:By Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed in $S$.
Thus the properties as listed above hold.
{{qed|lemma}}
Now, suppose the properties:
:$(1): \quad$ Any intersection of arbitrarily many closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
:$(2): \quad$ The union of any finite number of closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
:$(3): \quad S$ and $\O$ are both closed sets of $S$ under $\tau$.
all hold.
That means $\ds \bigcap \mathbb V$ is closed.
So $\ds S \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {S \setminus V}$ is open.
Thus we have that the union of arbitrarily many open sets of $S$ under $\tau$ is an open set of $S$ under $\tau$.
Similarly we deduce that the intersection of any finite number of open sets of $S$ under $\tau$ is an open set of $S$ under $\tau$.
And of course by Open and Closed Sets in Topological Space, $\O$ and $S$ are both open in $S$.
So $\tau$ is a topology on $S$.
{{qed}}
\end{proof}
|
22698
|
\section{Topology Defined by Neighborhood System}
Tags: Topology
\begin{theorem}
Let $S$ be a set.
Let $\family {\NN_x}_{x \mathop \in S}$ be an indexed family where $\NN_x$ is non-empty set of subsets of $S$.
Assume that
:$(\text N 1): \quad \forall x \in S, U \in \NN_x: x \in U$
:$(\text N 2): \quad \forall x \in S, U \in \NN_x, y \in U:\exists V \in \NN_y: V \subseteq U$
:$(\text N 3): \quad \forall x \in S, U_1, U_2 \in \NN_x: \exists U \in \NN_x: U \subseteq U_1 \cap U_2$
Then $T = \struct {S, \tau}$ is a topological space where:
:$\ds \tau = \set {\bigcup \GG: \GG \subseteq \bigcup_{x \mathop \in S} \NN_x}$
Moreover, $\family {\NN_x}_{x \mathop \in S}$ is a neighborhood system of $T$.
\end{theorem}
\begin{proof}
Define:
:$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$
According to Topology Defined by Basis it should be proved that
:$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$
:$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$
Ad $(\text B 1)$:
Let $A_1, A_2 \in \BB$.
Let $x \in A_1 \cap A_2$.
By definition of intersection:
:$x \in A_1 \land x \in A_2$
By definition of union:
:$\exists x_1 \in S: A_1 \in \NN_{x_1}$
and
:$\exists x_2 \in S: A_2 \in \NN_{x_2}$
By $(\text N 2)$:
:$\exists V_1 \in \NN_x: V_1 \subseteq A_1$
and
:$\exists V_2 \in \NN_x: V_2 \subseteq A_2$
By $(\text N 3)$:
:$\exists U \in \NN_x: U \subseteq V_1 \cap V_2$
By definition of union:
:$U \in \BB$
By Set Intersection Preserves Subsets:
:$V_1 \cap V_2 \subseteq A_1 \cap A_2$
Thus by Subset Relation is Transitive:
:$U \subseteq A_1 \cap A_2$
Ad $(\text B 2)$:
Let $x \in S$.
By definition of empty set:
:$\exists U: U \in \NN_x$
By $(\text N 1)$:
:$x \in U$
By definition of union:
:$U \in \BB$
Thus:
:$\exists A \in \BB: x \in A$
By definition of $\tau$ and $\BB$:
:$\tau = \set {\bigcup \GG: \GG \subseteq \BB}$
Thus by Topology Defined by Basis:
:$T = \struct {S, \tau}$ is a topological space
It remains to prove that
:$\family {\NN_x}_{x \mathop \in S}$ is a neighborhood system
Let $x \in S$.
Let $U \in \tau$.
By definition of $\tau$:
:$\exists \GG \subseteq \BB: U = \bigcup \GG$
Let $x \in U$.
By definition of union:
:$\exists V \in \GG: x \in V$
By definition of subset:
:$V \in \BB$
By definition of union:
:$\exists y \in S: V \in \NN_y$
By $(\text N 2)$:
:$\exists W \in \NN_x: W \subseteq V$
By Set is Subset of Union/Set of Sets:
:$V \subseteq U$
Thus by Subset Relation is Transitive:
:$\exists W \in \NN_x: W \subseteq U$
Thus by definition:
:$\NN_x$ is a local basis.
{{qed}}
\end{proof}
|
22699
|
\section{Topology as Magma of Sets}
Tags: Topology, Magmas of Sets
\begin{theorem}
The concept of a topology is an instance of a magma of sets.
\end{theorem}
\begin{proof}
It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a topology.
Let $X$ be any set, and let $\powerset X$ be its power set.
Define:
:$\phi_1: \powerset X \to \powerset X: \map {\phi_1} S := X$
:$\phi_2: \powerset X^2 \to \powerset X: \map {\phi_2} {S, T} := S \cap T$
For each index set $I$, define:
:$\ds \phi_I: \powerset X^I \to \powerset X: \map {\phi_I} {\family {S_i}_{i \mathop \in I} } := \bigcup_{i \mathop \in I} S_i$
It is blatantly obvious that these partial mappings capture the axioms for a topology.
{{finish}}
{{qed}}
Category:Topology
Category:Magmas of Sets
\end{proof}
|
22700
|
\section{Topology forms Complete Lattice}
Tags: Topology, Complete Lattices
\begin{theorem}
Let $\struct {X, \tau}$ be a topological space.
Then $\struct {\tau, \subseteq}$ is a complete lattice.
\end{theorem}
\begin{proof}
To show that $\struct {\tau, \subseteq}$ is a complete lattice, we must show that every subset of $\tau$ has a supremum and an infimum.
Let $S \subseteq \tau$.
By the definition of a topology:
:$\ds \bigcup S \in \tau$
By Union is Smallest Superset, $\ds \bigcup S$ is the supremum of $S$.
Let $I$ be the interior of $\ds \bigcap S$, where by Intersection of Empty Set $\ds \bigcap \O$ is conventionally taken to be $X$.
Then by the definition of interior and Intersection is Largest Subset:
:$I \in \tau$
and
:$I \subseteq U$
for each $U \in S$.
Let $V \in \tau$ with $V \subseteq U$ for each $U \in S$.
By Intersection is Largest Subset:
:$\ds V \subseteq \bigcap S$.
Then by the definition of interior:
:$V \subseteq I$
Thus $I$ is the infimum of $S$.
So each subset of $\tau$ has a supremum and an infimum.
Thus, by definition, $\struct {\tau, \subseteq}$ is a complete lattice.
{{qed}}
Category:Topology
Category:Complete Lattices
\end{proof}
|
22701
|
\section{Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete}
Tags: Metric Spaces, Discrete Topology
\begin{theorem}
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Then the metric topology for $\delta$ is a discrete topology.
\end{theorem}
\begin{proof}
Let $\tau_\delta$ denote the metric topology for $\delta$.
In Scaled Euclidean Metric is Metric it is demonstrated that $\delta$ is indeed a metric on $\Z_{>0}$.
Let $a \in \Z_{>0}$.
Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, \delta}$:
:$\map {B_\epsilon} a := \set {x \in A: \map \delta {x, a} < \epsilon}$
Let $x \in \R_{>0}$.
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \dfrac 1 {a \paren {a + 1} }$.
But we have:
{{begin-eqn}}
{{eqn | q = \forall x \in \Z_{>0}, x \ne a
| l = \map \delta {x, a}
| r = \frac {\size {x - a} } {x a}
| c =
}}
{{eqn | r = \size {\frac 1 x - \frac 1 a}
| c =
}}
{{eqn | o = \ge
| r = \size {\frac 1 {a + 1} - \frac 1 a}
| c =
}}
{{eqn | r = \epsilon
| c =
}}
{{end-eqn}}
and so:
:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
It follows that:
:$\map {B_\epsilon} a := \set a$
Thus by definition of $\tau_d$:
:$\forall a \in \Z_{>0}: \set a \in \tau_\delta$
It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_\delta}$ is a discrete topological space.
{{qed}}
\end{proof}
|
22702
|
\section{Topology induced by Usual Metric on Positive Integers is Discrete}
Tags: Metric Spaces, Discrete Topology
\begin{theorem}
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Then the metric topology for $d$ is a discrete topology.
\end{theorem}
\begin{proof}
Let $\tau_d$ denote the metric topology for $d$.
Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$.
Let $a \in \Z_{>0}$.
Recall the definition of the open $\epsilon$-ball of $a$ in $\struct {\Z_{>0}, d}$:
:$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$
But we have:
:$\forall x \in \Z_{>0}, x \ne a: \map d {x, a} \ge 1$
and so:
:$\forall x \in \Z_{>0}, x \ne a: x \notin \map {B_\epsilon} a$
It follows that:
:$\map {B_\epsilon} a := \set a$
Thus by definition of $\tau_d$:
:$\forall a \in \Z_{>0}: \set a \in \tau_d$
It follows from Basis for Discrete Topology that $\struct {\Z_{>0}, \tau_d}$ is a discrete topological space.
{{qed}}
\end{proof}
|
22703
|
\section{Topology on Singleton is Indiscrete Topology}
Tags: Singletons, Trivial Topological Spaces, Indiscrete Space, Trivial Topological Space, Indiscrete Spaces, Indiscrete Topology, Topology
\begin{theorem}
Let $S$ be a singleton.
The only possible topology on $S$ is the indiscrete topology.
\end{theorem}
\begin{proof}
Let $S$ be a set containing exactly one element.
Suppose $S = \set x$ for some object $x$.
Then the power set of $S$ is the set:
:$\powerset S = \set {\O, \set x}$
That is:
:$\powerset S = \set {\O, S}$
Let $\tau$ be a topology on $S$.
$\tau$ is a subset of $\powerset S$, $\tau$ must equal one of the following sets:
:$\tau_1 = \O$
:$\tau_2 = \set \O$
:$\tau_3 = \set S$
or
:$\tau_4 = \set {\O, S}$.
By definition of a topology, $S \in \tau$.
Thus $\tau \ne \tau_1$ and $\tau \ne \tau_2$.
By Empty Set is Element of Topology, also $\O \in \tau$.
Therefore $\tau \ne \tau_3$.
By Indiscrete Topology is Topology, $\tau_4$ is a topology on $S$.
So if $S$ is a set containing exactly one element, the only possible topology on $S$ is the indiscrete topology.
{{qed}}
Category:Indiscrete Topology
Category:Singletons
\end{proof}
|
22704
|
\section{Torelli's Sum}
Tags: Rising Factorials
\begin{theorem}
:$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x \paren {x - k z + 1}^{\overline {k - 1} } \paren {y + k z}^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$x^{\overline k}$ denotes $x$ to the $k$ rising.
\end{theorem}
\begin{proof}
From Rising Factorial as Factorial by Binomial Coefficient:
:$\paren {x + y}^{\overline n} = n! \dbinom {x + y + n - 1} n$
Recall Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \paren {n - k} } {n - k}$ by $\dfrac r {r - t k}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
Let the following substitutions be made:
:$r \gets x$
:$t \gets -\paren {1 - z}$
:$s \gets y - 1 + n z$
and so to obtain:
:$\ds \dbinom {x + y + n - 1} n = \sum_k \dbinom {x + \paren {1 - z} k} k \dbinom {y - 1 + n z + \paren {n - k} \paren {1 - z} } {n - k} \dfrac x {x + \paren {1 - z} k}$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom {x + \paren {1 - z} k} k
| r = \dfrac {\paren {x - k z + 1}^{\overline k} } {k!}
| c = Rising Factorial as Factorial by Binomial Coefficient
}}
{{eqn | ll= \leadsto
| l = \dfrac x {x + \paren {1 - z} k} \dbinom {x + \paren {1 - z} k} k
| r = \dfrac {x \paren {x - k z + 1}^{\overline {k - 1} } } {k!}
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \dbinom {y - 1 + n z + \paren {n - k} \paren {1 - z} } {n - k}
| r = \dbinom {y - 1 + n z + n - k - n z + k z} {n - k}
| c =
}}
{{eqn | r = \dbinom {y + k z \paren {n - k} - 1} {n - k}
| c =
}}
{{eqn | r = \frac 1 {\paren {n - 1}!} \paren {y + k z}^{\overline {n - k} }
| c = Rising Factorial as Factorial by Binomial Coefficient
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = n! \dbinom {x + y + n - 1} n
| r = \sum_k \frac {n!} {k! \paren {n - k}!} x \paren {x - k z + 1}^{\overline {k - 1} } \paren {y + k z}^{\overline {n - k} }
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x + y}^{\overline n}
| r = \sum_k \binom n k x \paren {x - k z + 1}^{\overline {k - 1} } \paren {y + k z}^{\overline {n - k} }
| c =
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Ruggiero Torelli|cat = Torelli}}
\end{proof}
|
22705
|
\section{Torus as Surface of Revolution}
Tags: Induced Metric, Surfaces of Revolution, Solid Geometry
\begin{theorem}
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a circle defined by $x^2 + \paren {y - 2}^2 = 1$ in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {\sin t, 2 + \cos t}$
Then the induced metric on $S_C$ is:
:$g = d t^2 + \paren {2 + \cos t}^2 \, d \theta^2$
\end{theorem}
\begin{proof}
We have that:
:$\map {\gamma'} t = \tuple {\cos t, - \sin t}$
Furthermore:
:$\paren {\cos t}^2 + \paren {- \sin t}^2 = 1$
Hence, $\map \gamma t$ is a unit-speed curve.
By the corollary of the induced metric on the surface of revolution:
:$g = d t^2 + \paren {2 + \cos t}^2 \, d \theta^2$
{{qed}}
\end{proof}
|
22706
|
\section{Total Force on Charged Particle from 2 Charged Particles}
Tags: Electric Charge
\begin{theorem}
Let $p_1$, $p_2$ and $p_3$ be charged particles.
Let $q_1$, $q_2$ and $q_3$ be the electric charges on $p_1$, $p_2$ and $p_3$ respectively.
Let $\mathbf F_{ij}$ denote the force exerted on $q_j$ by $q_i$.
Let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of the other two charged particles.
Then the force $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:
{{begin-eqn}}
{{eqn | l = \mathbf F_1
| r = \mathbf F_{21} + \mathbf F_{31}
| c =
}}
{{eqn | r = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}
| c =
}}
{{end-eqn}}
where:
:$\mathbf F_{21} + \mathbf F_{31}$ denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
:$\mathbf r_{ij}$ denotes the displacement from $p_i$ to $p_j$
:$r_{ij}$ denotes the distance between $p_i$ and $p_j$
:$\varepsilon_0$ denotes the vacuum permittivity.
\end{theorem}
\begin{proof}
:500px
By definition, the force $\mathbf F_{21}$ and $\mathbf F_{31}$ are vector quantities.
Hence their resultant can be found by using the Parallelogram Law.
The result follows from Coulomb's Law of Electrostatics.
{{qed}}
\end{proof}
|
22707
|
\section{Total Force on Charged Particle from Multiple Charged Particles}
Tags: Electric Charge
\begin{theorem}
Let $p_1, p_2, \ldots, p_n$ be charged particles.
Let $q_1, q_2, \ldots, q_n$ be the electric charges on $p_1, p_2, \ldots, p_n$ respectively.
For all $i$ in $\set {1, 2, \ldots, n}$ where $i \ne j$, let $\mathbf F_{i j}$ denote the force exerted on $q_j$ by $q_i$.
For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q_i$ by the combined action of all the other charged particles.
Then the force $\mathbf F_i$ exerted on $q_i$ by the combined action of all the other charged particles is given by:
{{begin-eqn}}
{{eqn | l = \mathbf F_i
| r = \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \mathbf F_{j i}
| c =
}}
{{eqn | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}
| c =
}}
{{end-eqn}}
where:
:the summation denotes the vector sum of $\mathbf F_{21}$ and $\mathbf F_{31}$
:$\mathbf r_{ij}$ denotes the displacement from $p_i$ to $p_j$
:$r_{ij}$ denotes the distance between $p_i$ and $p_j$
:$\varepsilon_0$ denotes the vacuum permittivity.
\end{theorem}
\begin{proof}
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
$\map P 2$ is the case:
$\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$
which is Coulomb's Law of Electrostatics.
Thus $\map P 0$ is seen to hold.
\end{proof}
|
22708
|
\section{Total Number of Set Partitions}
Tags: Set Partitions
\begin{theorem}
Let $S$ be a finite set of cardinality $n$.
Then the number of different partitions of $S$ is $B_n$, where $B_n$ is the $n$th Bell number.
\end{theorem}
\begin{proof}
The number of different partitions of $S$ is '''defined''' as $B_n$.
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind:
:$B_n = \ds \sum_{k \mathop = 0}^n {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind.
From Number of Set Partitions by Number of Components, $\ds {n \brace k}$ is the number of partitions of $S$ into $k$ components.
Hence the result.
{{qed}}
\end{proof}
|
22709
|
\section{Total Ordering is Total Relation}
Tags: Total Relations, Total Orderings
\begin{theorem}
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a total ordering.
Then $\RR$ is a total relation.
\end{theorem}
\begin{proof}
By definition of total ordering:
:$\RR$ is a reflexive relation on the strength of being an ordering
:$\RR$ is a connected relation on the strength of being a total ordering.
{{qed}}
Category:Total Orderings
Category:Total Relations
\end{proof}
|
22710
|
\section{Total Ordering on Field of Quotients is Unique}
Tags: Fields of Quotients, Fields, Quotient Fields, Integral Domains
\begin{theorem}
Let $\struct {K, +, \circ}$ be a field of quotients of an ordered integral domain $\struct {D, +, \circ, \le}$.
Then there is one and only one total ordering $\le'$ on $K$ which is compatible with its ring structure and induces on $D$ its given total ordering $\le$.
That ordering is the one defined by:
:$P = \set {\dfrac x y \in K: x \in D_+, y \in D_+^*}$
\end{theorem}
\begin{proof}
First, note that from Divided by Positive Element of Field of Quotients:
:$\forall z \in K: \exists x, y \in R: z = \dfrac x y, y \in R_+^*$
Now we show that $P$ satistfies conditions $(1)$ to $(4)$ of Positive Elements of Ordered Ring.
From Addition of Division Products and Product of Division Products, it is clear that $P$ satisfies $(1)$ and $(3)$.
{{qed|lemma}}
Next we establish $(2)$.
Let $z \in P \cap \paren {-P}$.
Then $z \in P$ and $z \in \paren {-P}$. Thus:
:$\exists x_1, x_2 \in D_+, y_1, y_2 \in D_+^*: z = x_1 / y_1, -z = x_2 / y_2$
So:
:$x_1 / y_1 = -\paren {x_2 / y_2} \implies x_1 \circ y_2 = -\paren {x_2 \circ y_1}$
But $0 \le x_1 \circ y_2$ and $-\paren {x_2 \circ y_1} \le 0$.
So $x_1 \circ y_2 = 0$.
As $0 < y_2$, this means $x = 0$ and therefore $z = 0$.
Thus $(2)$ has been established.
{{qed|lemma}}
Now to show that $(4)$ holds.
Let $z = x / y$ where $x \in D, y \in D_+$.
Suppose $0 \le x$. Then $z \in P$.
However, suppose $x < 0$. Then $0 < \paren {-x}$ so $-z = \paren {-x} / y \in P$.
Thus:
:$z = -\paren {-z} \in -P$
So:
:$P \cup \paren {-P} = K$
So by Positive Elements of Ordered Ring, the relation $\le'$ on $K$ defined by $P$ is a total ordering on $K$ compatible with its ring structure.
{{qed|lemma}}
Now we need to show that the ordering induced on $D$ by $\le'$ is indeed $\le$.
Let $z \in D_+$. Then $z = z / 1_D \in P$, as $1_D \in D_+^*$.
Thus $D_+ \subseteq P$ and $D_+ \subseteq D$ so $D_+ \subseteq D \cap P$ from Intersection is Largest Subset.
Conversely, let $z \in D \cap P$. Then:
:$\exists x \in D_+, y \in D_+^*: z = x / y$
If $x = 0$ then $z = 0$, and if $0 < x$ then as $z \circ y = x$ and $0 < y$, it follows that $0 < z$ by item 1 of Properties of a Totally Ordered Ring.
So $\forall z \in D: 0 \le z \iff z \in P$.
Thus it follows that $z \in D \cap P \implies z \in D_+$, i.e. $D \cap P \subseteq D_+$.
Thus $D_+ = D \cap P$.
By item $(2)$ of Properties of Ordered Ring, we have:
:$x \le y \iff 0 \le y + \left({-x}\right)$
and thus it follows that the ordering induced on $D$ by $\le'$ is $\le$.
{{qed|lemma}}
Now we need to show uniqueness.
Let $\preceq$ be a total ordering on $K$ which is compatible with its ring structure and induces on $D$ the ordering $\le$.
Let $Q = \set {z \in K: 0 \preceq z}$.
We now show that $Q = P$.
Let $x \in D_+, y \in D_+^*$.
Then $0 \preceq x$ and $0 \prec 1 / y$ from item 4 of Properties of a Totally Ordered Ring.
Thus by compatibility with ring structure, $0 \preceq x / y$.
Hence $P \subseteq Q$.
Conversely, let $z \in Q$.
Let $z = x / y$ where $x \in D, y \in D_+^*$.
Then $x = z \circ y$ and by compatibility with ring structure, $0 \preceq x$.
Thus $0 \le x$ and hence $z \in P$, and so $Q \subseteq P$.
So $Q = P$.
Therefore, by item $(2)$ of Properties of Ordered Ring, it follows that $\preceq$ is the same as $\le$.
{{Qed}}
{{improve|Put this in the context of Ordered Integral Domain.}}
\end{proof}
|
22711
|
\section{Total Probability Theorem/Conditional Probabilities}
Tags: Total Probability Theorem
\begin{theorem}
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\set {B_1, B_2, \ldots}$ be a partition of $\Omega$ such that $\forall i: \map \Pr {B_i} > 0$.
Let $C \in \Sigma$ be an event independent to any of the $B_i$.
Then:
:$\ds \forall A \in \Sigma: \condprob A C = \sum_i \condprob A {C \cap B_i} \, \map \Pr {B_i}$
\end{theorem}
\begin{proof}
First define $Q_C := \condprob {\, \cdot} C$.
Then, from Conditional Probability Defines Probability Space, $\struct {\Omega, \Sigma, Q_C}$ is a probability space.
Therefore the Total Probability Theorem also holds true.
Hence we have:
{{begin-eqn}}
{{eqn | l = \map {Q_C} A
| r = \sum_i \map {Q_C} {A \mid B_i} \, \map {Q_C} {B_i}
| c = Total Probability Theorem
}}
{{eqn | r = \sum_i \condprob {\paren {A \mid B_i} } C \, \condprob {B_i} C
| c = Definition of $Q_C$
}}
{{eqn | r = \sum_i \frac {\condprob {\paren {A \cap B_i} } C } {\condprob {B_i} C} \, \condprob {B_i} C
| c = {{Defof|Conditional Probability}} for $\condprob {\paren {A \mid B_i} } C$
}}
{{eqn | r = \sum_i \frac {\map \Pr {A \cap B_i \cap C} } {\map \Pr C} \frac {\map \Pr C} {\map \Pr {B_i \cap C} } \, \condprob {B_i} C
| c = {{Defof|Conditional Probability}} for $\condprob {A \cap B_i} C$
}}
{{eqn | r = \sum_i \frac {\map \Pr {A \cap B_i \cap C} } {\map \Pr {B_i \cap C} } \, \condprob {B_i} C
| c = simplifying thanks to the Multiplicative Inverse for Real Numbers
}}
{{eqn | r = \sum_i \condprob A {B_i \cap C} \, \condprob {B_i} C
| c = {{Defof|Conditional Probability}} for $\condprob A {B_i \cap C}$
}}
{{eqn | r = \sum_i \condprob A {B_i \cap C} \, \map \Pr {B_i}
| c = $C$ and $B_i$ are independent
}}
{{eqn | r = \sum_i \condprob A {C \cap B_i} \, \map \Pr {B_i}
| c = Intersection is Commutative
}}
{{end-eqn}}
{{qed}}
Category:Total Probability Theorem
\end{proof}
|
22712
|
\section{Total Solid Angle Subtended by Spherical Surface}
Tags: Solid Angles, Spheres
\begin{theorem}
The total solid angle subtended by a spherical surface is $4 \pi$.
\end{theorem}
\begin{proof}
:400px
Let $\d S$ be an element of a surface $S$.
Let $\mathbf n$ be a unit normal on $\d S$ positive outwards.
From a point $O$, let a conical pencil touch the boundary of $S$.
Let $\mathbf r_1$ be a unit vector in the direction of the position vector $\mathbf r = r \mathbf r_1$ with respect to $O$.
Let spheres be drawn with centers at $O$ of radii $1$ and $r$.
Let $\d \omega$ be the area cut from the sphere of radius $1$.
We have:
:$\dfrac {\d \omega} {1^2} = \dfrac {\d S \cos \theta} {r^2}$
where $\theta$ is the angle between $\mathbf n$ and $\mathbf r$.
Thus $\d \omega$ is the solid angle subtended by $\d S$ at $O$.
Hence by definition of solid angle subtended:
:$\ds \Omega = \iint_S \frac {\mathbf {\hat r} \cdot \mathbf {\hat n} \rd S} {r^2}$
{{WIP|I've lost the thread of the source work. I will restart vector analysis from a different viewpoint, as the coverage of vector integration by Hague is not easy to interpret.}}
\end{proof}
|
22713
|
\section{Total Variation is Non-Negative}
Tags: Total Variation
\begin{theorem}
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Let $V_f$ be the total variation of $f$ on $\closedint a b$.
Then:
:$V_f \ge 0$
with equality {{iff}} $f$ is constant.
\end{theorem}
\begin{proof}
We use the notation from the definition of bounded variation.
Note that by the definition of absolute value, we have:
:$\size x \ge 0$
for all $x \in \R$.
Let $P$ be a finite subdivision of $\closedint a b$.
Then:
:$\ds \map {V_f} P = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$
So, by the definition of supremum, we have:
:$\ds V_f = \sup_P \paren {\map {V_f} P} \ge 0$
Let $f$ be constant.
Then:
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 0$
for all $x_{i - 1}, x_i \in \closedint a b$.
So, for any finite subdivision $P$ we have:
{{begin-eqn}}
{{eqn | l = \map {V_f} P
| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
}}
{{eqn | r = \sum_{i \mathop = 1}^{\size P - 1} 0
}}
{{eqn | r = 0
}}
{{end-eqn}}
We then have:
{{begin-eqn}}
{{eqn | l = V_f
| r = \sup_P \paren {\map {V_f} P}
| c = {{Defof|Total Variation}}
}}
{{eqn | r = \sup \set {0}
}}
{{eqn | r = 0
| c = {{Defof|Supremum of Subset of Real Numbers}}
}}
{{end-eqn}}
So if $f$ is constant, then $f$ is of bounded variation with:
:$V_f = 0$
It remains to show that if $V_f = 0$, then $f$ is constant.
It suffices to show that if $f$ is not constant then $V_f > 0$.
Since $f$ is non-constant, we can pick $x \in \openint a b$ such that either:
:$\map f x \ne \map f a$
or:
:$\map f x \ne \map f b$
So that either:
:$\size {\map f x - \map f a} > 0$
or:
:$\size {\map f b - \map f x} > 0$
Note that we then have:
{{begin-eqn}}
{{eqn | l = \map {V_f} {\set {a, x, b} }
| r = \size {\map f x - \map f a} + \size {\map f b - \map f x}
}}
{{eqn | o = >
| r = 0
}}
{{end-eqn}}
We must then have:
:$V_f \ge \map {V_f} {\set {a, x, b} } > 0$
Hence the claim.
{{qed}}
\end{proof}
|
22714
|
\section{Total Vector Area of Polyhedron is Zero}
Tags: Vector Area, Polyhedra
\begin{theorem}
Let $P$ be a polyhedron.
Let the positive direction be defined as outward.
Let $\mathbf T$ be the total vector area of all the faces of $P$.
Then:
:$\mathbf T = \mathbf 0$
\end{theorem}
\begin{proof}
$P$ can be geometrically divided into a finite number of tetrahedra.
Every face of these tetrahedra which are internal to $P$ appears twice: once with a positive vector area, and once with a negative normal.
Hence for any polyhedron, the total vector area is the sum of the vector areas of all the tetrahedra.
The result follows from Total Vector Area of Tetrahedron is Zero.
{{qed}}
\end{proof}
|
22715
|
\section{Total Vector Area of Tetrahedron is Zero}
Tags: Tetrahedra, Vector Area
\begin{theorem}
Let $T$ be a tetrahedron whose faces have vector area $\mathbf S_1$, $\mathbf S_2$, $\mathbf S_3$ and $\mathbf S_4$.
Let the positive direction be defined as outward.
:300px
Then the total vector area is zero:
:$\mathbf S_1 + \mathbf S_2 + \mathbf S_3 + \mathbf S_4 = \mathbf 0$
\end{theorem}
\begin{proof}
Let the vector areas be resolved upon the faces of $T$.
Some of the projections will be positive and some negative.
Let $T$ be imagined in a fluid which is in equilibrium under pressure.
Each face experiences a force which is normal to its plane and proportional to its area.
Because the fluid inside is in equilibrium with the fluid outside, the resultant of the forces on its faces will equal zero.
Hence the sum of all the vector areas is also zero, as the pressure is the same on all faces.
Hence the result.
{{qed}}
\end{proof}
|
22716
|
\section{Totally Bounded Metric Space is Bounded}
Tags: Totally Bounded Metric Spaces, Bounded Metric Spaces, Metric Spaces, Totally Bounded Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a totally bounded metric space.
Then $M$ is bounded.
\end{theorem}
\begin{proof}
Let $M = \struct {A, d}$ be totally bounded.
Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
:$\ds \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
for all $x \in A$.
Let us set:
:$a := x_0$
:$\ds D := \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}$
:$K := D + 1$
Now let $x \in A$ be arbitrary.
Then by assumption there exists $i$ such that $\map d {x_i, x} \le 1$.
Hence:
:$\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$
So $M$ is bounded, as claimed.
{{qed}}
\end{proof}
|
22717
|
\section{Totally Bounded Metric Space is Separable}
Tags: Totally Bounded Metric Spaces, Separable Spaces, Totally Bounded Spaces, Second-Countable Spaces, Countability Axioms, Metric Spaces
\begin{theorem}
A totally bounded metric space is separable.
\end{theorem}
\begin{proof}
Let $M = \struct {A, d}$ be a totally bounded metric space.
By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\sequence {F_n}_{n \mathop \ge 1}$ such that:
:For all natural numbers $n \ge 1$, $F_n$ is a finite $\paren {1/n}$-net for $M$.
Let $\ds S = \bigcup_{n \mathop \ge 1} F_n$.
From Countable Union of Countable Sets is Countable, it follows that $S$ is countable.
It suffices to prove that $S$ is everywhere dense in $M$.
Let $S^-$ denote the closure of $S$.
Let $x \in X$.
Let $U \subseteq X$ be open in $M$ such that $x \in U$.
By definition, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.
That is, the open $\epsilon$-ball of $x$ in $M$ is contained in $U$.
By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$.
That is, $\dfrac 1 n < \epsilon$, and so $\map {B_{1 / n} } x \subseteq \map {B_\epsilon} x$.
Since $\subseteq$ is a transitive relation, we have $\map {B_{1/n} } x \subseteq U$.
By the definition of a net, there exists a $y \in F_n$ such that $x \in \map {B_{1/n} } y$.
By {{Metric-space-axiom|3}}, it follows from the definition of an open ball that $y \in \map {B_{1/n} } x$.
Since $y \in S \cap U$, it follows that $x$ is an adherent point of $S$.
By definition of adherent point, we have $x \in S^-$.
That is, $X \subseteq S^-$, and so $S$ is everywhere dense in $M$.
{{qed}}
{{ACC}}
Category:Separable Spaces
Category:Metric Spaces
Category:Totally Bounded Metric Spaces
\end{proof}
|
22718
|
\section{Totally Disconnected Space is Punctiform}
Tags: Connectedness, Totally Disconnected Spaces, Punctiform Spaces, Continua
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is punctiform.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be totally disconnected.
Then by definition its components are singletons.
Thus by definition each of its connected sets are degenerate.
{{qed}}
\end{proof}
|
22719
|
\section{Totally Disconnected Space is T1}
Tags: T1 Spaces, Totally Disconnected Spaces, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be totally disconnected.
Then as its components are singletons, it follows that each of its points is closed.
From Equivalence of Definitions of $T_1$ Space, it follows that $T$ is a $T_1$ (Fréchet) space.
{{qed}}
\end{proof}
|
22720
|
\section{Totally Disconnected Space is Totally Pathwise Disconnected}
Tags: Totally Disconnected Spaces, Totally Pathwise Disconnected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a totally pathwise disconnected space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then by definition $T$ contains no non-degenerate connected sets.
{{AimForCont}} $T$ is not a totally pathwise disconnected space.
That is, there exist two points $x, y \in S$ such that there exists a path between $x$ and $y$.
That is, $x$ and $y$ are path-connected.
Thus they are in the same path component.
From Path-Connected Space is Connected, $x$ and $y$ are therefore in the same component.
But that contradicts the definition of $T$ having no non-degenerate connected sets.
Hence, by Proof by Contradiction, $T$ is a totally pathwise disconnected space.
{{qed}}
\end{proof}
|
22721
|
\section{Totally Disconnected and Locally Connected Space is Discrete}
Tags: Totally Disconnected Spaces, Locally Connected Spaces, Discrete Topology, Local Connectedness, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is both totally disconnected and locally connected.
Then $T$ is the discrete space on $S$.
\end{theorem}
\begin{proof}
So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected.
As $T$ is totally disconnected, every point is a component and therefore closed.
As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$.
In order for $T$ to be covered by $\BB$, every singleton subset of $T$ must be in $\BB$.
From the definition of topology, the union of any number of these singleton sets is an open set of $T$.
That is, every subset of $S$ is open in $T$.
That is, every element of the power set of $S$ is open in $T$.
This is precisely the definition of the discrete space on $S$.
{{qed}}
\end{proof}
|
22722
|
\section{Totally Disconnected but Connected Set must be Singleton}
Tags: Connectedness, Connected Spaces, Singletons, Totally Disconnected Spaces, Connected Sets
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both totally disconnected and connected.
Then $H$ is a singleton.
\end{theorem}
\begin{proof}
If $H$ is totally disconnected then its individual points are separated.
If $H$ is connected it can not be represented as the union of two (or more) separated sets.
So $H$ can have only one point in it.
{{qed}}
\end{proof}
|
22723
|
\section{Totally Ordered Abelian Group Isomorphism}
Tags: Morphisms, Abelian Groups, Isomorphisms, Group Theory
\begin{theorem}
Let $\struct {\Z', +', \le'}$ be a totally ordered abelian group.
Let $0'$ be the identity of $\struct {\Z', +', \le'}$.
Let $\N' = \set {x \in \Z': x \ge' 0'}$.
Let $\Z'$ contain at least two elements.
Let $\N'$ be well-ordered for the ordering induced on $\N'$ by $\le'$.
Then the mapping $g: \Z \to \Z'$ defined by:
:$\forall n \in \Z: \map g n = \paren {+'}^n 1'$
is an isomorphism from $\struct {\Z, +, \le}$ onto $\struct {\Z', +', \le'}$, where $1'$ is the smallest element of $\N' \setminus \set {0'}$.
\end{theorem}
\begin{proof}
First we establish that $g$ is a homomorphism.
Suppose $z \in \Z'$ such that $z \ne 0'$.
Then by Ordering of Inverses in Ordered Monoid, either $z >' 0'$ or $-z >' 0'$.
Thus either:
:$z \in \N' \setminus \set {0'}$
or:
:$-z \in \N' \setminus \set {0'}$
and thus $\N' \setminus \set {0'}$ is not empty.
Therefore $\N' \setminus \set {0'}$ has a minimal element.
Call this minimal element $1'$.
It is clear that $\N'$ is an ordered semigroup satisfying:
:{{NOSAxiom|1}}
:{{NOSAxiom|2}}
:{{NOSAxiom|4}}.
Also:
{{begin-eqn}}
{{eqn | o =
| r = 0' \le' x \le' y
| c =
}}
{{eqn | o = \leadsto
| r = 0' \le' y - x
| c =
}}
{{eqn | o = \leadsto
| r = y - x \in \N' \land x +' \paren {y - x} = y
| c =
}}
{{end-eqn}}
Thus $\N'$ also satisfies {{NOSAxiom|3}}.
So $\struct {\N', +', \le'}$ is a naturally ordered semigroup.
So, by Naturally Ordered Semigroup is Unique, the restriction to $\N$ of $g$ is an isomorphism from $\struct {\N, +, \le}$ to $\struct {\N', +', \le'}$.
By Index Law for Sum of Indices, $g$ is a homomorphism from $\struct {\Z, +}$ into $\struct {\Z', +'}$.
Next it is established that $g$ is surjective.
Let $y \in \Z': y <' 0'$.
{{begin-eqn}}
{{eqn | o =
| r = y <' 0'
| c =
}}
{{eqn | o = \leadsto
| r = -y >' 0'
| c =
}}
{{eqn | o = \leadsto
| r = \exists n \in \N: -y = \map g n
| c =
}}
{{eqn | o = \leadsto
| r = y = -\map g n = \map g {-n}
| c = Homomorphism with Identity Preserves Inverses
}}
{{end-eqn}}
Therefore $g$ is a surjection.
Now to show that $g$ is a monomorphism, that is, it is injective.
Let $n < m$.
{{begin-eqn}}
{{eqn | o =
| r = n < m
| c =
}}
{{eqn | o = \leadsto
| r = m - n \in \N_{>0}
| c =
}}
{{eqn | o = \leadsto
| r = \map g m - \map g n - \map g {m - n} \in \N' \setminus \set {0'}
| c =
}}
{{eqn | o = \leadsto
| r = \map g n <' \paren {\map g m - \map g n} +' \map g n = \map g m
| c = Strict Ordering Preserved under Product with Cancellable Element
}}
{{end-eqn}}
Therefore it can be seen that $g$ is strictly increasing.
It follows from Monomorphism from Total Ordering that $g$ is a monomorphism from $\struct {\Z, +, \le}$ to $\struct {\Z', +', \le'}$.
A surjective monomorphism is an isomorphism, and the result follows.
{{Qed}}
\end{proof}
|
22724
|
\section{Totally Ordered Ring Zero Precedes Element or its Inverse}
Tags: Ordered Rings
\begin{theorem}
Let $\struct {R, +, \circ, \preceq}$ be an ordered ring.
From the definition of ordered ring, $\preceq$ is compatible with $+$.
Let $0_R$ be the zero element of $R$.
Let $x \ne 0_R$ be a non-zero element of $R$.
Let $-x$ be the ring negative of $x$.
Then:
:$0_R \prec x \lor 0_R \prec -x$
but not both.
\end{theorem}
\begin{proof}
By the definition of total ordering, $\preceq$ is connected.
As $x \ne 0_R$, one of the following is true, but not both:
:$(1): \quad 0_R \prec x$
:$(2): \quad x \prec 0_R$
If $(2)$, because $\prec$ is compatible with $+$:
{{begin-eqn}}
{{eqn | l = x
| o = \prec
| r = 0_R
}}
{{eqn | ll= \leadsto
| l = x + \paren {-x}
| o = \prec
| r = 0_R + \paren {-x}
}}
{{eqn | ll= \leadsto
| l = 0_R
| o = \prec
| r = -x
| c = {{Defof|Ring Zero}}, {{Defof|Ring Negative}}
}}
{{end-eqn}}
{{qed}}
Category:Ordered Rings
\end{proof}
|
22725
|
\section{Totally Ordered Set is Lattice}
Tags: Lattice Theory, Orderings, Total Orderings
\begin{theorem}
Every totally ordered set is a lattice.
\end{theorem}
\begin{proof}
Let $\struct {S, \preceq}$ be a totally ordered set.
Then we have:
:$\forall x, y \in S: x \preceq y \lor y \preceq x$
:$\forall x, y \in S: x \preceq y \implies \sup \set {x, y} = y \land \inf \set {x, y} = x$
:$\forall x, y \in S: y \preceq x \implies \sup \set {x, y} = x \land \inf \set {x, y} = y$
Thus the conditions for $\struct {S, \preceq}$ to be a lattice are fulfilled.
{{qed}}
\end{proof}
|
22726
|
\section{Totally Separated Space is Completely Hausdorff and Urysohn}
Tags: Completely Hausdorff Spaces, Totally Separated Spaces, Urysohn Spaces, Discrete Topology, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is completely Hausdorff and Urysohn.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a totally separated space.
Then for every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$.
Consider a mapping $f: S \to \closedint 0 1$ such that:
:$\Img {f {\restriction_U} } = \set 0$
:$\Img {f {\restriction_V} } = \set 1$
where:
:$f {\restriction_U}$ and $f {\restriction_V}$ denote the restrictions of $f$ to $U$ and $V$ respectively
:$\Img g$ denotes the image set of a mapping $g$.
From Constant Mapping is Continuous, both $f {\restriction_U}$ and $f {\restriction_V}$ are continuous on $U$ and $V$ respectively.
From Continuous Mapping of Separation, it follows that $f$ is continuous on $S$ itself.
By definition, $f$ is then an Urysohn function on $x$ and $y$.
Hence $T$ is Urysohn.
Having proved that $T$ is an Urysohn space, it follows from Urysohn Space is Completely Hausdorff Space that $T$ is also completely Hausdorff.
{{qed}}
\end{proof}
|
22727
|
\section{Totally Separated Space is Totally Disconnected}
Tags: Totally Separated Spaces, Totally Disconnected Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is totally separated.
Then $T$ is totally disconnected.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a totally separated space.
Then for every $x, y \in S: x \ne y$ there exists a partition $U \mid V$ of $T$ such that $x \in U, y \in V$.
Suppose $T$ were not totally disconnected.
Then $\exists H \subseteq S: x, y \in U$ such that $H$ is connected.
But by definition of connected, there can be no partition $U \mid V$ of $T$ such that $x \in U, y \in V$.
Hence the result.
{{qed}}
\end{proof}
|
22728
|
\section{Tower Law for Subgroups}
Tags: Subgroups, Tower Law for Subgroups, Index of Subgroups, Group Theory, Named Theorems, Index of Subgroup of Subgroup
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$.
\end{theorem}
\begin{proof}
Each coset $g H$ of $G$ modulo $H$ contains all the cosets $g \left({h K}\right) = \left({g h}\right) K$ of $G$ modulo $K$, where $h K$ runs through all the cosets of $H$ modulo $K$.
Alternatively (assuming $G$ is finite):
{{begin-eqn}}
{{eqn | l=\left[{G : H}\right]
| r=\frac {\left\vert{G}\right\vert} {\left\vert{H}\right\vert}
| c=Lagrange's Theorem
}}
{{eqn | l=\left[{G : K}\right]
| r=\frac {\left\vert{G}\right\vert} {\left\vert{K}\right\vert}
| c=Lagrange's Theorem
}}
{{eqn | ll=\implies
| l=\left[{G : K}\right]
| r=\frac {\left\vert{H}\right\vert} {\left\vert{K}\right\vert} \times \left[{G : H}\right]
| c=
}}
{{end-eqn}}
Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\left\vert{H}\right\vert} {\left\vert{K}\right\vert} = \left[{H : K}\right]$. Hence the result.
{{qed}}
\end{proof}
|
22729
|
\section{Trace Sigma-Algebra is Sigma-Algebra}
Tags: Trace Sigma-Algebras, Sigma-Algebras
\begin{theorem}
Let $X$ be a set.
Let $\Sigma$ be a $\sigma$-algebra on $X$.
Let $E \subseteq X$ be a subset of $X$.
Then the trace $\sigma$-algebra $\Sigma_E$ is a $\sigma$-algebra on $E$.
\end{theorem}
\begin{proof}
Verifying the axioms for a $\sigma$-algebra in turn:
\end{proof}
|
22730
|
\section{Trace Sigma-Algebra of Generated Sigma-Algebra}
Tags: Trace Sigma-Algebras, Sigma-Algebras
\begin{theorem}
Let $X$ be a Set, and let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $A \subseteq X$ be a subset of $X$.
Then the following equality holds:
:$A \cap \map \sigma \GG = \map \sigma {A \cap \GG}$
where
:$\map \sigma \GG$ denotes the smallest $\sigma$-algebra on $X$ that contains $\GG$
:$\map \sigma {A \cap \GG}$ denotes the smallest $\sigma$-algebra on $A$ that contains ${A \cap \GG}$
:$A \cap \map \sigma \GG$ denotes the trace $\sigma$-algebra on $A$
:$A \cap \GG$ is a shorthand for $\set {A \cap G: G \in \GG}$
\end{theorem}
\begin{proof}
By definition of generated $\sigma$-algebra:
:$\GG \subseteq \map \sigma \GG$
whence from Set Intersection Preserves Subsets:
:$A \cap \GG \subseteq A \cap \map \sigma \GG$
and therefore, by definition of generated $\sigma$-algebra:
:$\map \sigma {A \cap \GG} \subseteq A \cap \map \sigma \GG$
For the reverse inclusion, define $\Sigma$ by:
:$\Sigma := \set {E \subseteq X: A \cap E \in \map \sigma {A \cap \GG} }$
We will show that $\Sigma$ is a $\sigma$-algebra on $X$.
Since $A \in \map \sigma {A \cap \GG}$:
:$A \cap X = A \in \map \sigma {A \cap \GG}$
and therefore $X \in \Sigma$.
Suppose that $E \in \Sigma$.
Then by Set Intersection Distributes over Set Difference and Intersection with Subset is Subset:
:$\paren {X \setminus E} \cap A = \paren {X \cap A} \setminus \paren {E \cap A} = A \setminus \paren {E \cap A}$
Since $E \cap A \in \map \sigma {A \cap \GG}$ and this is a $\sigma$-algebra on $A$:
:$A \setminus \paren {E \cap A} \in \map \sigma {A \cap \GG}$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.
Then by Intersection Distributes over Union:
:$\ds \paren {\bigcup_{n \mathop \in \N} E_n} \cap A = \bigcup_{n \mathop \in \N} \paren {E_n \cap A}$
The latter expression is a countable union of elements of $\map \sigma {A \cap \GG}$, hence again in $\map \sigma {A \cap \GG}$.
Therefore, $\Sigma$ is a $\sigma$-algebra.
It is also apparent that $\GG \subseteq \Sigma$ since:
:$A \cap \GG \subseteq \map \sigma {A \cap \GG}$
by definition of generated $\sigma$-algebra.
Thus, as $\Sigma$ is a $\sigma$-algebra:
:$\map \sigma \GG \subseteq \Sigma$
and therefore:
:$A \cap \map \sigma \GG \subseteq \map \sigma {A \cap \GG}$
Hence the result, by definition of set equality.
{{qed}}
Category:Sigma-Algebras
Category:Trace Sigma-Algebras
\end{proof}
|
22731
|
\section{Trace in Terms of Dual Basis}
Tags: Linear Algebra
\begin{theorem}
Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of dimension $n$.
Let $\tuple {e_1, \ldots, e_n}$ be a basis of $M$.
Let $\tuple {e_1^*,\ldots, e_n^*}$ be its dual basis
Let $f: M \to M$ be a linear operator.
Then its trace equals:
:$\map \tr f = \ds \sum_{i \mathop = 1}^n e_i^* \paren {\map f {e_i} }$
\end{theorem}
\begin{proof}
Let $\ds \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$
Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.
Then by the above assumption:
:$A_{ij} = c_{ij}$
Then:
{{begin-eqn}}
{{eqn | l = \map \tr f
| r = \map \tr A
| c = {{Defof|Trace of Linear Operator}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n A_{ii}
| c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n c_{ii}
| c = from above
}}
{{end-eqn}}
Now it remains to show that $c_{ii} = e_i^* \paren {\map f {e_i} }$:
{{begin-eqn}}
{{eqn | l = e_i^* \paren {\map f {e_i} }
| r = e_i^* \paren {\sum_{j \mathop = 1}^n c_{ij} e_j}
| c = from above assumption
}}
{{eqn | r = \sum_{j \mathop = 1}^n c_{ij} e_i^* \paren {e_j}
| c = $e_i$ is a linear form
}}
{{eqn | r = \map {c_{ii} } 1
| c = {{Defof|Ordered Dual Basis}}: other terms vanish
}}
{{eqn | r = c_{ii}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22732
|
\section{Trace in Terms of Orthonormal Basis}
Tags: Linear Algebra
\begin{theorem}
Let $\mathbb K \subset \C$ be a field.
Let $\struct {V, \innerprod {\, \cdot \,} {\, \cdot \,} }$ be an inner product space over $\mathbb K$ of dimension $n$.
Let $\tuple {e_1, \ldots, e_n}$ be an orthonormal basis of $V$.
Let $f: V \to V$ be a linear operator.
Then its trace equals:
:$\map \tr f = \ds \sum_{i \mathop = 1}^n \innerprod {\map f {e_i} } {e_i}$
\end{theorem}
\begin{proof}
Let $\ds \map f {e_i} = \sum_{j \mathop = 1}^n c_{ij} e_j$
Let $A$ be the matrix relative to the basis $\tuple {e_1, \ldots, e_n}$.
Then by the above assumption, $A_{ij} = c_{ij}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \tr f
| r = \map \tr A
| c = {{Defof|Trace of Linear Operator}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n A_{ii}
| c = {{Defof|Trace of Matrix}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n c_{ii}
| c = From above
}}
{{end-eqn}}
Now it remains to show that $c_{ii} = \innerprod {\map f {e_i} } {e_i}$:
{{begin-eqn}}
{{eqn | l = \innerprod {\map f {e_i} } {e_i}
| r = \innerprod {\sum_{j \mathop = 1}^n c_{ij} e_j} {e_i}
| c = From above assumption
}}
{{eqn | r = \sum_{j \mathop = 1}^n c_{ij} \innerprod {e_j} {e_i}
| c = Axiom $(2)$ of Inner Product: Bilinearity
}}
{{eqn | r = c_{ii} (1)
| c = {{Defof|Orthonormal Subset}}: other terms vanish
}}
{{eqn | r = c_{ii}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22733
|
\section{Trace of Alternating Product of Matrices and Almost Zero Matrices}
Tags: Matrices, Matrix Theory
\begin{theorem}
Let $R$ be a ring with unity.
Let $n, m$ be positive integers.
Let $E_{ij}$ denote the $n \times n$ matrix with only zeroes except a $1$ at the $\tuple {i, j}$th element.
Let $A_1, \ldots, A_m \in R^{n \times n}$.
Let $i_k, j_k \in \set {1, \ldots, n}$ for $k \in \set {1, \ldots, m}$.
Let $i_0 = i_m$ and $j_0 = j_m$.
Then:
:$\map \tr {A_1 E_{i_1, j_1} A_2 E_{i_2, j_2} \cdots A_m E_{i_m, j_m} } = \ds \prod_{k \mathop = 1}^m \sqbrk {A_k}_{j_{k - 1} i_k}$
\end{theorem}
\begin{proof}
Use induction and the facts $E_{i j} A E_{k l} = A_{j k} E_{i l}$ and $\map \tr {A E_{i j} } = A_{j i}$ (induction basis).
{{ProofWanted}}
Category:Matrix Theory
\end{proof}
|
22734
|
\section{Trace of Matrix Product}
Tags: Matrix Algebra
\begin{theorem}
Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\ds \map \tr {\mathbf A \mathbf B} = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
where $\map \tr {\mathbf A \mathbf B}$ denotes the trace of $\mathbf A \mathbf B$.
Using the Einstein summation convention, this can be expressed as:
:$\map \tr {\mathbf A \mathbf B} = a_{i j} b_{j i}$
\end{theorem}
\begin{proof}
Let $\mathbf C := \mathbf A \mathbf B$.
By definition of matrix product:
:$\ds c_{i k} = \sum_{j \mathop = 1}^n a_{i j} b_{j k}$
Thus for the diagonal elements:
:$\ds c_{i i} = \sum_{j \mathop = 1}^n a_{i j} b_{j i}$
By definition of trace:
:$\ds \map \tr {\mathbf C} = \sum_{i \mathop = 1}^n c_{i i}$
Hence the result.
{{qed}}
\end{proof}
|
22735
|
\section{Trace of Matrix Product/General Result}
Tags: Matrix Algebra
\begin{theorem}
Let $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$ be square matrices of order $n$.
Let $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$ be the (conventional) matrix product of $\mathbf A_1, \mathbf A_2, \ldots, \mathbf A_m$.
Then:
:$(1): \quad \ds \map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
where:
:$\map {a_1} {i_1, i_2}$ (for example) denotes the element of $\mathbf A_1$ whose indices are $i_1$ and $i_2$
:$\map \tr {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m}$ denotes the trace of $\mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$.
In $(1)$, the Einstein summation convention is used, with the implicit understanding that a summation is performed over each of the indices $i_1$ to $i_m$.
\end{theorem}
\begin{proof}
Let $\mathbf C = \mathbf A_1 \mathbf A_2 \cdots \mathbf A_m$
From Product of Finite Sequence of Matrices, the general element of $\mathbf C$ is given in the Einstein summation convention by:
:$\map c {i_1, j} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, j}$
Thus for the diagonal elements:
:$\ds \map c {i_1, i_1} = \map {a_1} {i_1, i_2} \map {a_2} {i_2, i_3} \cdots \map {a_{m - 1} } {i_{m - 1}, i_m} \map {a_m} {i_m, i_1}$
which is the Einstein summation convention for the trace of $\mathbf C$.
{{qed}}
\end{proof}
|
22736
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\section{Trace of Unit Matrix}
Tags: Matrix Algebra
\begin{theorem}
Let $\mathbf I_n$ be the unit matrix of order $n$.
Then:
:$\map \tr {\mathbf I_n} = n$
where $\map \tr {\mathbf I_n}$ denotes the trace of $\mathbf I_n$.
\end{theorem}
\begin{proof}
By definition:
:$\mathbf I_n := \sqbrk a_n: a_{i j} = \delta_{i j}$
That is: each of the elements on the main diagonal is equal to $1$.
There are $n$ such elements.
Hence the result.
{{qed}}
\end{proof}
|
22737
|
\section{Transcendence of Sum or Product of Transcendentals}
Tags: Transcendental Numbers
\begin{theorem}
Let $a$ and $b$ be two transcendental numbers.
Then at least one of $a + b$ and $a \times b$ is transcendental.
\end{theorem}
\begin{proof}
Proof by Contradiction:
{{AimForCont}} $a + b$ and $a \times b$ are both not transcendental.
Hence by definition, they are both algebraic.
Hence, $\left({z - a}\right) \left({z - b}\right)$ is a polynomial with algebraic coefficients.
Therefore, $a$ and $b$ must both be algebraic.
However, this contradicts with the assumption that $a$ and $b$ are both transcendental.
Hence by Proof by Contradiction it must follow that at least one of $a + b$ and $a \times b$ is transcendental.
{{qed}}
Category:Transcendental Numbers
\end{proof}
|
22738
|
\section{Transcendental Numbers are Uncountable}
Tags: Transcendental Numbers
\begin{theorem}
The set of transcendental real numbers is uncountable.
\end{theorem}
\begin{proof}
By definition, a transcendental number (in this context) is a real number which is not an algebraic number.
Recall that the Real Numbers are Uncountable.
Also recall that the Algebraic Numbers are Countable.
The result follows from Uncountable Set less Countable Set is Uncountable.
{{qed}}
\end{proof}
|
22739
|
\section{Transcendental Slope}
Tags: Transcendental Number Theory, Transcendental Number Theory
\begin{theorem}
The slope of a line may be transcendental.
\end{theorem}
\begin{proof}
The slope form of any number $x$ may be produced by:
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = \frac {x} {1}
|c = (Slope Form of $x$)
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn|l = {\mathrm m}
|r = {x}
|c =
}}
{{end-eqn}}
If $x$ is transcendental, then the slope of a line $\mathrm m$ is transcendental.
\end{proof}
|
22740
|
\section{Transfinite Induction/Principle 1}
Tags: Ordinals
\begin{theorem}
Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For all elements $x$ of $\On$, if $x$ is a subset of $A$, then $x$ is an element of $A$.
Then $\On \subseteq A$.
\end{theorem}
\begin{proof}
{{NotZFC}}
{{AimForCont}} that $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a strictly minimal element $y$ under $\in$.
By Element of Ordinal is Ordinal, $y$ must be a subset of $\On$, the class of all ordinals.
However, from the fact that $y$ is a strictly minimal element under $\in$ of $\On \setminus A$:
:$\paren {\On \setminus A} \cap y = \O$
So by its subsethood of $\On$:
:$\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$
Therefore $y \subseteq A$.
However, by the hypothesis, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\On \setminus A$.
Therefore $\On \subseteq A$.
{{qed}}
{{LEM|Reductio ad Absurdum|4}}
\end{proof}
|
22741
|
\section{Transfinite Induction/Principle 1/Proof 2}
Tags: Ordinals
\begin{theorem}
Let $\On$ denote the class of all ordinals.
Let $A$ denote a class.
Suppose that:
:For each element $x$ of $\On$, if $\forall y \in \On: \paren {y < x \implies y \in A}$ then $x$ is an element of $A$.
Then $\On \subseteq A$.
\end{theorem}
\begin{proof}
{{AimForCont}} $\neg \On \subseteq A$.
Then:
:$\paren {\On \setminus A} \ne \O$
From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals.
By Ordinal Class is Strongly Well-Ordered by Subset, $\On \setminus A$ must have a smallest element $y$.
Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$.
This contradicts the fact that $y$ is an element of $\On \setminus A$.
Therefore $\On \subseteq A$.
{{qed}}
{{LEM|Reductio ad Absurdum|4}}
Category:Ordinals
\end{proof}
|
22742
|
\section{Transfinite Induction/Principle 2}
Tags: Ordinals
\begin{theorem}
Let $A$ be a class satisfying the following conditions:
* $\O \in A$
* $\forall x \in A: x^+ \in A$
* If $y$ is a limit ordinal, then $\paren {\forall x < y: x \in A} \implies y \in A$
where $x^+$ denotes the successor of $x$.
Then $\On \subseteq A$.
\end{theorem}
\begin{proof}
{{NotZFC}}
We shall prove this using the first principle of transfinite induction.
Assume $y$ is an ordinal such that $y \subseteq A$.
This implies $\forall x < y: x \in A$.
By definition, $y$ is a limit ordinal, $y = \O$, or $\exists x: y = x^+$.
If $y$ is a limit ordinal, then, since:
:$\forall x < y: x \in A$
it follows that $y \in A$.
If $y = \O$, then $y \in A$ {{hypothesis}}.
Finally, assume $y = z^+$ for some ordinal $z$.
Since $z < z^+ = y$ and $\forall x < y: x \in A$, it follows that $z \in A$.
By hypothesis, we conclude $y = z^+ \in A$.
In any case, $y \in A$.
Therefore, for all ordinals $y$, if $y \subseteq A$, then $y \in A$.
By the first principle of transfinite induction, $\On \subseteq A$.
{{qed}}
Category:Ordinals
\end{proof}
|
22743
|
\section{Transfinite Induction/Schema 1}
Tags: Ordinals
\begin{theorem}
Let $\map P x$ be a property
Suppose that:
: If $\map P x$ holds for all ordinals $x$ less than $y$, then $\map P y$ also holds.
Then $\map P x$ holds for all ordinals $x$.
\end{theorem}
\begin{proof}
Since an ordinal $x$ is less than an ordinal $y$ if and only if $x \in y$, the premise of the theorem can be written
:$\forall y \in \operatorname{On}: \left({\left({\forall x \in \operatorname{On}: \left({x \in y \implies P \left({x}\right)}\right)}\right) \implies P \left({y}\right)}\right)$.
The statement:
:$\forall x \in \operatorname{On}: \left({x \in y \implies P \left({x}\right)}\right)$
is equivalent to:
:$y \subseteq \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\}$ {{explain}}
Since $\left({\forall x \in \operatorname{On}: \left({x \in y \implies P \left({x}\right)}\right)}\right) \implies P \left({y}\right)$,
:$y \subseteq \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\} \implies y \in \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\}$ by the above equivalence.
By the Principle of Transfinite Induction (above):
: $\operatorname{On} \subseteq \left\{{x \in \operatorname{On} : P \left({x}\right)}\right\}$
Therefore:
: $x \in \operatorname{On} \implies P \left({x}\right)$
for all $x$.
{{qed}}
Category:Ordinals
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\end{proof}
|
22744
|
\section{Transfinite Induction/Schema 2}
Tags: Ordinals
\begin{theorem}
Let $\map \phi x$ be a property satisfying the following conditions:
:$(1): \quad \map \phi \O$ is true
:$(2): \quad$ If $x$ is an ordinal, then $\map \phi x \implies \map \phi {x^+}$
:$(3): \quad$ If $y$ is a limit ordinal, then $\paren {\forall x < y: \map \phi x} \implies \map \phi y$
where $x^+$ denotes the successor of $x$.
Then, $\map \phi x$ is true for all ordinals $x$.
\end{theorem}
\begin{proof}
Defint the class:
:$A := \left\{{x \in \operatorname{On}: \phi \left({x}\right) = \mathrm T}\right\}$.
Then $\phi \left({x}\right) = \mathrm T$ is equivalent to the statement:
:that $x \in A$
The three conditions in the hypothesis become:
:$(1a): \quad \varnothing \in A$
:$(2a): \quad x \in A \implies x^+ \in A$
:$(3a): \quad \left({\forall x < y: x \in A}\right) \implies y \in A$
These are precisely the conditions for the class $A$ in the second principle of transfinite induction.
Therefore, $\operatorname{On} \subseteq A$.
Thus, $\phi \left({x}\right)$ holds for all $x \in \operatorname{On}$.
{{qed}}
In proofs by '''transfinite induction''' using this particular schema, the following terms are used.
\end{proof}
|
22745
|
\section{Transfinite Recursion/Corollary}
Tags: Ordinals
\begin{theorem}
Let $x$ be an ordinal.
Let $G$ be a mapping
There exists a unique mapping $f$ that satisfies the following properties:
:The domain of $f$ is $x$
:$\forall y \in x: \map f y = \map G {f \restriction y}$
\end{theorem}
\begin{proof}
Construct $K$ and $F$ as in the First Principle of Transfinite Recursion.
Set $f = \paren {F \restriction x}$.
Then since $x \subseteq \Dom F$, the domain of $f$ is $x$.
{{begin-eqn}}
{{eqn | q = \forall y \in x
| l = \map f y
| r = \map F y
| c = from the fact that $f$ is a restriction
}}
{{eqn | r = \map G {F \restriction y}
| c = Transfinite Recursion: Theorem 1
}}
{{eqn | r = \map G {f \restriction y}
| c =
}}
{{end-eqn}}
Thus such a mapping $f$ exists.
Suppose there are two mappings $f$ and $g$ that satisfy these conditions.
We will use the first principle of transfinite induction to show that $f = g$.
Take $y \in x$.
Suppose $\forall z \in y: \map f z = \map g z$.
Then:
:$\paren {f \restriction y} = \paren {g \restriction y}$
and so:
{{begin-eqn}}
{{eqn | l = \map f y
| r = \map G {f \restriction y}
| c = Assumption
}}
{{eqn | r = \map G {g \restriction y}
| c = Substitutivity of Equality
}}
{{eqn | r = \map g y
| c = {{Hypothesis}}
}}
{{end-eqn}}
So $\map f y = \map g y$ for all $y \in x$ by transfinite induction.
Since $x$ is the domain of $f$ and $g$, it follows that $f = g$.
Thus the mapping $f$ is unique.
{{qed}}
\end{proof}
|
22746
|
\section{Transfinite Recursion/Theorem 2}
Tags: Ordinals
\begin{theorem}
Let $\Dom x$ denote the domain of $x$.
Let $\Img x$ denote the image of the mapping $x$.
{{explain|We infer that $x$ is a mapping, but what is its context?}}
Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions:
:$(1): \quad x = \O$ and $y = a$
{{explain|What is $a$?}}
:$(2): \quad \exists \beta: \Dom x = \beta^+$ and $y = \map H {\map x {\bigcup \Dom x} }$
{{explain|What is $H$?}}
:$(3): \quad \Dom x$ is a limit ordinal and $y = \bigcup \Rng x$.
{{explain|Is this invoking well-founded recursion?}}
Let $\map F \alpha = \map G {F \restriction \alpha}$ for all ordinals $\alpha$.
Then:
:$F$ is a mapping and the domain of $F$ is the ordinals, $\On$.
:$\map F \O = a$
:$\map F {\beta^+} = \map H {\map F \beta}$
:For limit ordinals $\beta$, $\ds \map F \beta = \bigcup_{\gamma \mathop \in \beta} \map F \gamma$
:$F$ is unique.
::That is, if there is another function $A$ satisfying the above three properties, then $A = F$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map F \O
| r = \map G {F \restriction \O}
| c = {{Hypothesis}}
}}
{{eqn | r = \map G \O
| c = Restriction of $\O$
}}
{{eqn | r = a
| c = Definition of $G$
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \map F {\beta^+}
| r = \map G {F \restriction \beta^+}
| c = {{Hypothesis}}
}}
{{eqn | r = \map H {\map {F \restriction \beta^+} {\bigcup \beta^+} }
| c = Definition of $G$
}}
{{eqn | r = \map H {\map F \beta}
| c = Union of successor set is the original set by Union of Ordinals is Least Upper Bound
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = \map F \beta
| r = \map G {F \restriction \beta}
| c = {{Hypothesis}}
}}
{{eqn | r = \bigcup \Img {F \restriction \beta}
| c = Definition of $G$
}}
{{eqn | r = \bigcup_{\gamma \mathop \in \beta} \map F \gamma
| c =
}}
{{end-eqn}}
{{qed|lemma}}
We can proceed in the fourth part by Transfinite Induction.
\end{proof}
|
22747
|
\section{Transfinite Recursion/Uniqueness of Transfinite Recursion}
Tags: Ordinals
\begin{theorem}
Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal.
Let $f$ satisfy the condition that:
:$\forall x \in y: \map f x = \map G {f \restriction x}$
where $f \restriction x$ denotes the restriction of $f$ to $x$.
{{explain|What is $G$?}}
Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal.
Let $g$ satisfy the condition that:
:$\forall x \in z: \map g x = \map G {g \restriction x}$
Let $y \subseteq z$.
Then:
:$\forall x \in y: \map f x = \map g x$
\end{theorem}
\begin{proof}
Proof by transfinite induction:
Suppose that:
:$\forall x \in \alpha: \map f x = \map g x$
for some arbitrary ordinal $\alpha < y$.
Then $\alpha < z$.
{{explain|Find the link to the result proving this.}}
Hence:
{{begin-eqn}}
{{eqn | q = \forall x \in \alpha
| l = \map f x
| r = \map g x
}}
{{eqn | ll= \leadsto
| l = f \restriction \alpha
| r = g \restriction \alpha
| c = Equality of Restrictions
}}
{{eqn | ll= \leadsto
| l = \map G {f \restriction \alpha}
| r = \map G {g \restriction \alpha}
| c = Substitution
}}
{{eqn | ll= \leadsto
| l = \map f \alpha
| r = \map g \alpha
| c = {{Hypothesis}}
}}
{{end-eqn}}
So applying induction:
: $\forall \alpha < y: \map f \alpha = \map g \alpha$
{{qed}}
\end{proof}
|
22748
|
\section{Transformation of P-Norm}
Tags: P-Sequence Metrics, P-Norms, Norm Theory, Functional Analysis
\begin{theorem}
Let $p, q \ge 1$ be real numbers.
Let $\ell^p$ denote the $p$-sequence space.
Let $\norm {\mathbf x}_p$ denote the $p$-norm.
Let $\mathbf x = \sequence {x_n} \in \ell^{p q}$.
Suppose further that $\mathbf x^p = \sequence { {x_n}^p} \in \ell^q$.
Then:
:$\norm {\mathbf x^p}_q = \norm {\mathbf x}_{p q}^p$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \norm {\mathbf x^p}_q
| r = \paren {\sum_{n \mathop = 0}^\infty \size { {x_n}^p}^q}^{1 / q}
| c = {{Defof|P-Norm|$p$-Norm}}
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / q}
| c = Power of Power
}}
{{eqn | r = \paren {\paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / p q} }^p
| c = Power of Power
}}
{{eqn | r = \norm {\mathbf x}_{p q}^p
| c = {{Defof|P-Norm|$p$-Norm}}
}}
{{end-eqn}}
{{qed}}
Category:Functional Analysis
Category:Norm Theory
Category:P-Sequence Metrics
Category:P-Norms
\end{proof}
|
22749
|
\section{Transformation of Unit Matrix into Inverse}
Tags: Unit Matrices, Inverse Matrices, Matrix Algebra, Elementary Row Operations
\begin{theorem}
Let $\mathbf A$ be a square matrix of order $n$ of the matrix space $\map {\MM_\R} n$.
Let $\mathbf I$ be the unit matrix of order $n$.
Suppose there exists a sequence of elementary row operations that reduces $\mathbf A$ to $\mathbf I$.
Then $\mathbf A$ is invertible.
Futhermore, the same sequence, when performed on $\mathbf I$, results in the inverse of $\mathbf A$.
\end{theorem}
\begin{proof}
For ease of presentation, let $\breve {\mathbf X}$ be the inverse of $\mathbf X$.
We have that $\mathbf A$ can be transformed into $\mathbf I$ by a sequence of elementary row operations.
By repeated application of Elementary Row Operations as Matrix Multiplications, we can write this assertion as:
{{begin-eqn}}
{{eqn | l = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf A
| r = \mathbf I
}}
{{end-eqn}}
From Elementary Row Matrix is Invertible:
:$\mathbf E_1, \dotsc, \mathbf E_t \in \GL {n, \R}$
{{MissingLinks|$\GL {n, \R}$, and explain the significance of this. It's General Linear Group, clearly.}}
We can multiply on the left both sides of this equation by:
{{begin-eqn}}
{{eqn | l = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf A
| r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
}}
{{eqn | ll= \leadsto
| l = \mathbf {I I} \cdots \mathbf {I I A}
| r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t \mathbf I
| c = {{Defof|Inverse Matrix}}
}}
{{eqn | ll= \leadsto
| l = \mathbf A
| r = \breve {\mathbf E}_1 \breve {\mathbf E}_2 \cdots \breve {\mathbf E}_{t - 1} \breve {\mathbf E}_t
| c = {{Defof|Unit Matrix}}
}}
{{eqn | ll= \leadsto
| l = \breve {\mathbf A}
| r = \breve {\breve {\mathbf E} }_t \breve {\breve {\mathbf E} }_{t - 1} \cdots \breve {\breve {\mathbf E} }_2 \breve {\breve {\mathbf E} }_1
| c = Inverse of Matrix Product, Leibniz's Law
}}
{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1
| c = Inverse of Group Inverse
}}
{{eqn | r = \mathbf E_t \mathbf E_{t - 1} \cdots \mathbf E_2 \mathbf E_1 \mathbf I
| c = {{Defof|Unit Matrix}}
}}
{{end-eqn}}
By repeated application of Elementary Row Operations as Matrix Multiplications, each $\mathbf E_n$ on the {{RHS}} corresponds to an elementary row operation.
Hence the result.
{{qed}}
{{proofread}}
Category:Unit Matrices
Category:Inverse Matrices
Category:Elementary Row Operations
\end{proof}
|
22750
|
\section{Transitive Chaining}
Tags: Transitive Relations
\begin{theorem}
Let $\RR$ be a transitive relation on a set $S$.
Let $n \in \N$ be a natural number.
Let $n \ge 2$.
Let $\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ be a sequence of $n$ terms.
For each $k \in \set {1, 2, \dots, n - 1}$, let $x_k \mathrel \RR x_{k + 1}$.
That is, let $x_1 \mathrel \RR x_2$, $x_2 \mathrel \RR x_3, \dotsc, x_n - 1 \mathrel \RR x_n$.
Then:
:$x_1 \mathrel \RR x_n$
\end{theorem}
\begin{proof}
The proof proceeds by induction on $n$, the number of terms in the sequence.
We first define a propositional function, $P$, as follows:
For each $n \in \N$ such that $n \ge 2$, let $\map P n$ be the proposition that if both of the following hold:
:$\sequence {x_k}_{k \mathop \in \set {1, 2, \dots, n} }$ is a sequence of $n$ terms
:$\forall k \in \set {1, 2, \dots, n - 1}: x_k \mathrel \RR x_{k + 1}$
then:
:$x_1 \mathrel \RR x_n$
\end{proof}
|
22751
|
\section{Transitive Closure Always Exists (Set Theory)}
Tags: Relational Closures, Relational Closure, Transitive Classes
\begin{theorem}
Let $S$ be a set.
Let $G$ be a mapping such that $\map G x = x \cup \bigcup x$.
{{explain|Domain and range of $G$ needed}}
Let $F$ be defined using the Principle of Recursive Definition:
:$\map F 0 = S$
:$\map F {n^+} = \map G {\map F n}$
Let $\ds T = \bigcup_{n \mathop \in \omega} \map F n$.
Then:
:$T$ is a set and is transitive
:$S \subseteq T$
:If $R$ is transitive and $S \subseteq R$, then $T \subseteq R$.
That is, given any set $S$, there is an explicit construction for its transitive closure.
\end{theorem}
\begin{proof}
$\omega$ is a set by the Axiom of Infinity.
Thus by the Axiom of Replacement, the image of $\omega$ under $F$ is also a set.
Since $T$ is the union of $\map F \omega$, it is thus a set by the axiom of unions.
Furthermore:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = T
| c = Assumption
}}
{{eqn | ll= \leadsto
| q = \exists n
| l = x
| o = \in
| r = \map F n
| c = {{Defof|Union of Family}}
}}
{{eqn | ll= \leadsto
| q = \exists n
| l = x
| o = \subseteq
| r = \bigcup \map F n
| c = Set is Subset of Union
}}
{{eqn | l = y
| o = \in
| r = x
| c = Assumption
}}
{{eqn | ll= \leadsto
| q = \exists n
| l = y
| o = \in
| r = \bigcup \map F n
| c =
}}
{{eqn | ll= \leadsto
| q = \exists n
| l = y
| o = \in
| r = \map F {n + 1}
| c = Definition of $F$
}}
{{eqn | ll= \leadsto
| l = y
| o = \in
| r = T
| c = Definition of $T$
}}
{{end-eqn}}
{{explain|domain of $n$}}
By the above equations:
:$x \in T \land y \in x \implies y \in T$
Thus by definition $T$ is transitive.
{{qed|lemma}}
We have that:
:$S = \map F 0$
By Set is Subset of Union:
:$\ds S \subseteq \bigcup_{n \mathop \in \omega} \map F n$
So $S \subseteq T$.
{{qed|lemma}}
Finally, suppose that $S \subseteq R$ and $R$ is transitive.
$T \subseteq R$ follows by finite induction:
For all $n \in \omega$, let $\map P n$ be the proposition:
:$\map F n \subseteq R$
\end{proof}
|
22752
|
\section{Transitive Closure of Reflexive Relation is Reflexive}
Tags: Reflexive Relations, Transitive Closures, Transitive Closure
\begin{theorem}
Let $S$ be a set.
Let $\RR$ be a reflexive relation on $S$.
Let $\TT$ be the transitive closure of $\RR$.
Then $\TT$ is reflexive.
\end{theorem}
\begin{proof}
Let $a \in S$.
Since $\RR$ is reflexive:
:$\tuple {a, a} \in \RR$
By the definition of transitive closure:
:$\RR \subseteq \TT$
Thus by the definition of subset:
:$\tuple {a, a} \in \TT$
Since this holds for all $a \in S, \TT$ is reflexive.
{{qed}}
Category:Reflexive Relations
Category:Transitive Closures
\end{proof}
|
22753
|
\section{Transitive Closure of Reflexive Symmetric Relation is Equivalence}
Tags: Transitive Closures, Equivalence Relations, Transitive Closure
\begin{theorem}
Let $S$ be a set.
Let $\RR$ be a symmetric and reflexive relation on $S$.
Then the transitive closure of $\RR$ is an equivalence relation.
\end{theorem}
\begin{proof}
Let $\sim$ be the transitive closure of $\RR$.
Checking in turn each of the criteria for equivalence:
\end{proof}
|
22754
|
\section{Transitive Closure of Set-Like Relation is Set-Like}
Tags: Set Theory, Transitive Closures
\begin{theorem}
Let $A$ be a class.
Let $\RR$ be a set-like endorelation on $A$.
Let $\RR^+$ be the transitive closure of $\RR$.
Then $\RR^+$ is also a set-like relation.
\end{theorem}
\begin{proof}
Let $x \in A$.
Let $A'$ be the class of all subsets of $A$.
For each $s \in A'$, $\RR^{-1}$ is a subset of $A$.
Hence by Inverse Image of Set under Set-Like Relation is Set and the definition of endorelation:
:$\RR^{-1} \in A'$
Define a mapping $G: A' \to A'$ as:
:$\forall s \in A': \map G s = \map {\RR^{-1} } s$
Recursively define a mapping $f: \N \to A'$ as follows:
:$\map f n = \begin {cases} \set x & : n = 0 \\ \map G {\map f {n - 1} } & : n > 0 \end {cases}$
By the Axiom of Infinity and the Axiom of Replacement:
:$\map f \N$ is a set.
Thus by the Axiom of Unions:
:$\ds \bigcup \map f \N$ is a set.
Let $y \in \map {\paren {\RR^+}^{-1} } x$.
By the definition of transitive closure:
:for some $n \in \N_{>0}$ there are $a_0, a_1, \dots, a_n$ such that $y = a_0 \mathrel \RR a_1 \mathrel \RR \cdots \mathrel \RR a_n = x$.
{{explain|it's a finite sort of induction, and a simple and common pattern.}}
Then by induction (working from $n$ to $0$), $\ds a_n, a_{n - 1}, \dots, a_0 \in \bigcup \map f \N$.
As this holds for all such $y$:
:$\ds \map {\paren {\RR^+}^{-1} } x \subseteq \bigcup \map f \N$
By the Axiom of Specification:
:$\map {\paren {\RR^+}^{-1} } x$ is a set.
As this holds for all $x \in A$:
:$\RR^+$ is a set-like relation.
{{qed}}
Category:Set Theory
Category:Transitive Closures
\end{proof}
|
22755
|
\section{Transitive Closure of Symmetric Relation is Symmetric}
Tags: Transitive Closures, Transitive Closure, Symmetric Relations
\begin{theorem}
Let $S$ be a set.
Let $\RR$ be a symmetric relation on $S$.
Let $\TT$ be the transitive closure of $\RR$.
The $\TT$ is symmetric.
\end{theorem}
\begin{proof}
Let $a, b \in S$ with $a \mathrel \TT b$.
By the definition of transitive closure, there is an $n \in \N$ such that $a \mathrel {\RR^n} b $.
Thus there are $x_0, x_1, \dots x_n \in S$ such that:
:$x_0 = a$
:$x_n = b$
:For $k = 0, \dots, n-1$: $x_k \mathrel \RR x_{k + 1}$
For $k = 0, \dots, n$, let $y_k = x_{n - k}$.
Then:
:$y_0 = x_n = b$
:$y_n = x_0 = a$
For $k = 0, \dots, n - 1$::
:$x_{n - k - 1} \mathrel \RR x_{n - k}$
Thus:
:$y_{k + 1} \mathrel \RR y_k$
Since $\RR$ is symmetric:
:For $k = 0, \dots, n - 1$: $y_k \mathrel \RR y_{k + 1}$
Thus $b \mathrel {\RR^n} a$, so $b \mathrel \TT a$.
{{qed}}
Category:Symmetric Relations
Category:Transitive Closures
\end{proof}
|
22756
|
\section{Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements}
Tags: Compatible Relations, Semigroups
\begin{theorem}
Let $\struct {S, \circ}$ be a semigroup.
Let $\RR$ be a transitive relation on $S$ which is compatible with $\circ$.
Let $x, y \in S$ such that $x \mathrel \RR y$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \mathrel \RR y^n$
where $x^n$ is the $n$th power of $x$.
\end{theorem}
\begin{proof}
We proceed by mathematical induction.
By definition of power:
:$x^1 = x$
:$y^1 = y$
Hence, by assumption:
:$x^1 \mathrel \RR y^1$
Suppose now that for $n \ge 1$:
:$x^n \mathrel \RR y^n$
Recall the assumption that $x \mathrel \RR y$.
Applying Operating on Transitive Relationships Compatible with Operation to these relations yields:
:$x^n \circ x \mathrel \RR y^n \circ y$
By definition of power:
:$x^{n + 1} = x^n \circ x$
and:
:$y^{n + 1} = y^n \circ y$
The result follows by the Principle of Mathematical Induction.
{{qed}}
Category:Compatible Relations
Category:Semigroups
\end{proof}
|
22757
|
\section{Transitive Relation whose Symmetric Closure is not Transitive}
Tags: Transitive Relations, Symmetric Closures, Relation Theory
\begin{theorem}
Let $S = \set {p, q}$, where $p$ and $q$ are distinct elements.
Let $\RR = \set {\tuple {p, q} }$.
Then $\RR$ is transitive but its symmetric closure is not.
\end{theorem}
\begin{proof}
$\RR$ is vacuously transitive because there are no elements $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then $\RR^\leftrightarrow = \RR \cup \RR^{-1} = \set {\tuple {p, q}, \tuple {q, p} }$.
Then:
:$p \mathrel {\RR^\leftrightarrow} q$ and $q \mathrel {\RR^\leftrightarrow} p$
but:
:$p \not \mathrel {\RR^\leftrightarrow} p$
Therefore $\RR^\leftrightarrow$ is not transitive.
{{qed}}
Category:Transitive Relations
Category:Symmetric Closures
\end{proof}
|
22758
|
\section{Transitive Set Contained in Von Neumann Hierarchy Level}
Tags: Von Neumann Hierarchy, Axiom of Foundation
\begin{theorem}
Let $G$ be a transitive set.
Then for some ordinal $i$, $G \subseteq V_i$.
\end{theorem}
\begin{proof}
{{NotZFC}}
{{AimForCont}} for each ordinal $i$ the set $G \setminus V_i$ is non-empty.
Let $i$ be any ordinal.
Then by the axiom of foundation:
$\exists x: x \in G\setminus V_i \text{ and } x \cap \paren {G \setminus V_i} = \O$
Since $G$ is transitive, $x \subseteq G$.
Since $x \subseteq G$ and $x \cap \paren {G \setminus V_i} = \O$:
:$x \subseteq V_i$
By the definition of $V$, $x \in V_{i + 1}$, so:
:$x \notin G \setminus V_{i + 1}$
Thus $G \setminus V_{i + 1} \subsetneqq G \setminus V_i$ for each ordinal $i$.
Let $H: \On \to \powerset G$ be defined by:
:$\map H i = G \setminus V_i$ for each ordinal $i$
Since the class of ordinals is well-ordered, applying Strictly Increasing Mapping on Well-Ordered Class proves that for any ordinals $p$ and $q$,
:$p < q$ implies $\map H q \subsetneq \map H p$.
Since the class of ordinals is totally ordered, for any two distinct ordinals $i$ and $j$:
: $\map H i \ne \map H j$
so $H$ is injective.
But that contradicts the fact that $\powerset G$ is a set.
{{qed}}
Category:Von Neumann Hierarchy
\end{proof}
|
22759
|
\section{Transitive Subgroup of Prime containing Transposition}
Tags: Transitive Subgroups
\begin{theorem}
Let $p$ be a prime number.
Let $S_p$ denote the symmetric group on $p$ letters.
Let $H$ be a transitive subgroup of $S_p$.
If $H$ contains a transposition, then $H = S_p$.
\end{theorem}
\begin{proof}
{{WLOG}}, let $\tuple {1, 2}$ be the transposition contained by $H$.
Let us define an equivalence relation $\sim$ on the set $\N_p = \set {1, 2, \ldots, p}$ as:
:$i \sim j \iff \tuple {i, j} \in H$
Because $H$ is a transitive subgroup it follows that each $\sim$-equivalence class has the same number of elements.
In fact, if $\phi \in H$ and $\phi_1 := \map \phi 1 = i$, then $\phi$ yields a bijection from the $\sim$-equivalence class of $1$ to that of $i$, because:
:$\tuple {1, k} \in H \iff \tuple {i, \phi_k} = \tuple {\phi_1, \phi_k} = \phi \circ \tuple {1, k} \circ \phi^{-1} \in H$
The number $s$ of elements of any given $\sim$-equivalence class must be a divisor of $p$.
Thus $s = 1$ or $s = p$.
However, the $\sim$-equivalence class of $1$ contains at least both $1$ and $2$.
So there can be only one $\sim$-equivalence class which then contains $p$ elements.
In other words, $H$ contains all the transpositions of $S_p$.
From Existence and Uniqueness of Cycle Decomposition, every permutation is a composition of transpositions.
Hence:
:$H = S_p$
{{qed}}
\end{proof}
|
22760
|
\section{Transitive and Antitransitive Relation is Asymmetric}
Tags: Transitive Relations, Symmetric Relations
\begin{theorem}
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Let $\RR$ be both transitive and antitransitive.
Then $\RR$ is asymmetric.
\end{theorem}
\begin{proof}
Let $\tuple {x, y} \in \RR$ for some $x, y \in S$.
Then as $\RR$ is antitransitive:
:$\tuple {x, x} \notin \RR$
and so as $\RR$ is transitive and $\tuple {x, x} \notin \RR$:
:$\tuple {y, x} \notin \RR$
That is, $\RR$ is asymmetric.
{{qed}}
\end{proof}
|
22761
|
\section{Transitivity of Algebraic Extensions}
Tags: Field Extensions
\begin{theorem}
Let $E / F / K$ be a tower of field extensions.
Let $E$ be algebraic over $F$.
Let $F$ be algebraic over $K$.
Then $E$ is algebraic over $K$.
\end{theorem}
\begin{proof}
Let $x \in E$.
There are $a_0, \ldots, a_n \in F$ such that $a_0 + \cdots + a_n x^n = 0$.
Let $L = \map K {a_0, \ldots, a_n}$.
We have that $L / K$ is finitely generated and algebraic.
Therefore by Finitely Generated Algebraic Extension is Finite this extension is finite.
We have that $\map L x / L$ is simple and algebraic.
So by Structure of Simple Algebraic Field Extension, this extension is also finite.
Therefore, by the Tower Law:
:$\map L x / K$ is finite.
That is, $x$ is contained in a finite extension of $K$.
Therefore because a Finite Field Extension is Algebraic, it follows that $x$ is algebraic over $K$, as was to be proved.
{{qed}}
\end{proof}
|
22762
|
\section{Transitivity of Big-O Estimates/General}
Tags: Asymptotic Notation
\begin{theorem}
Let $X$ be a topological space.
Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\norm {\,\cdot\,}$.
Let $f, g, h: X \to V$ be functions.
Let $x_0 \in X$.
Let $f = \map \OO g$ and $g = \map \OO h$ as $x \to x_0$, where $\OO$ denotes big-O notation.
Then $f = \map \OO h$ as $x \to x_0$.
\end{theorem}
\begin{proof}
Because $f = \map \OO g$ and $g = \map \OO h$, there exist neighborhoods $U$ and $V$ of $x_0$ and real numbers $c, d \ge 0$ such that:
:$\norm {\map f x} \le c \cdot \norm {\map g x}$ for all $x \in U$
:$\norm {\map g x} \le d \cdot \norm {\map h x}$ for all $x \in V$.
By Intersection of Neighborhoods in Topological Space is Neighborhood, $U\cap V$ is a neighborhood of $x_0$.
For $x \in U \cap V$, we have:
:$\norm {\map f x} \le c \cdot \norm {\map g x} \le c d \cdot \norm {\map h x}$
Thus $f = \map \OO h$ for $x \to x_0$.
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22763
|
\section{Transitivity of Big-O Estimates/Sequences}
Tags: Asymptotic Notation
\begin{theorem}
Let $\sequence {a_n}$, $\sequence {b_n}$ and $\sequence {c_n}$ be sequences of real or complex numbers.
Let $a_n = \map \OO {\sequence {b_n} }$ and $b_n = \map \OO {\sequence {c_n} }$, where $O$ denotes big-O notation.
Then $a_n = \map \OO {\sequence {c_n} }$.
\end{theorem}
\begin{proof}
Because $a_n = \map \OO {\sequence {b_n} }$, there exists $K \ge 0$ and $n_0 \in \N$ such that $\size {a_n} \le K \cdot \size {b_n}$ for $n \ge n_0$.
Because $b_n = \map \OO {\sequence {c_n} }$, there exists $L \ge 0$ and $n_1 \in \N$ such that $\size {b_n} \le L \cdot \size {c_n}$ for $n \ge n_1$.
Then $\size {a_n} \le K L \cdot \size {c_n}$ for $n \ge \max \set {n_0, n_1}$.
Thus $a_n = \map \OO {\sequence {c_n} }$.
{{qed}}
Category:Asymptotic Notation
\end{proof}
|
22764
|
\section{Transitivity of Finite Generation}
Tags: Commutative Algebra, Algebraic Number Theory
\begin{theorem}
Let $A \subseteq B \subseteq C$ be rings.
Let $B$ be a finitely generated $A$-module.
Let $C$ be a finitely generated $B$-module.
{{explain|Are $B$ and $C$ rings or modules?}}
Then $C$ is a finitely generated $A$-module.
\end{theorem}
\begin{proof}
Let $b_1, \ldots, b_n$ generate $B$ over $A$.
Let $c_1, \ldots, c_m$ generate $C$ over $B$.
Then for any $x \in C$ there are $\beta_k \in B$, $k = 1, \ldots, m$ such that:
:$\ds x = \sum_{k \mathop = 1}^m \beta_k c_k$
For any $i \in \set {1, \ldots, m}$ there exists $\alpha_{i j} \in A$, $j = 1, \ldots, n$ such that:
:$\ds \beta_i = \sum_{j \mathop = 1}^n \alpha_{i j} b_j$
So
:$\ds x = \sum_{k \mathop = 1}^m \sum_{j \mathop = 1}^n \alpha_{k j} b_j c_k$
Therefore, $\set {b_j c_k : j = 1, \ldots, n, k = 1, \ldots, m}$ generates $C$ over $A$.
{{qed}}
Category:Commutative Algebra
Category:Algebraic Number Theory
\end{proof}
|
22765
|
\section{Transitivity of Integrality}
Tags: Commutative Algebra, Transitivity of Integrality, Algebraic Number Theory
\begin{theorem}
Let $A \subseteq B \subseteq C$ be extensions of commutative rings with unity.
Suppose that $C$ is integral over $B$, and $B$ is integral over $A$.
Then $C$ is integral over $A$.
\end{theorem}
\begin{proof}
First, a lemma:
\end{proof}
|
22766
|
\section{Translation-Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure}
Tags: Measure Theory
\begin{theorem}
Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Suppose that $\mu$ is translation-invariant.
Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\infty$.
Then $\mu = \kappa \lambda^n$, where $\lambda^n$ is the $n$-dimensional Lebesgue measure.
\end{theorem}
\begin{proof}
From Characterization of Euclidean Borel Sigma-Algebra, we have:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\JJ^n_{ho, \text {rat} }$ denotes the collection of half-open $n$-rectangles with rational endpoints.
So let $J = \horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho, \text {rat} }$.
Let $M \in \N$ be a common denominator of the $a_i, b_i$ (which are rational by assumption).
We may then cover $J$ by finitely many pairwise disjoint half-open $n$-rectangles $\hointr 0 {\dfrac 1 M}^n$, in that:
:$\ds J = \bigcup_{i \mathop = 1}^{\map k J} \mathbf x_i + \hointr 0 {\dfrac 1 M}^n$
for some $\map k J \in \N$ and suitable $\mathbf x_i$, where:
:$\mathbf x_i + \hointr 0 {\dfrac 1 M}^n := \horectr {\mathbf x_i} {\mathbf x_i + \dfrac 1 M}$
Using that $\mu$ is translation-invariant, this means:
:$\map \mu J = \map k J \, \map \mu {\hointr 0 {\dfrac 1 M}^n}$
Also, by Lebesgue Measure is Translation-Invariant:
:$\map {\lambda^n} J = \map k J \, \map {\lambda^n} {\hointr 0 {\dfrac 1 M}^n}$
A moment's thought shows us that the half-open $n$-rectangle $I = \hointr 0 1^n$ may be covered by $M^n$ copies of $\hointr 0 {\dfrac 1 M}^n$, so that:
:$\map k I = M^n$
For brevity, write $I / M$ for $\hointr 0 {\dfrac 1 M}^n$.
Now, using that:
:$\ds \map {\lambda^n} {I / M} = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$
and $\map \mu I = \kappa$, compute:
{{begin-eqn}}
{{eqn | l = \map \mu J
| r = \map k J \map \mu {I / M}
| c = Definition of $\map k J$
}}
{{eqn | r = \frac {\map k J} {M^n} \paren {M^n \map \mu {I / M} }
| c = $M^n / M^n = 1$
}}
{{eqn | r = \frac {\map k J} {M^n} \map \mu I
| c = $M^n = \map k I$
}}
{{eqn | r = \frac {\kappa \, \map k J} {M^n}
| c = Definition of $\kappa$
}}
{{eqn | r = \kappa \, \map k J \, \map {\lambda^n} {I / M}
}}
{{eqn | r = \kappa \, \map {\lambda^n} J
| c = Definition of $\map k J$
}}
{{end-eqn}}
Therefore, $\map \mu J = \kappa \, \map {\lambda^n} J$ for all $J \in \JJ^n_{ho, \text {rat} }$.
Let us quickly verify the other conditions for Uniqueness of Measures.
For $(1)$, we have Half-Open Rectangles Closed under Intersection.
For $(2)$, observe the exhausting sequence $\hointr {-k} k^n \mathop \uparrow \R^n$
Finally, for $(4)$, we recall that $\kappa < +\infty$.
Thus, by Uniqueness of Measures, $\mu = \kappa \lambda^n$.
{{qed}}
\end{proof}
|
22767
|
\section{Translation Mapping is Isometry}
Tags: Isometries, Translation Mappings
\begin{theorem}
Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions.
Let $\tau_\mathbf x$ be a translation on $\Gamma$:
:$\forall \mathbf y \in \R^n: \map {\tau_\mathbf x} {\mathbf y} = \mathbf y - \mathbf x$
where $\mathbf x$ is a vector in $\R^n$.
Then $\tau_\mathbf x$ is an isometry.
\end{theorem}
\begin{proof}
From Translation Mapping is Bijection, $\tau_\mathbf x$ is a bijection.
From Euclidean Metric on Real Number Space is Translation Invariant, $\tau_\mathbf x$ is distance-preserving on $\Gamma$.
The result follows by definition of isometry.
{{qed}}
Category:Isometries
Category:Translation Mappings
\end{proof}
|
22768
|
\section{Translation in Euclidean Space is Measurable Mapping}
Tags: Measure Theory, Translation Mappings
\begin{theorem}
Let $\BB$ be the Borel $\sigma$-algebra on $\R^n$.
Let $\mathbf x \in \R^n$, and denote with $\tau_{\mathbf x}: \R^n \to \R^n$ translation by $\mathbf x$.
Then $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable.
\end{theorem}
\begin{proof}
By Characterization of Euclidean Borel Sigma-Algebra, $\BB = \map \sigma {\JJ_{ho}^n}$.
Here, $\JJ_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra.
Now, for any half-open $n$-rectangle $\horectr {\mathbf a} {\mathbf b}$, it is trivial that:
:$\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \horectr {\mathbf a + \mathbf x} {\mathbf b + \mathbf x}$
That is, the preimage of a half-open $n$-rectangle under $\tau_{\mathbf x}$ is again a half-open $n$-rectangle.
In particular, since $\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho}^n} = \BB$, Mapping Measurable iff Measurable on Generator applies.
Thus it follows that $\tau_{\mathbf x}$ is $\BB \, / \, \BB$-measurable.
{{qed}}
\end{proof}
|
22769
|
\section{Translation of Index Variable of Summation}
Tags: Summations
\begin{theorem}
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R j} a_j$ denote a summation over $R$.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {c \mathop + j} } a_{c \mathop + j} = \sum_{\map R {c \mathop - j} } a_{c \mathop - j}$
where $c$ is an integer constant which is not dependent upon $j$.
\end{theorem}
\begin{proof}
First we consider the case where the fiber of truth of $R$ is finite.
{{ProofWanted}}
\end{proof}
|
22770
|
\section{Translation of Index Variable of Summation/Corollary}
Tags: Summations
\begin{theorem}
:$\ds \sum_{j \mathop = m}^n a_j = \sum_{j \mathop = m + c}^{n + c} a_{j - c}$
where $c$ is an integer constant which is not dependent upon $j$.
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Summations
\end{proof}
|
22771
|
\section{Translation of Integer Interval is Bijection}
Tags: Set Theory
\begin{theorem}
Let $a, b, c \in \Z$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Then the mapping $T: \closedint a b \to \closedint {a + c} {b + c}$ defined as:
:$\map T k = k + c$
is a bijection.
\end{theorem}
\begin{proof}
Note that if $k \in \closedint a b$, then indeed $k + c \in \closedint {a + c} {b + c}$.
\end{proof}
|
22772
|
\section{Translation of Open Set in Normed Vector Space is Open}
Tags: Normed Vector Spaces, Open Sets (Normed Vector Spaces)
\begin{theorem}
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open.
\end{theorem}
\begin{proof}
Let:
:$v \in U + x$
then:
:$v = u + x$
for some $u \in U$.
So:
:$v - x \in U$
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:
:$\norm {\paren {v - x} - \paren {v' - x} } < \epsilon$
we have $v' - x \in U$.
That is:
:$v' \in U + x$
Note that:
:$\norm {\paren {v - x} - \paren {v' - x} } = \norm {v - v'}$
So, whenever $v \in U + x$ and $v' \in X$ is such that:
:$\norm {v - v'} < \epsilon$
we have:
:$v' \in U + x$
Since $u$ was arbitrary:
:$U + x$ is open.
{{qed}}
Category:Open Sets (Normed Vector Spaces)
\end{proof}
|
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