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22473
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\section{Surjection from Finite Set to Itself is Permutation}
Tags: Permutation Theory, Surjections
\begin{theorem}
Let $S$ be a finite set.
Let $f: S \to S$ be an surjection.
Then $f$ is a permutation.
\end{theorem}
\begin{proof}
From Surjection iff Right Inverse, $f$ has a right inverse $g: S \to S$.
From Right Inverse Mapping is Injection, $g$ is an injection.
From Injection from Finite Set to Itself is Permutation, $g$ is a permutation and so a bijection.
From Inverse of Bijection is Bijection, $f$ is also a bijection.
Thus as $f$ is a bijection to itself, it is by definition a permutation.
{{qed}}
\end{proof}
|
22474
|
\section{Surjection from Natural Numbers iff Countable/Corollary 1}
Tags: Countably Infinite, Natural Numbers, Countable Sets, Surjections
\begin{theorem}
Let $T$ be a countably infinite set.
Let $S$ be a non-empty set.
Then $S$ is countable {{iff}} there exists a surjection $f: T \to S$.
\end{theorem}
\begin{proof}
Let $g: T \to \N$ be a bijection from $T$ to $\N$.
By Inverse of Bijection is Bijection, $g^{-1}: \N \to T$ is a bijection from $\N$ to $T$.
\end{proof}
|
22475
|
\section{Surjection from Natural Numbers iff Countable/Corollary 2}
Tags: Countable Sets, Surjections
\begin{theorem}
Let $T$ be a countably infinite set.
Let $S$ be an uncountable set.
Let $f:T \to S$ be a mapping.
Then $f$ is not a surjection.
\end{theorem}
\begin{proof}
By Corollary 1 no mapping from $T$ to $S$ is a surjection.
{{qed}}
Category:Countable Sets
Category:Surjections
\end{proof}
|
22476
|
\section{Surjection if Composite is Surjection}
Tags: Surjections, Mappings, Composite Mappings
\begin{theorem}
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection.
Then $g$ is a surjection.
\end{theorem}
\begin{proof}
Let $g \circ f$ be surjective.
Fix $z \in S_3$.
Now find an $x \in S_1: \map {g \circ f} x = z$.
The surjectivity of $g \circ f$ guarantees this can be done.
Now find an $y \in S_2: f \paren x = y$.
$f$ is a mapping and therefore a left-total relation; which guarantees this too can be done.
It follows that:
{{begin-eqn}}
{{eqn | l = \map g y
| r = \map g {\map f x}
}}
{{eqn | r = \map {g \circ f} x
| c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = z
| c = Choice of $x$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22477
|
\section{Surjection iff Right Inverse/Non-Uniqueness}
Tags: Surjections, Surjection iff Right Inverse
\begin{theorem}
Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a surjection.
A right inverse of $f$ is in general not unique.
Uniqueness occurs {{iff}} $f$ is a bijection.
\end{theorem}
\begin{proof}
If $f$ is not an injection then:
:$\exists y \in T: \exists x_1, x_2 \in S: \map f {x_1} = y = \map f {x_2}$
Hence we have more than one choice in $\map {f^{-1} } {\set y}$ for how to map $\map g y$.
That is, $\map g y$ is not unique.
This does not happen {{iff}} $f$ is an injection.
Hence the result.
{{qed}}
\end{proof}
|
22478
|
\section{Surjection that Preserves Inner Product is Linear}
Tags: Linear Transformations on Hilbert Spaces, Hilbert Spaces
\begin{theorem}
Let $H, K$ be Hilbert spaces, and denote by ${\innerprod \cdot \cdot}_H$ and ${\innerprod \cdot \cdot}_K$ their respective inner products.
Let $U: H \to K$ be a surjection such that:
:$\forall g, h \in H: {\innerprod g h}_H = {\innerprod {Ug} {Uh} }_K$
Then $U$ is a linear map, and hence an isomorphism.
\end{theorem}
\begin{proof}
Let $x, y \in H$.
Let $\alpha \in \GF$.
By surjectivity of $U$, choose $z \in H$ such that $Uz = \map U {\alpha x + y} - \paren { \alpha Ux + Uy }$.
Then:
{{begin-eqn}}
{{eqn | l = {\innerprod {Uz} {Uz} }_K
| r = {\innerprod {\map U {\alpha x + y} - \paren{\alpha Ux + Uy } } {Uz} }_K
}}
{{eqn | r = {\innerprod {\map U {\alpha x + y} } {Uz} }_K - \paren{ \alpha {\innerprod {Ux} {Uz} }_K + {\innerprod {Uy} {Uz} }_K}
| c = by linearity in the first coordinate
}}
{{eqn | r = {\innerprod {\alpha x + y} z }_H - \paren{ \alpha {\innerprod x z }_H + {\innerprod y z }_H}
| c = as $U$ preserves the inner product
}}
{{eqn | r = 0
| c = by linearity in the first coordinate
}}
{{end-eqn}}
By positivity, $Uz = {\bf 0}_K$.
Hence:
:$\map U {\alpha x + y} = \alpha U x + U y$
Thus, $U$ is linear.
{{qed}}
\end{proof}
|
22479
|
\section{Surjective Field Homomorphism is Field Isomorphism}
Tags: Field Homomorphisms, Field Isomorphisms
\begin{theorem}
Let $E$ and $F$ be fields.
Let $\phi: E \to F$ be a (field) homomorphism.
Let $\phi$ be a surjection.
Then $\phi$ is an isomorphism.
\end{theorem}
\begin{proof}
As asserted, let $\phi$ be a surjection.
From Field Homomorphism is either Trivial or Injection, $\phi$ is either an injection or the trivial homomorphism.
If $\phi$ is an injection, then, by definition, $\phi$ is a bijection.
Hence, again by definition, $\phi$ is an isomorphism.
If $\phi$ is not an injection, then $\phi$ is the trivial homomorphism.
But from Field Contains at least 2 Elements, $\Img \phi$ cannot in that case be a field.
Hence if $\phi$ is not an injection, then $\phi$ cannot be a surjection.
Hence the result.
{{qed}}
\end{proof}
|
22480
|
\section{Surjective Monotone Function is Continuous}
Tags: Monotone Real Functions, Real Analysis, Continuity, Continuity, Real Analysis
\begin{theorem}
Let $X$ be an open set of $\R$.
Let $Y$ be a real interval.
Let $f: X \to Y$ be a surjective monotone real function.
Then $f$ is continuous on $X$.
\end{theorem}
\begin{proof}
{{Tidy}}
{{MissingLinks}}
{{WLOG}}, let $f$ be increasing.
Let $c \in X$.
From Limit of Monotone Real Function: Corollary, the one sided limits of monotone functions exist:
{{begin-eqn}}
{{eqn | l = L^-_c
| m = \lim_{x \mathop \to c^-} \map f x
| mo= =
| r = \sup_{x \mathop < c} \map f x
}}
{{eqn | l = L^+_c
| m = \lim_{x \mathop \to c^+} \map f x
| mo= =
| r = \inf_{x \mathop > c} \map f x
}}
{{end-eqn}}
and satisfy:
:$L^-_c, L^+_c \in Y$
:$L^-_c \le \map f c \le L^+_c$
Suppose that $\ds L = \lim_{x \mathop \to c} \map f x$ exists.
From Limit iff Limits from Left and Right:
:$L = L^-_c$
This leads to:
:$L \le \map f c$
Similarly:
:$L = L^+_c$
which leads to:
:$L \ge \map f c$
Hence:
:$\ds \lim_{x \mathop \to c} \map f x = \map f c$
proving continuity at $c$.
By assumption, $f$ is increasing.
Suppose $\ds \lim_{x \mathop \to c} \map f x$ does not exist.
Then from Discontinuity of Monotonic Function is Jump Discontinuity, there is a jump discontinuity at $c$.
{{AimForCont}} $f$ has a jump discontinuity at $c$.
From Real Numbers are Densely Ordered:
:$L^-_c < y < L^+_c$
for some $y \in Y$.
By surjectivity, $y = \map f a$ for some $a \in X$.
Hence:
:$L^-_c < \map f a < L^+_c$
If $a < c$ then $\map f a \le L^-_c$.
This contradicts the previous inequality.
There is a similar contradiction if $a \ge c$.
{{finish}}
Category:Monotone Real Functions
Category:Real Analysis
Category:Continuity
\end{proof}
|
22481
|
\section{Surjective Restriction of Real Exponential Function}
Tags: Exponential Function, Surjections
\begin{theorem}
Let $\exp: \R \to \R$ be the exponential function:
:$\map \exp x = e^x$
Then the restriction of the codomain of $\exp$ to the strictly positive real numbers:
:$\exp: \R \to \R_{>0}$
is a surjective restriction.
Hence:
:$\exp: \R \to \R_{>0}$
is a bijection.
\end{theorem}
\begin{proof}
We have Exponential on Real Numbers is Injection.
Let $y \in \R_{> 0}$.
Then $\exists x \in \R: x = \map \ln y$
That is:
:$\exp x = y$
and so $\exp: \R \to \R_{>0}$ is a surjection.
Hence the result.
{{qed}}
\end{proof}
|
22482
|
\section{Sylow Subgroup is Hall Subgroup}
Tags: Group Theory, Hall Subgroups, Sylow p-Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H$ be a Sylow $p$-subgroup of $G$.
Then $H$ is a Hall subgroup of $G$.
\end{theorem}
\begin{proof}
Let $p$ be prime.
Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$.
By definition, a Sylow $p$-subgroup $H$ of $G$ is a subgroup of $G$ of order $p^n$.
By Lagrange's Theorem, the index of $H$ in $G$ is given by:
:$\index G H = \dfrac {\order G} {\order H}$
So in this case:
:$\index G H = \dfrac {k p^n} {p^n} = k$
As $p \nmid k$ it follows from Prime not Divisor implies Coprime that $k \perp p$.
The result follows from the definition of Hall subgroup.
{{qed}}
Category:Sylow p-Subgroups
Category:Hall Subgroups
\end{proof}
|
22483
|
\section{Sylow p-Subgroup is Unique iff Normal}
Tags: Normal Subgroups, Sylow p-Subgroups
\begin{theorem}
A group $G$ has exactly one Sylow $p$-subgroup $P$ {{iff}} $P$ is normal.
\end{theorem}
\begin{proof}
If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal.
Suppose a Sylow $p$-subgroup $P$ is normal.
Then it equals its conjugates.
Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup.
{{qed}}
\end{proof}
|
22484
|
\section{Sylow p-Subgroups of Group of Order 2p}
Tags: Groups of Order 2p, Sylow p-Subgroups of Group of Order 2p, Groups of Order 2 p, Sylow p-Subgroups
\begin{theorem}
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.
\end{theorem}
\begin{proof}
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides 2 p$
that is:
:$n_p \in \set {1, 2, p, 2 p}$
But $p$ and $2 p$ are congruent to $0$ modulo $p$
So:
:$n_p \notin \set {p, 2 p}$
Also we have that $p > 2$.
Hence:
:$2 \not \equiv 1 \pmod p$
and so it must be that $n_p = 1$.
It follows from Normal Sylow P-Subgroup is Unique that this Sylow $p$-subgroup is normal.
{{qed}}
\end{proof}
|
22485
|
\section{Symmetric Bilinear Form can be Diagonalized}
Tags: Bilinear Forms
\begin{theorem}
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension $n>0$.
Let $f$ be a symmetric bilinear form on $V$.
Then there exists an ordered basis for which the relative matrix of $f$ is diagonal.
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Bilinear Forms
\end{proof}
|
22486
|
\section{Symmetric Bilinear Form is Reflexive}
Tags: Bilinear Forms
\begin{theorem}
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$.
Let $b$ be a bilinear form on $V$.
Let $b$ be symmetric.
Then $b$ is reflexive.
\end{theorem}
\begin{proof}
Let $\left({v, w}\right) \in V \times V$ with $b \left({v, w}\right) = 0$.
Because $b$ is symmetric, $b \left({w, v}\right) = 0$.
Because $\left({v, w}\right)$ was arbitrary, $b$ is reflexive.
{{qed}}
\end{proof}
|
22487
|
\section{Symmetric Closure of Ordering may not be Transitive}
Tags: Orderings, Closure Operators, Symmetric Closures, Transitive Relations, Symmetric Closure
\begin{theorem}
Let $\struct {S, \preceq}$ be an ordered set.
Let $\preceq^\leftrightarrow$ be the symmetric closure of $\preceq$.
Then it is not necessarily the case that $\preceq^\leftrightarrow$ is transitive.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $S = \set {a, b, c}$ where $a$, $b$, and $c$ are distinct.
Let:
:${\preceq} = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {b, c} }$:
Then $\preceq$ is an ordering, but $\preceq^\leftrightarrow$ is not transitive, as follows:
$\preceq$ is reflexive because it contains the diagonal relation on $S$.
That $\preceq$ is transitive and antisymmetric can be verified by inspecting all ordered pairs of its elements.
Thus $\preceq$ is an ordering.
Now consider $\preceq^\leftrightarrow$, the symmetric closure of $\preceq$:
:${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \set{\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, c}, \tuple {c, a}, \tuple {b, c}, \tuple {c, b} }$
by inspection.
Now $\tuple {a, c} \in {\preceq^\leftrightarrow}$ and $\tuple {c, b} \in {\preceq^\leftrightarrow}$, but $\tuple {a, b} \notin {\preceq^\leftrightarrow}$.
Thus $\preceq^\leftrightarrow$ is not transitive.
{{qed}}
Category:Orderings
Category:Symmetric Closures
Category:Transitive Relations
\end{proof}
|
22488
|
\section{Symmetric Closure of Relation Compatible with Operation is Compatible}
Tags: Compatible Relations, Symmetric Closures
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a relation compatible with $\circ$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then $\RR^\leftrightarrow$ is compatible with $\circ$.
\end{theorem}
\begin{proof}
By the definition of symmetric closure:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$.
Here $\RR^{-1}$ is the inverse of $\RR$.
By Inverse of Relation Compatible with Operation is Compatible, $\RR^{-1}$ is compatible with $\circ$.
Thus by Union of Relations Compatible with Operation is Compatible:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$ is compatible with $\circ$.
{{qed}}
Category:Compatible Relations
Category:Symmetric Closures
\end{proof}
|
22489
|
\section{Symmetric Closure of Symmetric Relation}
Tags: Definitions: Relation Theory, Symmetric Relations, Symmetric Closures, Relation Theory
\begin{theorem}
Let $\RR$ be a symmetric relation on a set $S$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then:
:$\RR = \RR^\leftrightarrow$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \RR^\leftrightarrow
| r = \RR \cup \RR^{-1}
| c = {{Defof|Symmetric Closure}}
}}
{{eqn | r = \RR \cup \RR
| c = Inverse of Symmetric Relation is Symmetric
}}
{{eqn | r = \RR
| c = Union is Idempotent
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Relations
Category:Symmetric Closures
\end{proof}
|
22490
|
\section{Symmetric Difference is Subset of Union of Symmetric Differences}
Tags: Set Union, Symmetric Difference, Union
\begin{theorem}
Let $R, S, T$ be sets.
Then:
:$R \symdif S \subseteq \paren {R \symdif T} \cup \paren {S \symdif T}$
where $R \symdif S$ denotes the symmetric difference between $R$ and $S$.
\end{theorem}
\begin{proof}
From the definition of symmetric difference, we have:
:$R \symdif S = \paren {R \setminus S} \cup \paren {S \setminus R}$
Then from Set Difference is Subset of Union of Differences, we have:
:$R \setminus S \subseteq \paren {R \setminus T} \cup \paren {T \setminus S}$
:$S \setminus R \subseteq \paren {S \setminus T} \cup \paren {T \setminus R}$
Thus:
{{begin-eqn}}
{{eqn | l = \paren {R \setminus S} \cup \paren {S \setminus R}
| o = \subseteq
| r = \paren {R \setminus T} \cup \paren {T \setminus S} \cup \paren {S \setminus T} \cup \paren {T \setminus R}
| c = Set Union Preserves Subsets
}}
{{eqn | r = \paren {R \setminus T} \cup \paren {T \setminus R} \cup \paren {S \setminus T} \cup \paren {T \setminus S}
| c = Union is Commutative
}}
{{eqn | r = \paren {R \symdif T} \cup \paren {S \symdif T}
| c = {{Defof|Symmetric Difference}}
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Difference
Category:Set Union
\end{proof}
|
22491
|
\section{Symmetric Difference of Unions}
Tags: Set Union, Symmetric Difference, Union
\begin{theorem}
Let $R$, $S$ and $T$ be sets.
Then:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
where:
:$\symdif$ denotes the symmetric difference
:$\setminus$ denotes set difference
:$\cup$ denotes set union
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {R \cup T} \symdif \paren {S \cup T}
| r = \paren {\paren {R \cup T} \setminus \paren {S \cup T} } \cup \paren {\paren {S \cup T} \setminus \paren {R \cup T} }
| c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {\paren {R \setminus S} \setminus T} \cup \paren {\paren {S \setminus R} \setminus T}
| c = Set Difference with Union
}}
{{eqn | r = \paren {\paren {R \setminus S} \cup \paren {S \setminus R} } \setminus T
| c = Set Difference is Right Distributive over Union
}}
{{eqn | r = \paren {R \symdif S} \setminus T
| c = {{Defof|Symmetric Difference|index = 1}}
}}
{{end-eqn}}
{{Qed}}
Category:Symmetric Difference
Category:Set Union
\end{proof}
|
22492
|
\section{Symmetric Difference of Unions is Subset of Union of Symmetric Differences}
Tags: Set Union, Symmetric Difference, Union
\begin{theorem}
Let $I$ be an indexing set.
Let $S_\alpha, T_\alpha$ be sets, for all $\alpha \in I$.
Then:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}$
where $S \symdif T$ is the symmetric difference between $S$ and $T$.
\end{theorem}
\begin{proof}
From Difference of Unions is Subset of Union of Differences, we have:
:$\ds \bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha}$
:$\ds \bigcup_{\alpha \mathop \in I} T_i \setminus \bigcup_{\alpha \mathop \in I} S_\alpha \subseteq \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}$
where $\setminus$ denotes set difference.
Thus we have:
{{begin-eqn}}
{{eqn | l = \bigcup_{\alpha \mathop \in I} S_\alpha \symdif \bigcup_{\alpha \mathop \in I} T_\alpha
| r = \paren {\bigcup_{\alpha \mathop \in I} S_\alpha \setminus \bigcup_{\alpha \mathop \in I} T_\alpha} \cup \paren {\bigcup_{\alpha \mathop \in I} T_\alpha \setminus \bigcup_{\alpha \mathop \in I} S_\alpha}
| c = {{Defof|Symmetric Difference}}
}}
{{eqn | o = \subseteq
| r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \setminus T_\alpha} \cup \bigcup_{\alpha \mathop \in I} \paren {T_\alpha \setminus S_\alpha}
| c = Difference of Unions is Subset of Union of Differences
}}
{{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {\paren {S_\alpha \setminus T_\alpha} \cup \paren {T_\alpha \setminus S_\alpha} }
| c = Union is Associative and Union is Commutative
}}
{{eqn | r = \bigcup_{\alpha \mathop \in I} \paren {S_\alpha \symdif T_\alpha}
| c = {{Defof|Symmetric Difference}}
}}
{{end-eqn}}
{{qed}}
Category:Set Union
Category:Symmetric Difference
\end{proof}
|
22493
|
\section{Symmetric Difference on Power Set forms Abelian Group}
Tags: Abelian Groups, Symmetric Difference, Power Set, Group Theory
\begin{theorem}
Let $S$ be a set such that $\O \subset S$ (that is, $S$ is non-empty).
Let $A \symdif B$ be defined as the symmetric difference between $A$ and $B$.
Let $\powerset S$ denote the power set of $S$.
Then the algebraic structure $\struct {\powerset S, \symdif}$ is an abelian group.
\end{theorem}
\begin{proof}
From Power Set is Closed under Symmetric Difference, we have that $\struct {\powerset S, \symdif}$ is closed.
The result follows directly from Set System Closed under Symmetric Difference is Abelian Group.
{{Qed}}
\end{proof}
|
22494
|
\section{Symmetric Difference with Intersection forms Boolean Ring}
Tags: Set Intersection, Power Set, Symmetric Difference, Idempotent Rings
\begin{theorem}
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a Boolean ring.
\end{theorem}
\begin{proof}
From Symmetric Difference with Intersection forms Ring:
:$\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity.
From Intersection is Idempotent, $\cap$ is an idempotent operation on $S$.
Hence the result by definition of Boolean ring.
{{qed}}
\end{proof}
|
22495
|
\section{Symmetric Difference with Intersection forms Ring}
Tags: Symmetric Difference, Power Set, Intersection, Set Intersection, Examples of Rings with Unity, Examples of Commutative and Unitary Rings, Symmetric Difference with Intersection forms Ring, Commutative Algebra, Rings, Set Difference
\begin{theorem}
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cap$ denote the set intersection operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cap}$ is a commutative ring with unity, in which the unity is $S$.
This ring is not an integral domain.
\end{theorem}
\begin{proof}
* It has been established that $\left({\mathcal P \left({S}\right), *}\right)$ is an abelian group, where $\varnothing$ is the identity and each element is self-inverse.
* From Power Set with Intersection is a Monoid, we know that $\left({\mathcal P \left({S}\right), \cap}\right)$ is a commutative monoid whose identity is $S$.
* We have that Intersection Distributes over Symmetric Difference.
* Thus $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is a commutative ring with a unity which is $S$.
* Next we find that $\forall A \in \mathcal P \left({S}\right): A \cap \varnothing = \varnothing = \varnothing \cap A$. Thus $\varnothing$ is indeed the zero.
As set intersection is not cancellable, it follows that $\left({\mathcal P \left({S}\right), *, \cap}\right)$ is not an integral domain.
{{qed}}
\end{proof}
|
22496
|
\section{Symmetric Difference with Self is Empty Set}
Tags: Symmetric Difference, Empty Set
\begin{theorem}
The symmetric difference of a set with itself is the empty set:
:$S \symdif S = \O$
\end{theorem}
\begin{proof}
This follows directly from Symmetric Difference of Equal Sets:
:$S \symdif T = \O \iff S = T$
substituting $S$ for $T$.
{{Qed}}
\end{proof}
|
22497
|
\section{Symmetric Difference with Union does not form Ring}
Tags: Set Union, Symmetric Difference, Power Set, Examples of Rings
\begin{theorem}
Let $S$ be a set.
Let:
:$\symdif$ denote the symmetric difference operation
:$\cup$ denote the set union operation
:$\powerset S$ denote the power set of $S$.
Then $\struct {\powerset S, \symdif, \cup}$ does not form a ring.
\end{theorem}
\begin{proof}
For $\struct {S, \symdif, \cup}$ to be a ring, it is a necessary condition that $\cup$ be distributive over $*$.
Also, the identity element for set union and symmetric difference must be different.
However:
:$(1): \quad$ the identity for union and symmetric difference is $\O$ for both operations
:$(2): \quad$ set union is not distributive over symmetric difference:
From Symmetric Difference of Unions:
:$\paren {R \cup T} \symdif \paren {S \cup T} = \paren {R \symdif S} \setminus T$
The result follows.
{{qed}}
\end{proof}
|
22498
|
\section{Symmetric Group has Non-Normal Subgroup}
Tags: Symmetric Groups, Symmetry Group, Normal Subgroups, Symmetric Group, Symmetry Groups
\begin{theorem}
Let $S_n$ be the (full) symmetric group on $n$ elements, where $n \ge 3$.
Then $S_n$ contains at least one subgroup which is not normal.
\end{theorem}
\begin{proof}
Let $S_n$ act on the set $S$.
Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$.
As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$.
Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$.
$\rho$ can be described in cycle notation as $\paren {a \ b}$.
From Transposition is Self-Inverse it follows that $\set {e, \rho}$ is a subgroup of $S_n$.
Let $\pi$ be the permutation on $S$ described in cycle notation as $\paren {a \ b \ c}$.
By inspection it is found that $\pi^{-1} = \paren {a \ c \ b}$.
Then we have:
{{begin-eqn}}
{{eqn | l = \pi^{-1} \rho \pi
| r = \paren {a \ c \ b} \paren {a \ b} \paren {a \ b \ c}
| c =
}}
{{eqn | r = \paren {a \ c}
| c = by evaluation
}}
{{eqn | o = \notin
| r = \set {e, \rho}
| c =
}}
{{end-eqn}}
So, by definition, $\set {e, \rho}$ is not a normal subgroup.
{{qed}}
\end{proof}
|
22499
|
\section{Symmetric Group is Generated by Transposition and n-Cycle}
Tags: Examples of Generators of Groups, Symmetric Groups
\begin{theorem}
Let $n \in \Z: n > 1$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then the set of cyclic permutations:
:$\set {\begin {pmatrix} 1 & 2 \end{pmatrix}, \begin {pmatrix} 1 & 2 & \cdots & n \end{pmatrix} }$
is a generator for $S_n$.
\end{theorem}
\begin{proof}
Denote:
:$s = \begin {pmatrix} 1 & 2 \end{pmatrix}$
:$r = \begin {pmatrix} 1 & 2 & \cdots & n \end{pmatrix}$
By Cycle Decomposition of Conjugate,:
:$r s r^{-1} = r \begin {pmatrix} 1 & 2 \end{pmatrix} r^{-1} = \begin {pmatrix} \map r 1 & \map r 2 \end{pmatrix} = \begin {pmatrix} 2 & 3 \end{pmatrix}$.
By repeatedly using Cycle Decomposition of Conjugate:
:$r^2 s r^{-2} = \begin {pmatrix} 3 & 4 \end{pmatrix}$
:$r^3 s r^{-3} = \begin {pmatrix} 4 & 5 \end{pmatrix}$
:$\cdots$
:$r^{n - 2} s r^{-\paren {n - 2} } = \begin {pmatrix} n - 1 & n \end{pmatrix}$
The result then follows from Transpositions of Adjacent Elements generate Symmetric Group.
{{qed}}
\end{proof}
|
22500
|
\section{Symmetric Group is Group}
Tags: Symmetric Group is Group, Symmetric Groups, Symmetric Group, Group Examples
\begin{theorem}
Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group.
\end{theorem}
\begin{proof}
The fact that $\struct {S_n, \circ}$ is a group follows directly from group of permutations.
By definition of cardinality, as $\card T = n$ we can find a bijection between $T$ and $\N_n$.
From Number of Permutations, it is immediate that $\order {\paren {\Gamma \paren T, \circ} } = n! = \order {\struct {S_n, \circ} }$.
Again, we can find a bijection $\phi$ between $\struct {\Gamma \paren T, \circ}$ and $\struct {S_n, \circ}$.
The result follows directly from the Transplanting Theorem.
{{qed}}
\end{proof}
|
22501
|
\section{Symmetric Group is Subgroup of Monoid of Self-Maps}
Tags: Permutation Theory, Symmetric Group, Group of Permutations, Symmetric Groups
\begin{theorem}
Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself
Let $\struct {\Gamma \paren S, \circ}$ denote the symmetric group on $S$.
Let $\struct {S^S, \circ}$ be the monoid of self-maps under composition of mappings.
Then $\struct {\Gamma \paren S, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
\end{theorem}
\begin{proof}
By Symmetric Group is Group, $\struct {\Gamma \paren S, \circ}$ is a group.
Let $\phi \in \Gamma \paren S$ be a permutation on $S$.
As a permutation is a self-map, it follows that $\phi \in S^S$.
Thus by definition $\Gamma \paren S$ is a subset of $S^S$.
So by definition, $\Gamma \paren S$, is a subgroup of $\struct {S^S, \circ}$.
{{qed}}
Category:Symmetric Groups
Category:Permutation Theory
\end{proof}
|
22502
|
\section{Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3}
Tags: Dihedral Group D3, Examples of Group Isomorphisms, Symmetric Group on 3 Letters
\begin{theorem}
Let $S_3$ denote the Symmetric Group on 3 Letters.
Let $D_3$ denote the dihedral group $D_3$.
Then $S_3$ is isomorphic to $D_3$.
\end{theorem}
\begin{proof}
Consider $S_3$ as presented by its Cayley table:
{{:Symmetric Group on 3 Letters/Cayley Table}}
Consider $D_3$ as presented by its group presentation:
{{:Group Presentation of Dihedral Group D3}}
and its Cayley table:
{{:Dihedral Group D3/Cayley Table}}
Let $\phi: S_3 \to D_3$ be specified as:
{{begin-eqn}}
{{eqn | l = \map \phi {1 2 3}
| r = a
| c =
}}
{{eqn | l = \map \phi {2 3}
| r = b
| c =
}}
{{end-eqn}}
Then by inspection, we see:
{{begin-eqn}}
{{eqn | l = \map \phi {1 3 2}
| r = a^2
| c =
}}
{{eqn | l = \map \phi {1 3}
| r = a b
| c =
}}
{{eqn | l = \map \phi {1 2}
| r = a^2 b
| c =
}}
{{end-eqn}}
and the result follows.
{{qed}}
\end{proof}
|
22503
|
\section{Symmetric Group on Greater than 4 Letters is Not Solvable}
Tags: Symmetric Groups
\begin{theorem}
Let $n \in \N$ such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then $S_n$ is not a solvable group.
\end{theorem}
\begin{proof}
As stated, let $n > 4$ in the below.
Recall the definition of solvable group:
:A finite group $G$ is a '''solvable group''' {{iff}} it has a composition series in which each factor is a cyclic group.
:A '''composition series for $G$''' is a normal series for $G$ which has no proper refinement.
:A '''normal series''' for $G$ is a sequence of (normal) subgroups of $G$:
::$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
:where $G_{i - 1} \lhd G_i$ denotes that $G_{i - 1}$ is a proper normal subgroup of $G_i$.
Consider the alternating group on $n$ letters $A_n$.
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a proper normal subgroup of $S_n$.
But from Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group, the ''only'' normal subgroup of $S_n$ is $A_n$
From Alternating Group is Simple except on 4 Letters, $A_n$ is simple.
That is, $A_n$ has no proper normal subgroup.
It follows that the only composition series for $S_n$ is:
:$\set e \lhd A_n \lhd S_n$
From Alternating Group on More than 3 Letters is not Abelian, $A_n$ is not an abelian group.
From Cyclic Group is Abelian, $A_n$ is not cyclic.
Thus we have demonstrated that the only composition series for $S_n$ contains a factor which is not cyclic.
Hence the result.
{{qed}}
\end{proof}
|
22504
|
\section{Symmetric Group on n Letters is Isomorphic to Symmetric Group}
Tags: Symmetric Groups
\begin{theorem}
The symmetric group on $n$ letters $\struct {S_n, \circ}$ is isomorphic to the symmetric group on the $n$ elements of any set $T$ whose cardinality is $n$.
That is:
:$\forall T \subseteq \mathbb U, \card T = n: \struct {S_n, \circ} \cong \struct {\Gamma \paren T, \circ}$
\end{theorem}
\begin{proof}
The fact that $\struct {S_n, \circ}$ is a group is a direct implementation of the result Symmetric Group is Group.
By definition of cardinality, as $\card T = n$ we can find a bijection between $T$ and $\N_n$.
From Number of Permutations, it is immediate that $\order {\paren {\Gamma \paren T, \circ} } = n! = \order {\struct {S_n, \circ} }$.
Again, we can find a bijection $\phi$ between $\struct {\Gamma \paren T, \circ}$ and $\struct {S_n, \circ}$.
The result follows directly from the Transplanting Theorem.
{{qed}}
\end{proof}
|
22505
|
\section{Symmetric Groups of Same Order are Isomorphic}
Tags: Symmetric Groups of Same Order are Isomorphic, Symmetric Groups, Group Isomorphisms
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $T_1$ and $T_2$ be sets whose cardinality $\card {T_1}$ and $\card {T_2}$ are both $n$.
Let $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ be the symmetric group on $S$ and $T$ respectively.
Then $\struct {\map \Gamma {T_1}, \circ}$ and $\struct {\map \Gamma {T_2}, \circ}$ are isomorphic.
\end{theorem}
\begin{proof}
Consider the symmetric group on $n$ letters $S_n$.
From Symmetric Group on n Letters is Isomorphic to Symmetric Group we have that:
:$\struct {\Gamma \paren {T_1}, \circ}$ is isomorphic to $S_n$
:$\struct {\Gamma \paren {T_2}, \circ}$ is isomorphic to $S_n$
and hence from Isomorphism is Equivalence Relation:
:$\struct {\Gamma \paren {T_1}, \circ}$ is isomorphic to $\struct {\Gamma \paren {T_2}, \circ}$.
{{qed}}
\end{proof}
|
22506
|
\section{Symmetric Preordering is Equivalence Relation}
Tags: Preorder Theory, Equivalence Relations, Symmetric Relations
\begin{theorem}
Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be symmetric.
Then $\RR$ is an equivalence relation on $S$.
\end{theorem}
\begin{proof}
By definition, a preordering on $S$ is a relation on $S$ which is:
:$(1): \quad$ reflexive
and:
:$(2): \quad$ transitive.
Thus $\RR$ is a relation on $S$ which is reflexive, transitive and symmetric.
Thus by definition $\RR$ is an equivalence relation on $S$.
{{qed}}
\end{proof}
|
22507
|
\section{Symmetric Transitive and Serial Relation is Reflexive}
Tags: Reflexive Relations, Equivalence Relations, Symmetric Relations, Serial Relations, Transitive Relations, Relations
\begin{theorem}
Let $\RR$ be a relation which is:
:symmetric
:transitive
:serial.
Then $\RR$ is reflexive.
Thus such a relation is an equivalence.
\end{theorem}
\begin{proof}
Let $S$ be a set on which $\RR$ is a relation which is symmetric, transitive and serial.
As $\RR$ is symmetric:
:$x \mathrel \RR y \implies y \mathrel \RR x$
As $\RR$ is transitive:
:$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$
As $\RR$ is serial:
:$\forall x \in S: \exists y \in S: x \mathrel \RR y$
Let $x \in S$.
Then
{{begin-eqn}}
{{eqn | q = \exists y \in S
| l = \tuple {x, y}
| o = \in
| r = \RR
| c = as $\RR$ is serial
}}
{{eqn | ll= \leadsto
| l = \tuple {y, x}
| o = \in
| r = \RR
| c = as $\RR$ is symmetric
}}
{{eqn | ll= \leadsto
| l = \tuple {x, x}
| o = \in
| r = \RR
| c = as $\RR$ is transitive
}}
{{end-eqn}}
Thus:
:$\forall x: x \mathrel \RR x$
and by definition $\RR$ is reflexive.
It follows by definition that such a relation is an equivalence relation.
{{qed}}
\end{proof}
|
22508
|
\section{Symmetric and Antisymmetric Relation is Transitive}
Tags: Transitive Relations, Symmetric Relations
\begin{theorem}
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$ which is both symmetric and antisymmetric.
Then $\RR$ is transitive.
\end{theorem}
\begin{proof}
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
By Relation is Symmetric and Antisymmetric iff Coreflexive:
:$x = y, y = z$
and so trivially:
:$\tuple {x, z} = \tuple {x, x} \in \RR$
Thus $\RR$ is transitive.
{{qed}}
\end{proof}
|
22509
|
\section{Symmetry Group is Group}
Tags: Symmetry Groups, Group Examples
\begin{theorem}
Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
The symmetry group $\struct {S_P, \circ}$ is indeed a group.
\end{theorem}
\begin{proof}
By definition, a symmetry mapping is a bijection, and hence a permutation.
From Symmetric Group is Group, the set of all permutations on $P$ form the symmetric group $\struct {\map \Gamma P, \circ}$ on $P$.
Thus $S_P$ is a subset of $\struct {\map \Gamma P, \circ}$.
Let $A$ and $B$ be symmetry mappings on $P$.
From Composition of Symmetries is Symmetry, $A \circ B$ is also a symmetry mapping on $P$.
Also, by the definition of symmetry, $A^{-1}$ is also a symmetry mapping on $P$.
Thus we have:
:$A, B \in S_P \implies A \circ B \in S_P$
:$A \in S_P \implies A^{-1} \in \S_P$
The result follows from the Two-Step Subgroup Test.
{{qed}}
\end{proof}
|
22510
|
\section{Symmetry Group of Equilateral Triangle is Group}
Tags: Symmetric Group, Symmetry Group of Equilateral Triangle, Symmetry Groups, Group Examples
\begin{theorem}
The symmetry group of the equilateral triangle is a group.
\end{theorem}
\begin{proof}
Let us refer to this group as $D_3$.
Taking the group axioms in turn:
\end{proof}
|
22511
|
\section{Symmetry Group of Line Segment is Group}
Tags: Symmetry Group of Line Segment
\begin{theorem}
The symmetry group of the line segment is a group.
\end{theorem}
\begin{proof}
Let us refer to this group as $D_1$.
Taking the group axioms in turn:
\end{proof}
|
22512
|
\section{Symmetry Group of Rectangle is Klein Four-Group}
Tags: Klein Four-Group, Symmetry Group of Rectangle
\begin{theorem}
The symmetry group of the rectangle is the Klein $4$-group.
\end{theorem}
\begin{proof}
Comparing the Cayley tables of the symmetry group of the rectangle with the Klein $4$-group the isomorphism can be seen:
{{:Symmetry Group of Rectangle/Cayley Table}}
{{:Klein Four-Group/Cayley Table}}
Thus the required isomorphism $\phi$ can be set up as:
{{begin-eqn}}
{{eqn | l = \map \phi e
| r = e
}}
{{eqn | l = \map \phi r
| r = a
}}
{{eqn | l = \map \phi h
| r = b
}}
{{eqn | l = \map \phi v
| r = c
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22513
|
\section{Symmetry Group of Square is Group}
Tags: Symmetry Groups: Examples, Examples of Symmetry Groups, Groups of Order 8, Symmetry Group of Square, Symmetry Group Examples, Symmetry Groups
\begin{theorem}
The symmetry group of the square is a non-abelian group.
\end{theorem}
\begin{proof}
Let us refer to this group as $D_4$.
Taking the group axioms in turn:
\end{proof}
|
22514
|
\section{Symmetry Rule for Binomial Coefficients}
Tags: Discrete Mathematics, Symmetry Rule for Binomial Coefficients, Binomial Coefficients
\begin{theorem}
Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$
where $\dbinom n k$ is a binomial coefficient.
\end{theorem}
\begin{proof}
Follows directly from the definition, as follows.
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k > 0$.
In both cases $\displaystyle \binom n k = \binom n {n - k} = 0$.
Let $0 \le k \le n$.
{{begin-eqn}}
{{eqn | l=\binom n k
| r=\frac {n!} {k! \ \left({n - k}\right)!}
| c=
}}
{{eqn | r=\frac {n!} {\left({n - k}\right)! \ k!}
| c=
}}
{{eqn | r=\frac {n!} {\left({n - k}\right)! \ \left ({n - \left({n - k}\right)}\right)!}
| c=
}}
{{eqn | r=\binom n {n - k}
| c=
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22515
|
\section{Symmetry Rule for Binomial Coefficients/Complex Numbers}
Tags: Symmetry Rule for Binomial Coefficients
\begin{theorem}
For all $z, w \in \C$ such that it is not the case that $z$ is a negative integer and $w$ an integer:
:$\dbinom z w = \dbinom z {z - w}$
where $\dbinom z w$ is a binomial coefficient.
\end{theorem}
\begin{proof}
From the definition of the binomial coefficient:
:$\dbinom z w := \ds \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\zeta - \omega + 1} }$
where $\Gamma$ denotes the Gamma function.
{{begin-eqn}}
{{eqn | l = \dbinom z w
| r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\omega + 1} \map \Gamma {\zeta - \omega + 1} }
| c =
}}
{{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\zeta - \omega + 1} \map \Gamma {\omega + 1} }
| c =
}}
{{eqn | r = \lim_{\zeta \mathop \to z} \lim_{\omega \mathop \to w} \dfrac {\map \Gamma {\zeta + 1} } {\map \Gamma {\zeta - \omega + 1} \map \Gamma {\zeta - \paren {\zeta - \omega} + 1} }
| c =
}}
{{eqn | r = \dbinom z {z - w}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
22516
|
\section{Symmetry Rule for Gaussian Binomial Coefficients}
Tags: Symmetry Rule for Binomial Coefficients, Gaussian Binomial Coefficients
\begin{theorem}
Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = \dbinom n {n - k}_q$
where $\dbinom n k_q$ is a Gaussian binomial coefficient.
\end{theorem}
\begin{proof}
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k_q = \dbinom n {n - k}_q = 0$
Let $0 \le k \le n$.
Consider the case $k \le \dfrac n 2$.
Then $k \le n - k$.
{{begin-eqn}}
{{eqn | l = \binom n {n - k}_q
| r = \prod_{j \mathop = 0}^{\paren {n - k} - 1} \frac {1 - q^{n - j} } {1 - q^{j + 1} }
| c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{eqn | r = \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {\paren {n - k} - 1} } } {1 - q^{\paren {\paren {n - k} - 1} + 1} } }
| c =
}}
{{eqn | r = \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {k - 1} } } {1 - q^{\paren {k - 1} + 1} } } \paren {\frac {1 - q^{n - k} } {1 - q^{k + 1} } } \cdots \paren {\frac {1 - q^{k + 1} } {1 - q^{n - k} } }
| c =
}}
{{eqn | r = \paren {\frac {1 - q^{n - 0} } {1 - q^{0 + 1} } } \paren {\frac {1 - q^{n - 1} } {1 - q^{1 + 1} } } \paren {\frac {1 - q^{n - 2} } {1 - q^{2 + 1} } } \cdots \paren {\frac {1 - q^{n - \paren {k - 1} } } {1 - q^{\paren {k - 1} + 1} } }
| c = The tail cancels out
}}
{{eqn | r = \binom n k_q
| c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{end-eqn}}
The case $k \ge \dfrac n 2$ can be done by observing:
:$n - k \le \dfrac n 2$
and hence by the above:
:$\dbinom n k_q = \dbinom n {n - \paren {n - k} }_q = \dbinom n {n - k}_q$
{{qed}}
\end{proof}
|
22517
|
\section{Symmetry in Space Implies Conservation of Momentum}
Tags: Laws of Conservation
\begin{theorem}
The total derivative of the action $S_{12}$ from states $1$ to $2$ with regard to position is equal to the difference in momentum from states $1$ to $2$:
:$\dfrac {\d S_{1 2} } {\d x} = p_2 - p_1$
{{MissingLinks|Although we do have a page Definition:State, it refers to a concept in game theory and not physics.}}
\end{theorem}
\begin{proof}
From the definition of generalized momentum and the Euler-Lagrange Equations:
{{begin-eqn}}
{{eqn | l = 0
| r = \frac \d {\d t} \frac {\partial \LL} {\partial \dot x} - \frac {\partial \LL} {\partial x}
| c =
}}
{{eqn | r = \dot p_i - \frac {\partial \LL} {\partial x}
| c =
}}
{{eqn | ll= \leadsto
| l = \dot p_i
| r = \frac {\partial \LL} {\partial x}
| c =
}}
{{end-eqn}}
Therefore, via the definition of action, Definite Integral of Partial Derivative and the Fundamental Theorem of Calculus:
{{begin-eqn}}
{{eqn | l = \frac {\d S_{12} } {\d x}
| r = \frac \d {\d x} \int_{t_1}^{t_2} \LL \rd t
| c =
}}
{{eqn | r = \int_{t_1}^{t_2} \frac {\partial \LL} {\partial x} \rd t
| c =
}}
{{eqn | r = \int_{t_1}^{t_2} \dot p_i \rd t
| c =
}}
{{eqn | r = p_2 - p_1
| c =
}}
{{end-eqn}}
{{qed}}
Category:Laws of Conservation
\end{proof}
|
22518
|
\section{Symmetry of Bernoulli Polynomial}
Tags: Bernoulli Polynomials
\begin{theorem}
Let $\map {B_n} x$ denote the nth Bernoulli polynomial.
Then:
:$\map {B_n} {1 - x} = \paren {-1}^n \map {B_n} x$
\end{theorem}
\begin{proof}
Let $\map G {t, x}$ denote the Generating Function of Bernoulli Polynomials:
:$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map G {t, 1 - x}
| r = \frac {t e^{t \paren {1 - x} } } {e^t - 1}
}}
{{eqn | r = \frac {t e^{t - t x} } {e^t - 1}
}}
{{eqn | r = \frac {t e^{-t x} } {1 - e^{-t} }
}}
{{eqn | r = \frac {-t e^{-t x} } {e^{-t} - 1}
}}
{{eqn | r = \map G {-t, x}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \map G {t, 1 - x}
| r = \map G {-t, x}
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} {1 - x} } {k!} t^k
| r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \map {B_k} x} {k!} t^k
}}
{{eqn | ll= \leadsto
| l = \map {B_k} {1 - x}
| r = \paren {-1}^k \map {B_k} x
}}
{{end-eqn}}
{{qed}}
Category:Bernoulli Polynomials
\end{proof}
|
22519
|
\section{Symmetry of Relations is Symmetric}
Tags: Relations, Symmetric Relations
\begin{theorem}
Let $\RR$ be a relation on $S$ which is symmetric. Then:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$.
\end{theorem}
\begin{proof}
Let $\RR$ be symmetric.
{{begin-eqn}}
{{eqn | l = \tuple {x, y} \in \RR
| o = \implies
| r = \tuple {y, x} \in \RR
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | l = \tuple {y, x} \in \RR
| o = \implies
| r = \tuple {x, y} \in \RR
| c = {{Defof|Symmetric Relation}}
}}
{{eqn | ll= \leadsto
| l = \leftparen {\tuple {x, y} \in \RR}
| o = \iff
| r = \rightparen {\tuple {y, x} \in \RR}
| c = {{Defof|Biconditional}}
}}
{{end-eqn}}
{{qed}}
Category:Symmetric Relations
\end{proof}
|
22520
|
\section{Syndrome is Zero iff Vector is Codeword}
Tags: Linear Codes
\begin{theorem}
Let $C$ be a linear $\tuple {n, k}$-code whose master code is $\map V {n, p}$
Let $G$ be a (standard) generator matrix for $C$.
Let $P$ be a standard parity check matrix for $C$.
Let $w \in \map V {n, p}$.
Then the syndrome of $w$ is zero {{iff}} $w$ is a codeword of $C$.
\end{theorem}
\begin{proof}
Let $G = \paren {\begin{array} {c|c} \mathbf I & \mathbf A \end{array} }$.
Let $c \in \map V {n, p}$.
Then, by definition of $G$, $c$ is a codeword of $C$ {{iff}} $c$ is of the form $u G$, where $u \in \map V {k, p}$.
Thus $c \in C$ {{iff}}:
{{begin-eqn}}
{{eqn | l = c
| r = u G
| c =
}}
{{eqn | r = u \paren {\begin{array} {c {{!}} c} \mathbf I & \mathbf A \end{array} }
| c =
}}
{{eqn | r = \paren {\begin{array} {c {{!}} c} u & v \end{array} }
| c =
}}
{{end-eqn}}
where:
:$v = u \mathbf A$
:$\paren {\begin{array} {c|c} u & v \end{array} }$ denotes the $1 \times n$ matrix formed from the $k$ elements of $u$ and the $n - k$ elements of $v$.
Let $w \in \map V {n, p}$.
$w$ can be expressed in the form:
:$w = \paren {\begin{array} {c|c} u_1 & v_1 \end{array} }$
where $u_1 \in \map V {k, p}$.
The syndrome of $v$ is then calculated as:
{{begin-eqn}}
{{eqn | l = \map S v
| r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } w^\intercal
| c =
}}
{{eqn | r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } \paren {\begin{array} {c {{!}} c} u_1^\intercal & v_1^\intercal \end{array} }
| c =
}}
{{eqn | r = -\mathbf A^\intercal u_1^\intercal + v_1^\intercal
| c =
}}
{{end-eqn}}
It follows that the syndrome of $w$ is zero {{iff}} $w$ is the concatenation of $u_1$ and $v_1$, where:
:$v_1^\intercal = \mathbf A^\intercal u_1^\intercal = \paren {u_1 \mathbf A}^\intercal$
Thus the syndrome of $w$ is zero {{iff}} $w$ is a codeword of $C$.
{{qed}}
\end{proof}
|
22521
|
\section{Synthetic Basis formed from Synthetic Sub-Basis}
Tags: Sub-Bases, Topology, Topological Bases
\begin{theorem}
Let $X$ be a set.
Let $\SS$ be a synthetic sub-basis on $X$.
Define:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a synthetic basis on $X$.
\end{theorem}
\begin{proof}
We consider $X$ as the universe.
Thus, in accordance with Intersection of Empty Set, we take the convention that:
:$\ds \bigcap \O = X \in \BB$
By Set is Subset of Union: General Result, it follows that:
:$\ds X \subseteq \bigcup \BB$
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied.
We have that $\BB \subseteq \powerset X$.
Let $B_1, B_2 \in \BB$.
Then there exist finite $\FF_1, \FF_2 \subseteq \SS$ such that:
:$\ds B_1 = \bigcap \FF_1$
:$\ds B_2 = \bigcap \FF_2$
It follows that:
:$\ds B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$
{{explain|proof that $\ds \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\ds \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$}}
By Union is Smallest Superset, $\FF_1 \cup \FF_2 \subseteq \SS$.
We have that $\FF_1 \cup \FF_2$ is finite.
Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom $(\text B 2)$ for a synthetic basis is satisfied.
{{qed}}
\end{proof}
|
22522
|
\section{System of Simultaneous Equations may have Multiple Solutions}
Tags: Simultaneous Equations
\begin{theorem}
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.
\end{theorem}
\begin{proof}
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
| l = x_1 - 2 x_2 + x_3
| r = 1
}}
{{eqn | n = 2
| l = 2 x_1 - x_2 + x_3
| r = 2
}}
{{end-eqn}}
From its evaluation it has the following solutions:
{{begin-eqn}}
{{eqn | l = x_1
| r = 1 - \dfrac t 3
}}
{{eqn | l = x_2
| r = \dfrac t 3
}}
{{eqn | l = x_3
| r = t
}}
{{end-eqn}}
where $t$ is any number.
Hence the are as many solutions as the cardinality of the domain of $t$.
{{qed}}
\end{proof}
|
22523
|
\section{System of Simultaneous Equations may have No Solution}
Tags: Simultaneous Equations
\begin{theorem}
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is empty.
\end{theorem}
\begin{proof}
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
| l = x_1 + x_2
| r = 2
}}
{{eqn | n = 2
| l = 2 x_1 + 2 x_2
| r = 3
}}
{{end-eqn}}
From its evaluation it is seen to have no solutions.
Hence the result.
{{qed}}
\end{proof}
|
22524
|
\section{System of Simultaneous Equations may have Unique Solution}
Tags: Simultaneous Equations
\begin{theorem}
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.
\end{theorem}
\begin{proof}
Consider this system of simultaneous linear equations:
{{begin-eqn}}
{{eqn | n = 1
| l = x_1 - 2 x_2 + x_3
| r = 1
}}
{{eqn | n = 2
| l = 2 x_1 - x_2 + x_3
| r = 2
}}
{{eqn | n = 3
| l = 4 x_1 + x_2 - x_3
| r = 1
}}
{{end-eqn}}
From its evaluation it has the following unique solution:
{{begin-eqn}}
{{eqn | l = x_1
| r = -\dfrac 1 2
}}
{{eqn | l = x_2
| r = \dfrac 1 2
}}
{{eqn | l = x_3
| r = \dfrac 3 2
}}
{{end-eqn}}
Hence the result.
{{qed}}
\end{proof}
|
22525
|
\section{Szpilrajn Extension Theorem}
Tags: Order Theory
\begin{theorem}
Let $\struct {S, \prec}$ be a strictly ordered set.
{{Disambiguate|Definition:Strictly Ordered Set}}
Then there is a strict total ordering on $S$ of which $\prec$ is a subset.
\end{theorem}
\begin{proof}
{{proof wanted}}
{{Namedfor|Edward Szpilrajn|cat = Marczewski}}
Category:Order Theory
\end{proof}
|
22526
|
\section{T0 Property is Hereditary}
Tags: Topological Subspaces, T0 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_0$ (Kolmogorov) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_0$ (Kolmogorov) space.
\end{theorem}
\begin{proof}
Let $T$ be a $T_0$ (Kolmogorov) space.
That is:
:$\forall x, y \in S$ such that $x \ne y$, either:
::$\exists U \in \tau: x \in U, y \notin U$
:or:
::$\exists U \in \tau: y \in U, x \notin U$
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Then either:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
or:
:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
and so the $T_0$ axiom is satisfied.
{{qed}}
\end{proof}
|
22527
|
\section{T0 Space is Preserved under Closed Bijection}
Tags: Separation Axioms, T0 Spaces
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
\end{theorem}
\begin{proof}
Let $T_A$ be a $T_0$ (Kolmogorov) space.
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
By definition:
:$\forall x, y \in S_A$, either:
::$\exists U \in \tau_A: x \in U, y \notin U$
:or:
::$\exists U \in \tau_A: y \in U, x \notin U$
Suppose that:
:$\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$
That is:
:$\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$
From Image of Subset under Relation is Subset of Image: Corollary 3 it follows that:
:$\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$
that is:
:$\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$
But by definition of continuous mapping, $U = \phi^{-1} \sqbrk V$ is open in $T_A$ {{iff}} $V$ is open in $T_B$.
Thus:
:$\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$
This contradicts the condition that $T_A$ is a $T_0$ (Kolmogorov) space.
It follows that $T_B$ must also be a $T_0$ (Kolmogorov) space.
{{qed}}
\end{proof}
|
22528
|
\section{T0 Space is Preserved under Homeomorphism}
Tags: Separation Axioms, Homeomorphisms, T0 Spaces
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
\end{theorem}
\begin{proof}
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_0$ (Kolmogorov) Space is Preserved under Closed Bijection.
{{qed}}
\end{proof}
|
22529
|
\section{T1/2 Space is T0 Space}
Tags: T1: 2 Spaces, T0 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a $T_{\frac 1 2}$ topological space.
Then $T$ is $T_0$ space.
\end{theorem}
\begin{proof}
By Characterization of T0 Space by Closures of Singletons it suffices to prove that
:$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
where $\set y^-$ denotes the closure of $\set y$.
Let $x, y$ be points of $T$ such that:
:$x \ne y$
{{AimForCont}}:
:$x \in \set y^- \land y \in \set x^-$
We will prove that:
:$x \notin \set x'$
where $\set x'$ denotes the derivative of $\set x$.
{{AimForCont}}:
:$x \in \set x'$
As $S$ is open by definition of topological space
by Characterization of Derivative by Open Sets:
:$\exists z \in S: z \in \set x \cap S \land z \ne x$
By definition of intersection:
:$z \in \set x$
This contradicts $z \ne x$ by definition of singleton.
Thus $x \notin \set x'$.
We will prove that:
:$(1): \quad \lnot \forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
{{AimForCont}}:
:$\forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
As sublemma we will show that:
:$\forall U \in \tau: y \in U \implies \exists r \in S: r \in \set x \cap U \land y \ne r$
Let $U \in \tau$ such that:
:$y \in U$
Then by assumption:
:$\set x \cap U \ne \O$
By definition of empty set:
:$\exists z: z \in \set x \cap U$
By definition of intersection:
:$z \in \set x$
Then by definition of singleton:
:$z = x \ne y$
Thus:
:$\exists r \in S: r \in \set x \cap U \land y \ne r$
Then by Characterization of Derivative by Open Sets:
:$y \in \set x'$
By definition of relative complement:
:$y \notin \relcomp S {\set x'} \land x \in \relcomp S {\set x'}$
By definition of $T_{\frac 1 2}$ space:
:$\set x'$ is closed
By definition of closed set:
:$\relcomp S {\set x'}$ is open
By $x \in \set y^-$ and by Condition for Point being in Closure:
:$\set y \cap \relcomp S {\set x'} \ne \O$
Then by definition of empty set:
:$\exists z: z \in \set y \cap \relcomp S {\set x'}$
By definition of intersection:
:$z \in \set y \land z \in \relcomp S {\set x'}$
By definition of singleton:
:$z = y$
Thus $z \in \relcomp S {\set x'}$ contradicts $y \notin \relcomp S {\set x'}$.
This ends the proof of $(1)$.
Then by $(1)$ and Condition for Point being in Closure:
:$y \notin \set x^-$
This contradicts $y \in \set x^-$.
{{qed}}
\end{proof}
|
22530
|
\section{T1 Property is Hereditary}
Tags: T1 Spaces, Topological Subspaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_1$ (Fréchet) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
Let $T$ be a $T_1$ (Fréchet) space.
That is:
:$\forall x, y \in S$ such that $x \ne y$, both:
::$\exists U \in \tau: x \in U, y \notin U$
:and:
::$\exists U \in \tau: y \in U, x \notin U$
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$:
:$\exists U \in \tau: x \in U, y \notin U$
and:
:$\exists U \in \tau: y \in U, x \notin U$
Then both:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
and:
:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
and so the $T_1$ axiom is satisfied.
{{qed}}
\end{proof}
|
22531
|
\section{T1 Space is Preserved under Closed Bijection}
Tags: T1 Spaces, Separation Axioms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
\end{theorem}
\begin{proof}
Let $T_A$ be a $T_1$ (Fréchet) space.
By definition, all points in $T_A$ are closed.
Let $a \in S_A$.
Then $\set a$ is a closed set.
As $\phi$ is a closed mapping it follows directly that $\phi \sqbrk {\set a}$ is closed.
As $\phi$ is a bijection it follows that every point in $S_B$ is the image under $\phi$ of a single point in $S_A$.
Hence every point in $S_B$ is closed.
That is, $T_B$ is a $T_1$ (Fréchet) space.
{{qed}}
\end{proof}
|
22532
|
\section{T1 Space is Preserved under Homeomorphism}
Tags: T1 Spaces, Separation Axioms, Homeomorphisms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
\end{theorem}
\begin{proof}
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_1$ (Fréchet) Space is Preserved under Closed Bijection.
{{qed}}
\end{proof}
|
22533
|
\section{T1 Space is T0 Space}
Tags: T1 Spaces, Separation Axioms, T0 Spaces
\begin{theorem}
Let $\struct {S, \tau}$ be a Fréchet ($T_1$) space.
Then $\struct {S, \tau}$ is also a Kolmogorov ($T_0$) space.
\end{theorem}
\begin{proof}
Let $\struct {S, \tau}$ be a $T_1$ space.
Let $x, y \in S: x \ne y$.
From the definition of $T_1$ space:
:'''Both'''
::$\exists U \in \tau: x \in U, y \notin U$
:'''and''':
::$\exists V \in \tau: y \in V, x \notin V$
From the Rule of Simplification:
: $\exists U \in \tau: x \in U, y \notin U$
From the Rule of Addition:
:'''Either'''
::$\exists U \in \tau: x \in U, y \notin U$
:'''or''':
::$\exists V \in \tau: y \in V, x \notin V$
which is precisely the definition of a Kolmogorov ($T_0$) space.
{{qed}}
\end{proof}
|
22534
|
\section{T1 Space is T1/2 Space}
Tags: T1 Spaces, T1: 2 Spaces
\begin{theorem}
Let $T$ be a $T_1$ topological space.
Then $T$ is $T_{\frac 1 2}$ space.
\end{theorem}
\begin{proof}
By Closure of Derivative is Derivative in T1 Space:
:$\forall A \subseteq T: \left({A'}\right)^- = A'$
where
:$A'$ denotes the derivative of $A$
:$\left({A'}\right)^-$ denotes the closure of $A'$
Then by Topological Closure is Closed:
:$\forall A \subseteq T: A'$ is closed
Thus by definition:
:$T$ is $T_{\frac 1 2}$ space
{{qed}}
\end{proof}
|
22535
|
\section{T2 Property is Hereditary}
Tags: Hausdorff Spaces, Topological Subspaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_2$ (Hausdorff) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_2$ (Hausdorff) space.
That is, the property of being a $T_2$ (Hausdorff) space is hereditary.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
That is, for any two distinct elements $x, y \in S$ there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U, V \in \tau: x \in U, y \in V, U \cap V = \O$
As $x, y \in H$ we have that:
:$x \in U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_2$ axiom is satisfied in $H$.
{{qed}}
\end{proof}
|
22536
|
\section{T2 Space is Preserved under Closed Bijection}
Tags: Hausdorff Spaces, Separation Axioms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_2$ (Hausdorff) space, then so is $T_B$.
\end{theorem}
\begin{proof}
Let $T_A$ be a $T_2$ (Hausdorff) space.
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$
Suppose that $T_B$ is not Hausdorff.
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \O$
That is, there exists at least one pair of points $a$ and $b$ for which all the open sets containing $a$ and $b$ are not disjoint.
As $\tau_B$ is a topology, it follows that $W_B = U_B \cap V_B$ is also an open set.
Let $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$.
From Preimage of Intersection under Mapping, we have:
:$\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$
that is:
:$U_A \cap V_A = W_A$
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
As $\phi$ is continuous, all of $U_A, V_A, W_A$ are open in $T_A$.
From the bijective nature of $\phi$, we have that:
:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
Thus we have that:
:$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \O$
contradicting the fact that $T_A$ is a $T_2$ (Hausdorff) space.
Hence $T_B$ must after all be a $T_2$ (Hausdorff) space.
{{qed}}
\end{proof}
|
22537
|
\section{T2 Space is T1 Space}
Tags: Hausdorff Spaces, T1 Spaces, Separation Axioms, Definitions: Separation Axioms
\begin{theorem}
Let $\struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Then $\struct {S, \tau}$ is also a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
From the definition of $T_2$ (Hausdorff) space:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
As $U \cap V = \O$ it follows from the definition of disjoint sets that:
:$x \in U \implies x \notin V$
:$y \in V \implies y \notin U$
So if $x \in U, y \in V$ then:
:$\exists U \in \tau: x \in U, y \notin U$
:$\exists V \in \tau: y \in V, x \notin V$
which is precisely the definition of a $T_1$ (Fréchet) space.
{{qed}}
\end{proof}
|
22538
|
\section{T3 1/2 Property is Hereditary}
Tags: T3 1: 2 Spaces, Topological Subspaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_{3 \frac 1 2}$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_{3 \frac 1 2}$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
Then:
:For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_{3 \frac 1 2}$ space, we have that there exists an Urysohn function for $F$ and $\set y$:
That is, there exists a continuous mapping $f: S \to \closedint 0 1$, where $\closedint 0 1$ is the closed unit interval, such that:
:$f {\restriction_F} = 0, f {\restriction_{\set y} } = 1$
where $f {\restriction_F}$ denotes the restriction of $f$ to $F$.
That is:
:$\forall a \in F: \map f a = 0$
:$\forall b \in \set y: \map f b = 1$
From Continuity of Composite with Inclusion, as $f$ is continuous on $S$, then $f {\restriction_H}$ is continuous on $H$.
Thus $f {\restriction_H}$ is an Urysohn function for $F$ and $\set y$ in $H$.
So the $T_{3 \frac 1 2}$ axiom is satisfied in $H$.
{{qed}}
\end{proof}
|
22539
|
\section{T3 1/2 Space is Preserved under Homeomorphism}
Tags: T3 1: 2 Spaces, Separation Axioms, Homeomorphisms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
\end{theorem}
\begin{proof}
Let $F \subseteq S_B$ be closed, and let $y \in S_B$ such that $y \notin F$.
Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.
Since $T_A$ is a $T_{3 \frac 1 2}$ space, there exists an Urysohn function $f: X_A \to \closedint 0 1$ for $G$ and $\set z$.
Define $g: S_B \to \closedint 0 1$ by:
:$\map g x = \map f {\map {\phi^{-1} } x}$
By Composite of Continuous Mappings is Continuous, $g$ is continuous.
Also, for all $x \in F$:
:$\map g x = 0$
as $\map {\phi^{-1} } x \in G$.
Similarly, because $\map {\phi^{-1} } y = z$:
:$\map g y = 1$
Hence $g$ is an Urysohn function for $F$ and $\set y$.
Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a $T_{3 \frac 1 2}$ space.
{{qed}}
\end{proof}
|
22540
|
\section{T3 1/2 Space is T3 Space}
Tags: T3 1: 2 Spaces, T3 Spaces, Separation Axioms
\begin{theorem}
Let $T$ be a $T_{3 \frac 1 2}$ space.
Then $T$ is also a $T_3$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.
From the definition of $T_{3 \frac 1 2}$ space:
:For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$.
Let $F \subseteq S$ be a closed set in $T$ and let $y \in \relcomp S F$.
An Urysohn function for $F$ and $\set y$ is a continuous mapping $f: S \to \closedint 0 1$ where:
:$\forall a \in F: \map f a = 0$
:$\map f y = 1$
Let:
:$U = \map {f^{-1} } 0$
:$V = \map {f^{-1} } 1$
As $f$ is continuous, both $U$ and $V$ are open in $T$.
Suppose $x \in U \cap V$.
Then we would have:
:$x \in U \implies \map f x = 0$
:$x \in V \implies \map f x = 1$
which contradicts the many-to-one nature of a mapping.
So $U \cap V = \O$.
Thus we have:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
which is precisely the definition of a $T_3$ space.
{{qed}}
\end{proof}
|
22541
|
\section{T3 1/2 Space is not necessarily T2 Space}
Tags: Hausdorff Spaces, T3 1: 2 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a be a $T_{3 \frac 1 2}$ space.
Then it is not necessarily the case that $T$ is a $T_2$ (Hausdorff) space.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $S$ be a set and let $\PP$ be a partition on $S$ which is specifically not the (trivial) partition of singletons.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
From Partition Topology is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
From Partition Topology is not Hausdorff, $T$ is not a $T_2$ (Hausdorff) space.
The result follows.
{{qed}}
\end{proof}
|
22542
|
\section{T3 Property is Hereditary}
Tags: T3 Spaces, Topological Subspaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_3$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_3$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $F \subseteq H$ such that $F$ is closed in $H$.
Let $y \in H$ such that $y \notin F$.
From Closed Set in Topological Subspace $F$ is also closed in $T$.
Because $T$ is a $T_3$ space, we have that:
:$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$
As $F \subseteq H$ and $y \in H, y \notin F$ we have that:
:$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_3$ axiom is satisfied in $H$.
{{qed}}
\end{proof}
|
22543
|
\section{T3 Space is Preserved under Homeomorphism}
Tags: T3 Spaces, Separation Axioms, Homeomorphisms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_3$ space, then so is $T_B$.
\end{theorem}
\begin{proof}
Suppose that $T_A$ is a $T_3$ space.
Let $F$ be closed in $T_B$.
Let $y \in S_B$ such that $y \notin F$.
From Preimage of Intersection under Mapping it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are disjoint.
Also, as $\phi$ is a homeomorphism, it is a fortiori continuous.
Thus by Continuity Defined from Closed Sets, $\phi^{-1} \sqbrk F$ is closed.
Now as $T_A$ is a $T_3$ space, we find disjoint open sets $U_1$ containing $\phi^{-1} \sqbrk F$ and $U_2$ containing $\map {\phi^{-1}} y$.
From Image of Subset is Subset of Image, we have $F = \phi \sqbrk {\phi^{-1} \sqbrk F} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Mutatis mutandis, we deduce also $\set y \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $F$, and the other containing $y$.
Hence $T_B$ is shown to be a $T_3$ space as well.
{{qed}}
\end{proof}
|
22544
|
\section{T3 Space is Semiregular}
Tags: Regular Open Sets, T3 Spaces, Separation Axioms, Semiregular Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a $T_3$ space.
Then $T$ is a semiregular space.
\end{theorem}
\begin{proof}
Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$.
In other words, $\BB$ is the collection of regular open sets contained in $S$.
It follows immediately from the definition of a basis that our theorem is proved if we can show that:
:$\BB$ is a cover for $S$
:$\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$
First we show that $\BB$ is a cover for $S$.
Since $S$ is open:
:$S^\circ = S$
Since $S$ is closed:
:$S^- = S$
Therefore:
:$S^{- \circ} = S$
and so:
:$S \in \BB$
From Set is Subset of Union:
:$S \subseteq \bigcup \BB$
Thus $\BB$ covers $S$.
{{qed|lemma}}
Next we demonstrate the second condition for $\BB$ to be a basis.
Let $U, V \in \BB$, $x \in U \cap V$.
We will show that for some $W \in \BB$:
:$x \in W \subseteq U \cap V$
By General Intersection Property of Topological Space, $U \cap V \in \tau$.
Since $T$ is $T_3$, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$:
:$\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$
Let $W := {N_x}^{- \circ}$.
Then by Interior of Closure is Regular Open, $W^{- \circ} = W$ and $W \in \BB$.
On the other hand, since $N_x$ is closed:
:${N_x}^- = N_x$, and thus $W = {N_x}^\circ$
By definition of interior, we have $Q \subseteq W \subseteq N_x$.
Therefore we have $x \in W \subseteq N_x \subseteq U \cap V$, as desired.
It follows that $\BB$ is a basis for $\tau$.
Hence the result by definition of semiregular space.
{{qed}}
\end{proof}
|
22545
|
\section{T4 Property Preserved in Closed Subspace}
Tags: Topological Subspaces, Normal Spaces, Separation Axioms, Closed Sets, T4 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_K$ be a subspace of $T$ such that $K$ is closed in $T$.
If $T$ is a $T_4$ space then $T_K$ is also a $T_4$ space.
That is, the property of being a $T_4$ space is weakly hereditary.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_4$ space.
Then:
:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
We have that the set $\tau_K$ is defined as:
:$\tau_K := \set {U \cap K: U \in \tau}$
where $K$ is closed in $T$.
Let $A, B \subseteq K$ be closed in $K$ such that $A \cap B = \O$.
From Intersection with Subset is Subset we have that $A \subseteq K \iff A \cap K = A$.
As $K$ is itself closed in $T$, it follows that so is $A \cap K = A$ from Topology Defined by Closed Sets.
Similarly, so is $B \cap K = B$.
Because $T$ is a $T_4$ space, we have that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
As $A, B \subseteq K$ such that we have that:
:$A \subseteq U \cap K, B \subseteq V \cap K: \paren {U \cap K} \cap \paren {V \cap K} = \O$
From the definition of topological subspace, both $U \cap K$ and $V \cap K$ are open in $K$.
Thus the $T_4$ axiom is satisfied in $K$.
{{qed}}
\end{proof}
|
22546
|
\section{T4 Property Preserved in Closed Subspace/Corollary}
Tags: Closed Sets, Topological Subspaces, T4 Spaces, Normal Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_K$ be a subspace of $T$ such that $K$ is closed in $T$.
If $T$ is a normal space then $T_K$ is also a normal space.
That is, the property of being a normal space is weakly hereditary.
\end{theorem}
\begin{proof}
From the definition, $T = \struct {S, \tau}$ is a normal space {{iff}}:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
From Separation Properties Preserved in Subspace, any subspace of a $T_1$ space is also a $T_1$ space.
From T4 Property Preserved in Closed Subspace, any closed subspace of a $T_4$ space is also a $T_4$ space.
Hence the result.
{{qed}}
\end{proof}
|
22547
|
\section{T4 Property is not Hereditary}
Tags: T5 Spaces, Topological Subspaces, T4 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_4$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then it does not necessarily follow that $T_H$ is a $T_4$ space.
\end{theorem}
\begin{proof}
Let $T$ be the Tychonoff plank.
Let $T'$ be the deleted Tychonoff plank.
By definition, $T'$ is a subspace of $T$.
From Tychonoff Plank is Normal, $T$ is a normal space.
From Deleted Tychonoff Plank is Not Normal, $T'$ is not a normal space.
Thus it is seen that the property of being a normal space is not inherited by a subspace.
Hence the result.
{{qed}}
\end{proof}
|
22548
|
\section{T4 Space is Preserved under Homeomorphism}
Tags: T4 Spaces, Separation Axioms, Homeomorphisms
\begin{theorem}
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_4$ space, then so is $T_B$.
\end{theorem}
\begin{proof}
Suppose that $T_A$ is a $T_4$ space.
Let $B_1$ and $B_2$ be closed in $T_B$ that are disjoint.
Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the preimages of $B_1$ and $B_2$ under $\phi$, respectively.
From Preimage of Intersection under Mapping it follows that $A_1$ and $A_2$ are disjoint.
Also, as $\phi$ is a homeomorphism, it is a fortiori continuous.
Thus Continuity Defined from Closed Sets applies to yield that both $A_1$ and $A_2$ are closed.
Now as $T_A$ is a $T_4$ space, we find disjoint open sets $U_1$ containing $A_1$ and $U_2$ containing $A_2$.
From Image of Subset is Subset of Image, we have $B_1 = \phi \sqbrk {\phi^{-1} \sqbrk {B_1} } = \phi \sqbrk {A_1} \subseteq \phi \sqbrk {U_1}$.
Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.
Mutatis mutandis, we deduce also $B_2 \subseteq \phi \sqbrk {U_2}$.
From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.
Since $\phi$ is a homeomorphism, they are also both open in $T_B$.
Therewith, we have construed two disjoint open sets in $T_B$, one containing $B_1$, and the other containing $B_2$.
Hence $T_B$ is shown to be a $T_4$ space as well.
{{qed}}
\end{proof}
|
22549
|
\section{T4 and T3 Space is T 3 1/2}
Tags: T3 1: 2 Spaces, T3 Spaces, Separation Axioms, T4 Spaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be:
:a $T_4$ space
and also:
:a $T_3$ space.
Then $T$ is also a $T_{3 \frac 1 2}$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_4$ space which is also a $T_3$ space.
From it being $T_3$:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
Consider this $U \in \tau$, which is disjoint from $\set y$.
Then $\relcomp S U$ is a closed set which is disjoint from $F$ but such that $\set y \subseteq \relcomp S U$.
As $T$ is a $T_4$ space, we have that from Urysohn's Lemma there exists an Urysohn function $f$ for $F$ and $\relcomp S U$.
As $\set y \subseteq \relcomp S U$, this function $f$ is a Urysohn function for $F$ and $\set y$ as well.
So:
:For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$.
which is precisely the definition of a $T_{3 \frac 1 2}$ space.
{{qed}}
\end{proof}
|
22550
|
\section{T5 Property is Hereditary}
Tags: T5 Spaces, Topological Subspaces
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is a $T_5$ space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_5$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a $T_5$ space.
Then:
:$\forall A, B \subseteq S, \map {\cl_S} A \cap B = A \cap \map {\cl_S} B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $\map {\cl_S} A$ denotes the closure of $A$ in $S$.
That is:
:For any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $A, B \subseteq H$ such that $\map {\cl_H} A \cap B = A \cap \map {\cl_H} B = \O$.
That is, $A$ and $B$ are separated in $H$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\cl_S} A \cap B
| r = \map {\cl_S} A \cap \paren {H \cap B}
| c = as $B \subseteq H$
}}
{{eqn | r = \paren {\map {\cl_S} A \cap H} \cap B
| c = Intersection is Associative
}}
{{eqn | r = \map {\cl_H} A \cap B
| c = Closure of Subset in Subspace
}}
{{eqn | r = \O
| c = Assumption
}}
{{end-eqn}}
Similarly:
:$A \cap \map {\cl_S} B = \O$
So $A$ and $B$ are separated in $S$.
Because $T$ is a $T_5$ space, we have that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
It follows that:
:$\exists U \cap H, V \cap H \in \tau_H : A \subseteq U \cap H, B \subseteq V \cap H, \paren {U \cap H} \cap \paren {V \cap H} = \O$
and so the $T_5$ axiom is satisfied in $H$.
{{qed}}
\end{proof}
|
22551
|
\section{T5 Space is T4 Space}
Tags: T5 Spaces, Separation Axioms, T4 Spaces
\begin{theorem}
Let $\struct {S, \tau}$ be a $T_5$ space.
Then $\struct {S, \tau}$ is also a $T_4$ space.
\end{theorem}
\begin{proof}
Let $\struct {S, \tau}$ be a $T_5$ space.
From the definition of $T_5$ space:
:$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $A^-$ is the closure of $A$ in $T$.
Let $C, D \subseteq S$ be disjoint sets which are closed in $T$.
Thus $C, D \in \map \complement \tau$ from the definition of closed set.
From Topological Closure is Closed:
:$C^- = C, D^- = D$
and so from $C \cap D = \O$:
:$C^- \cap D = C \cap D^- = \O$
Thus from the definition of $T_5$ space:
:$\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$
which is precisely the definition of a $T_4$ space.
{{qed}}
\end{proof}
|
22552
|
\section{Tableau Confutation contains Finite Tableau Confutation}
Tags: Propositional Tableaus
\begin{theorem}
Let $\mathbf H$ be a countable set of WFFs of propositional logic.
Let $T$ be a tableau confutation of $\mathbf H$.
Then there exists a finite rooted subtree of $T'$ that is also a tableau confutation of $\mathbf H'$.
\end{theorem}
\begin{proof}
For each node $v \in T$, let $\map p v$ be the path from $v$ to $r_T$, the root of $T$.
This path is unique by Path in Tree is Unique.
Let $\VV$ be the subtree of $T$ consisting those nodes $v$ of $T$ such that $\map p v$ is not contradictory.
{{AimForCont}} that $\VV$ were infinite.
Then by König's Tree Lemma, $\VV$ has an infinite branch $\Gamma$.
Since $\VV \subseteq T$, it follows that $\Gamma$ is also a branch of $T$.
However, by construction, it is impossible that $\Gamma$ is contradictory.
This contradicts that $T$ is a tableau confutation.
Hence $\VV$ is finite.
Next, define a finite propositional tableau $T'$ by:
:$v \in T' \iff \map \pi v \in \VV$
that is, the rooted tree formed by $\VV$ and all its children.
Then by construction, for each leaf node $v$ of $T'$, we have that $v \notin \VV$.
That is, that $\map p v$ is a contradictory branch of $T'$.
By Leaf of Rooted Tree is on One Branch, every branch of $T'$ is contradictory.
Hence $T'$ is a tableau confutation of $\mathbf H'$, as desired.
{{qed}}
\end{proof}
|
22553
|
\section{Tableau Confutation implies Unsatisfiable}
Tags: Propositional Tableaus
\begin{theorem}
Let $\mathbf H$ be a collection of WFFs of propositional logic.
Suppose there exists a tableau confutation of $\mathbf H$.
Then $\mathbf H$ is unsatisfiable for boolean interpretations.
\end{theorem}
\begin{proof}
Let $\left({T, \mathbf H, \Phi}\right)$ be a tableau confutation of $\mathbf H$.
Suppose that $v$ were a boolean interpretation model for $\mathbf H$, i.e.:
:$v \models_{\mathrm{BI}} \mathbf H$
By Model of Root of Propositional Tableau is Model of Branch, it follows that:
:$v \models_{\mathrm{BI}} \Phi \left[{\Gamma}\right]$
for some branch $\Gamma$ of $T$.
Since $T$ is a tableau confutation, there is some WFF $\mathbf A$ such that:
:$\mathbf A, \neg\mathbf A \in \Phi \left[{\Gamma}\right]$
Hence $v \models_{\mathrm{BI}} \mathbf A$, i.e.:
:$v \left({\mathbf A}\right) = T$
But by the truth table for $\neg$, this means:
:$v \left({\neg\mathbf A}\right) = F$
which contradicts that $v \models_{\mathrm{BI}} \neg\mathbf A$.
Hence, no such boolean interpretation can exist.
That is, $\mathbf H$ is unsatisfiable for boolean interpretations.
{{qed}}
\end{proof}
|
22554
|
\section{Tableau Confutation is Finished}
Tags: Propositional Tableaus, Propositional Logic, Propositional Calculus
\begin{theorem}
Let $T$ be a tableau confutation.
Then $T$ is a finished tableau.
\end{theorem}
\begin{proof}
By definition of tableau confutation, every branch of $T$ is contradictory.
The result follows by definition of finished propositional tableau.
{{qed}}
\end{proof}
|
22555
|
\section{Tableau Extension Lemma/General Statement/Proof 1}
Tags: Propositional Tableaus
\begin{theorem}
Let $T$ be a finite propositional tableau.
Let its hypothesis set $\mathbf H$ be finite.
{{:Tableau Extension Lemma/General Statement}}
\end{theorem}
\begin{proof}
Let $T_{\mathbf H'}$ be the finite propositional tableau obtained by replacing the hypothesis set $\mathbf H$ of $T$ with $\mathbf H \cup \mathbf H'$.
By the Tableau Extension Lemma, $T_{\mathbf H'}$ has a finished extension $T'$.
By definition of extension, $T_{\mathbf H'}$ is a rooted subtree of $T'$.
But $T_{\mathbf H'}$ and $T$ are equal when considered as rooted trees.
The result follows.
{{qed}}
Category:Propositional Tableaus
\end{proof}
|
22556
|
\section{Tableau Extension Lemma/General Statement/Proof 2}
Tags: Propositional Tableaus
\begin{theorem}
Let $T$ be a finite propositional tableau.
Let its hypothesis set $\mathbf H$ be finite.
{{:Tableau Extension Lemma/General Statement}}
\end{theorem}
\begin{proof}
The proof uses induction on the number $n$ of elements of $\mathbf H$.
Suppose we are given the result for the case $n = 1$, that is, when $\mathbf H$ is a singleton.
Suppose also that we are given the result for all sets $\mathbf H'$ with $n$ elements.
Now, given a set $\mathbf H' = \left\{{\mathbf A_1, \ldots, \mathbf A_{n+1}}\right\}$ with $n+1$ elements.
Let $T$ be a finite propositional tableau.
By induction hypothesis, there is a finished finite propositional tableau $T'$ containing $T$ as a subgraph, and with root $\mathbf H \cup \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right)$.
Now apply the case $n = 1$ to this resulting propositional tableau $T'$ and the set $\left\{{\mathbf A_{n+1}}\right\}$.
This yields a finished finite propositional tableau $T''$ which:
$(1):\quad$ has root $\mathbf H \cup \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right\} \cup \left\{{\mathbf A_{n+1}}\right\} = \mathbf H \cup \mathbf H'$;
$(2):\quad$ contains $T'$ as a subgraph.
But then $T''$ also contains $T$ as a subgraph, proving the result for $\mathbf H'$.
It thus only remains to take care of the base cases $n = 0$ and $n = 1$.
First, the case $n = 0$.
Let $T$ be a finite propositional tableau.
To find the finite propositional tableau $T'$ with the desired properties, we use some of the tableau construction rules, starting with $T$.
Let $t$ be any leaf node of $T$, and let $\Gamma_t$ be the branch from Leaf of Rooted Tree is on One Branch.
Let $n \left({\Gamma_t}\right)$ be the number of non-basic WFFs that were not used to add any of the nodes of $\Gamma_t$ to $T$.
It is seen that for any application of the tableau construction rules on $t$:
:If $s$ is added by the rule, then $n \left({\Gamma_s}\right) \le n \left({\Gamma_t}\right)$.
Moreover, it is seen that any rule reduces the total count $m \left({\Gamma_t}\right)$ of logical connectives occurring in these non-basic, unused WFFs along $\Gamma_t$.
In conclusion:
:If $s$ is added by a rule, then $m \left({\Gamma_s}\right) < m \left({\Gamma_t}\right)$
By the Method of Infinite Descent applied to $m \left({\Gamma_t}\right)$, only finitely many rules can be applied, starting from $t$.
Since $T$ has only finitely many leaves and corresponding branches, only finitely many rules can be applied to $T$ in total.
Let $T'$ be the finite propositional tableau resulting from applying all these possible rules.
By construction of $T'$, it follows that every branch of $\Gamma$ is either contradictory or finished.
That is, $T'$ is finished.
Finally, the last case, $n = 1$.
Let $\mathbf A$ be a WFF of propositional logic.
Let $T$ be a finite propositional tableau.
First, using the case $n = 0$, extend $T$ to a finished finite propositional tableau $T'$.
Again using the case $n = 0$, let $T_{\mathbf A}$ be a finished finite propositional tableau with root $\left\{{\mathbf A}\right\}$.
Now add $\mathbf A$ to the root of $T'$.
Then at every leaf $t$ of $T'$, $\mathbf A$ is the only WFF that is not used yet.
As far as the rules for propositional tableaus are concerned, there is no difference between:
:$t$ as a leaf of $T'$, and
:the tableau consisting only of a root and with hypothesis set $\mathbf A$.
Therefore, the rules allow to "paste", as it were, the finished tableau $T_{\mathbf A}$ under every leaf $t$ of $T'$.
Denote the resulting tableau with $T'_{\mathbf A}$.
Then for any branch $\Gamma$ of $T'_{\mathbf A}$ and every non-basic WFF $\mathbf B$ along it:
:$\mathbf B$ is on $T'$, or:
:$\mathbf B$ is on a copy of $T_{\mathbf A}$.
In either case, the finished nature of these tableaus implies that:
:$\mathbf B$ is used at some node of $\Gamma$
Hence $\Gamma$ is contradictory or finished.
In conclusion, $T'_{\mathbf A}$ is finished, and contains $T$ as a subgraph.
The result follows from the Principle of Mathematical Induction.
{{qed}}
Category:Propositional Tableaus
\end{proof}
|
22557
|
\section{Tail of Convergent Sequence}
Tags: Convergence, Sequences
\begin{theorem}
Let $\left\langle{a_n}\right\rangle$ be a real sequence.
Let $N \in \N$ be a natural number.
Let $a \in R$ be a real number.
Then:
:$a_n \to a$
{{iff}}:
:$a_{n + N} \to a$
\end{theorem}
\begin{proof}
{{ProofWanted}}
Category:Convergence
Category:Sequences
\end{proof}
|
22558
|
\section{Tail of Convergent Series tends to Zero}
Tags: Series
\begin{theorem}
Let $\sequence {a_n}_{n \mathop \ge 1}$ be a sequence of real numbers.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a convergent series.
Let $N \in \N_{\ge 1}$ be a natural number.
Let $\ds \sum_{n \mathop = N}^\infty a_n$ be the tail of the series $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then:
:$\ds \sum_{n \mathop = N}^\infty a_n$ is convergent
:$\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
That is, the tail of a convergent series tends to zero.
\end{theorem}
\begin{proof}
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Let $\sequence {s'_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = N}^\infty a_n$.
It will be shown that $\sequence {s'_n}$ fulfils the Cauchy criterion.
That is:
:$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \size {s'_n - s'_m} < \epsilon$
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
As $\sequence {s_n}$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence.
Thus:
:$\exists N: \forall m, n > N: \size {s_n - s_m} < \epsilon$
Now:
{{begin-eqn}}
{{eqn | l = s_n
| r = \sum_{k \mathop = 1}^n a_k
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^{N - 1} a_k + \sum_{k \mathop = N}^n a_k
| c = Indexed Summation over Adjacent Intervals
}}
{{eqn | r = s_{N - 1} + s'_n
}}
{{end-eqn}}
and similarly:
:$s_m = s_{N - 1} + s'_m$
Thus:
:$s'_n = s_n - s_{N - 1}$
and:
:$s'_m = s_m - s_{N - 1}$
So:
{{begin-eqn}}
{{eqn | l = \size {s_n - s_m}
| o = <
| r = \epsilon
| c =
}}
{{eqn | ll= \leadsto
| l = \size {s_n - s_{N - 1} - s_m + s_{N - 1} }
| o = <
| r = \epsilon
| c =
}}
{{eqn | ll= \leadsto
| l = \size {\paren {s_n - s_{N - 1} } - \paren {s_m - s_{N - 1} } }
| o = <
| r = \epsilon
| c =
}}
{{eqn | ll= \leadsto
| l = \size {\size {s_n - s_{N - 1} } - \size {s_m - s_{N - 1} } }
| o = <
| r = \epsilon
| c = Triangle Inequality
}}
{{eqn | ll= \leadsto
| l = \size {s'_n - s'_m}
| o = <
| r = \epsilon
| c =
}}
{{end-eqn}}
So $\ds \sum_{n \mathop = N}^\infty a_n$ fulfils the Cauchy criterion.
By Convergent Sequence is Cauchy Sequence it follows that it is convergent.
Now it is shown that $\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
We have that $\sequence {s_n}$ is convergent,
Let its limit be $l$.
Thus we have:
:$\ds l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$
So:
:$\ds \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$
But $s_{N - 1} \to l$ as $N - 1 \to \infty$.
The result follows.
{{qed}}
\end{proof}
|
22559
|
\section{Tamref's Last Theorem}
Tags: Number Theory
\begin{theorem}
The Diophantine equation:
:$n^x + n^y = n^z$
has exactly one form of solutions in integers:
:$2^x + 2^x = 2^{x + 1}$
for all $x \in \Z$.
\end{theorem}
\begin{proof}
Since $n^z = n^x + n^y > n^x$ and $n^y$, $z > x,y$.
{{WLOG}} assume that $x \le y < z$.
{{begin-eqn}}
{{eqn | l = n^x + n^y
| r = n^z
}}
{{eqn | ll = \leadsto
| l = 1 + n^{y - x}
| r = n^{z - x}
| c = Divide both sides by $n^x$
}}
{{eqn | ll = \leadsto
| l = 1
| r = n^{y - x} \paren {n^{z - y} - 1}
}}
{{end-eqn}}
Since both $n^{y - x}$ and $n^{z - y} - 1$ are positive integers, both are equal to $1$.
This gives:
:$y = x$ and $n^{z - y} = 2$
which gives the integer solution:
:$n = 2$, $z - y = 1$
Thus the solutions are:
:$\tuple {n, x, y, z} = \tuple {2, x, x, x + 1}, x \in \Z$
{{qed}}
\end{proof}
|
22560
|
\section{Tangent Exponential Formulation/Formulation 1}
Tags: Tangent Exponential Formulation, Tangent Function
\begin{theorem}
Let $z$ be a complex number.
Let $\tan z$ denote the tangent function and $i$ denote the imaginary unit: $i^2 = -1$.
Then:
:$\tan z = i \dfrac {1 - e^{2 i z} } {1 + e^{2 i z} }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan z
| r = \frac {\sin z} {\cos z}
| c = {{Defof|Complex Tangent Function}}
}}
{{eqn | r = \frac {\frac 1 2 i \paren {e^{-i z} - e^{i z} } }
{\frac 1 2 \paren {e^{-i z} + e^{i z} } }
| c = Sine Exponential Formulation and Cosine Exponential Formulation
}}
{{eqn | r = i \frac {e^{-i z} - e^{i z} }
{e^{-i z} + e^{i z} }
}}
{{eqn | r = i \frac {1 - e^{2 i z} }
{1 + e^{2 i z} }
| c = multiplying numerator and denominator by $e^{i z}$
}}
{{end-eqn}}
{{qed}}
Category:Tangent Exponential Formulation
\end{proof}
|
22561
|
\section{Tangent Function is Periodic on Reals}
Tags: Tangent Function, Analysis
\begin{theorem}
The tangent function is periodic on the set of real numbers $\R$ with period $\pi$.
This can be written:
:$\tan x = \map \tan {x \bmod \pi}$
where $x \bmod \pi$ denotes the modulo operation.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \tan {x + \pi}
| r = \frac {\map \sin {x + \pi} } {\map \cos {x + \pi} }
| c = {{Defof|Real Tangent Function}}
}}
{{eqn | r = \frac {-\sin x} {-\cos x}
| c = Sine and Cosine are Periodic on Reals
}}
{{eqn | r = \tan x
| c=
}}
{{end-eqn}}
From Derivative of Tangent Function, we have that:
:$\map {D_x} {\tan x} = \dfrac 1 {\cos^2 x}$
provided $\cos x \ne 0$.
From Shape of Cosine Function, we have that $\cos > 0$ on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Derivative of Monotone Function, $\tan x$ is strictly increasing on that interval, and hence can not have a period of ''less'' than $\pi$.
Hence the result.
{{qed}}
\end{proof}
|
22562
|
\section{Tangent Inequality}
Tags: Trigonometry, Tangent Function, Inequalities
\begin{theorem}
:$x < \tan x$
for all $x$ in the interval $\left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
\end{theorem}
\begin{proof}
Let $f \left({x}\right) = \tan x - x$.
By Derivative of Tangent Function, $f' \left({x}\right) = \sec^2 x - 1$.
By Shape of Secant Function, $\sec^2 x > 1$ for $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
Hence $f' \left({x}\right) > 0$.
From Derivative of Monotone Function, $f \left({x}\right)$ is strictly increasing in this interval.
Since $f \left({0}\right) = 0$, it follows that $f \left({x}\right) > 0$ for all $x$ in $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
{{qed}}
Category:Tangent Function
Category:Inequalities
\end{proof}
|
22563
|
\section{Tangent Line to Convex Graph}
Tags: Analysis, Differential Calculus, Convex Real Functions, Tangents, Analytic Geometry
\begin{theorem}
Let $f$ be a real function that is:
:continuous on some closed interval $\closedint a b$
:differentiable and convex on the open interval $\openint a b$.
Then all the tangent lines to $f$ are below the graph of $f$.
{{explain|"below"}}
\end{theorem}
\begin{proof}
:500px
Let $\TT$ be the tangent line to $f$ at some point $\tuple {c, \map f c}$, $c \in \openint a b$.
Let the gradient of $\TT$ be $m$.
Let $\tuple {x_1, y_1}$ be an arbitrary point on $\TT$.
From the point-slope form of a straight line:
{{begin-eqn}}
{{eqn | l = y - y_1
| r = m \paren {x - x_1}
| c =
}}
{{eqn | ll= \leadsto
| l = y
| r = m \paren {x - x_1} + y_1
| c =
}}
{{end-eqn}}
For $\TT$:
:$y = \map \TT x$
:$y_1 = \map f c$
:$m = \map {f'} c$
:$x = x$
:$x_1 = c$
so:
:$\map \TT x = \map {f'} c \paren {x - c} + \map f c$
Consider the graph of $f$ to the right of $\tuple {c, \map f c}$, that is, any $x$ in $\openint c b$.
Let $d$ be the directed vertical distance from $\TT$ to the graph of $f$.
That is, if $f$ is above $\TT$ then $d > 0$.
If $f$ is below $\TT$, then $d < 0$.
(From the diagram, it is apparent that $\TT$ is below $f$, but we shall prove it analytically.)
$d$ can be evaluated by:
{{begin-eqn}}
{{eqn | l = d
| r = \map f x - \map \TT x
| c =
}}
{{eqn | r = \map f x - \map {f'} c \paren {x - c} - \map f c
| c =
}}
{{eqn | r = \map f x - \map f c - \map {f'} c \paren {x - c}
| c =
}}
{{end-eqn}}
By the Mean Value Theorem, there exists some constant $k$ in $\openint c b$ such that:
{{begin-eqn}}
{{eqn | l = \map {f'} k
| r = \frac {\map f x - \map f c} {x - c}
| c =
}}
{{eqn | ll= \leadsto
| l = \map {f'} k \paren {x - c}
| r = \map f x - \map f c
| c =
}}
{{end-eqn}}
Substitute this into the formula for $d$:
{{begin-eqn}}
{{eqn | l = d
| r = \map {f'} k \paren {x - c} - \map {f'} c \paren {x - c}
| c =
}}
{{eqn | r = \paren {\map {f'} k - \map {f'} c} \paren {x - c}
| c =
}}
{{end-eqn}}
Recall that $x$ lies in the interval $\openint c b$.
So $x > c$, and the quantity $x - c$ is (strictly) positive.
$k$ is also in the interval $\openint c b$ and so $k > c$.
By construction, $f$ is convex.
By the definition of convex:
:$k > c \implies \map {f'} k > \map {f'} c$
which means that:
:$\paren {\map {f'} k - \map {f'} c} > 0$
Then $d$ is the product of two (strictly) positive quantities and is itself (strictly) positive.
Similarly, consider the graph of $f$ to the left of $\tuple {c, \map f c}$, that is, any $x$ in $\openint a c$.
By the same process as above, we will have:
:$d = \paren {\map {f'} k - \map {f'} c} \paren {x - c}$
This time, $x < c$ and the quantity $x - c$ is (strictly) negative.
Further, $k < c$, and so by a similar argument as above:
:$k < c \implies \map {f'} k < \map {f'} c$
and the quantity $\paren {\map {f'} k - \map {f'} c}$ is also (strictly) negative.
Thus $d$ will be the product of two (strictly) negative quantities, and will again be (strictly) positive.
{{qed}}
\end{proof}
|
22564
|
\section{Tangent Secant Theorem}
Tags: Circles, Named Theorems, Tangent Secant Theorem, Tangents
\begin{theorem}
Let $D$ be a point outside a circle $ABC$.
Let $DB$ be tangent to the circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Then $DB^2 = AD \cdot DC$.
{{:Euclid:Proposition/III/36}}
\end{theorem}
\begin{proof}
Let $DA$ pass through the center $F$ of circle $ABC$.
Join $FB$.
From Radius at Right Angle to Tangent, $\angle FBD$ is a right angle.
:320px
We have that $F$ bisects $AC$ and that $CD$ is added to it.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
But $FC = FB$ and so:
:$AD \cdot DC + FB^2 = FD^2$
But from Pythagoras's Theorem we have that $FD^2 = FB^2 + DB^2$ and so:
:$AD \cdot DC + FB^2 = FB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
{{qed|lemma}}
Now let $DA$ be such that it does not pass through the center $E$ of circle $ABC$.
Draw $EF$ perpendicular to $DA$ and draw $EB, EC, ED$.
:320px
From Radius at Right Angle to Tangent, $\angle EBD$ is a right angle.
From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AC$.
So $AF = FC$.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
Let $FE^2$ be added to each:
:$AD \cdot DC + FC^2 + FE^2 = FD^2 + FE^2$
Now $\angle DFE$ is a right angle and so by Pythagoras's Theorem we have:
:$FD^2 + FE^2 = ED^2$
:$FC^2 + FE^2 = EC^2$
This gives us:
:$AD \cdot DC + EC^2 = ED^2$
But $EC = EB$ as both are radii of the circle $ABC$.
Next note that $\angle EBD$ is a right angle and so by Pythagoras's Theorem we have:
:$ED^2 = EB^2 + DB^2$
which gives us:
:$AD \cdot DC + EB^2 = EB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
{{qed}}
{{Euclid Note|36|III|{{EuclidNoteConverse|prop = 37|title = Converse of Tangent Secant Theorem}}}}
\end{proof}
|
22565
|
\section{Tangent Space is Vector Space}
Tags:
\begin{theorem}
Let $M$ be a smooth manifold of dimension $n \in \N$.
Let $m \in M$ be a point.
Let $\struct {U, \kappa}$ be a chart with $m \in U$.
Let $T_m M$ be the tangent space at $m$.
Then $T_m M$ is a real vector space of dimension $n$, spanned by the basis:
:$\set {\valueat {\dfrac \partial {\partial \kappa^i} } m : i \in \set {1, \dotsc, n} }$
that is, the set of partial derivatives with respect to the $i$th coordinate function $\kappa^i$ evaluated at $m$.
\end{theorem}
\begin{proof}
Let $V$ be an open neighborhood of $m$ with $V \subseteq U \subseteq M$.
Let $\map {C^\infty} {V, \R}$ be the set of smooth mappings $f: V \to \R$.
Let $X_m, Y_m \in T_m M$.
Let $\lambda \in \R$.
Then, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
:$X_m, Y_m$ are linear transformations on $\map {C^\infty} {V, \R}$.
Hence $\paren {X_m + \lambda Y_m}$ are also linear transformations.
Therefore, it is enough to show that $X_m + \lambda Y_m$ satisfies the Leibniz law.
Let $f, g \in \map {C^\infty} {V, \R}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {X_m + \lambda Y_m} } {f g}
| r = \map {X_m} {f g} + \lambda \map {Y_m} {f g}
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \map {X_m} f \map g m + \map f m \map {X_m} g + \lambda \paren {\map {Y_m} f \map g m + \map f m \map {Y_m} g}
| c = Leibniz law for $X_m, Y_m$
}}
{{eqn | r = \map {\paren {X_m + \lambda Y_m} } f \map g m + \map f m \map {\paren {X_m + \lambda Y_m} } g
| c = reordering summands
}}
{{end-eqn}}
It follows that:
:$X_m + \lambda Y_m \in T_m M$
Hence $T_m M$ is a real vector space.
Again, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
:for all $X_m \in T_m M$ there exists a smooth curve:
::$\gamma: I \subseteq \R \to M$
:where $\map \gamma 0 = m$ such that:
{{begin-eqn}}
{{eqn | l = \map {X_m} f
| r = \valueat {\map {\frac {\map \d {f \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}
}}
{{eqn | r = \valueat {\map {\frac {\map \d {f \circ \kappa^{-1} \circ \kappa \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}
| c = $f \circ \kappa^{-1} \circ \kappa = f$, as $\kappa$ is a homeomorphism, in particular a bijection.
}}
{{eqn | r = \sum_{i \mathop = 1}^n \valueat {\map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} \tau} \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}
| c = Chain Rule for Real-Valued Functions
}}
{{eqn | r = \sum_{i \mathop = 1}^n \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0 \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} 0}
| c = rearranging
}}
{{eqn | r = \sum_{i \mathop = 1}^n \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0 \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m}
| c = as $m = \map \gamma 0$
}}
{{end-eqn}}
We define:
:$X^i_m := \map {\dfrac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0$
and as above:
:$\valueat {\map {\dfrac \partial {\partial \kappa^i} } m} f := \map {\dfrac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m}$
Therefore:
:$\ds \map {X_m} f = \map {\paren {\sum_{i \mathop = 1}^n X^i_m \valueat {\dfrac \partial {\partial \kappa^i} } m} } f$
{{iff}}:
:$\ds X_m = \sum_{i \mathop = 1}^n X^i_m \valueat {\frac \partial {\partial \kappa^i} } m$
Hence:
:$\set {\valueat {\dfrac \partial {\partial \kappa^i} } m: i \in \set {1, \dotsc, n} }$
forms a basis.
Hence, by definition of dimension of vector space:
:$\dim T_m M = n = \dim M$
This completes the proof.
{{qed}}
\end{proof}
|
22566
|
\section{Tangent in terms of Secant}
Tags: Trigonometric Functions, Tangent Function, Secant Function
\begin{theorem}
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = \tan x
| r = +\sqrt {\sec^2 x - 1}
| c = if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$
}}
{{eqn | l = \tan x
| r = -\sqrt {\sec^2 x - 1}
| c = if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$
}}
{{end-eqn}}
where $\tan$ denotes the real tangent function and $\sec$ denotes the real secant function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sec^2 x - \tan^2 x
| r = 1
| c = Difference of Squares of Secant and Tangent
}}
{{eqn | ll= \leadsto
| l = \tan^2 x
| r = \sec^2 x - 1
}}
{{eqn | ll= \leadsto
| l = \tan x
| r = \pm \sqrt {\sec^2 x - 1}
}}
{{end-eqn}}
Also, from Sign of Tangent:
:If there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$, $\tan x > 0$.
:If there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$, $\tan x < 0$.
When $\cos x = 0$, $\tan x$ and $\sec x$ is undefined.
{{qed}}
\end{proof}
|
22567
|
\section{Tangent is Reciprocal of Cotangent}
Tags: Trigonometric Functions, Cotangent Function, Reciprocal, Tangent Function
\begin{theorem}
Let $\theta$ be an angle such that $\sin \theta \ne 0$ and $\cos \theta \ne 0$.
Then:
:$\tan \theta = \dfrac 1 {\cot \theta}$
where $\tan$ denotes the tangent function and $\cot$ denotes the cotangent function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \frac 1 {\tan \theta}
| r = \cot \theta
| c = Cotangent is Reciprocal of Tangent
}}
{{eqn | ll= \leadsto
| l = \tan \theta
| r = \frac 1 {\cot \theta}
}}
{{end-eqn}}
$\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$.
{{qed}}
\end{proof}
|
22568
|
\section{Tangent is Sine divided by Cosine}
Tags: Trigonometry, Sine Function, Tangent Function, Cosine Function
\begin{theorem}
Let $\theta$ be an angle such that $\cos \theta \ne 0$.
Then:
:$\tan \theta = \dfrac {\sin \theta} {\cos \theta}$
where $\tan$, $\sin$ and $\cos$ mean tangent, sine and cosine respectively.
\end{theorem}
\begin{proof}
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
{{begin-eqn}}
{{eqn | l = \frac {\sin \theta} {\cos \theta}
| r = \frac {y / r} {x / r}
| c = Sine of Angle in Cartesian Plane and Cosine of Angle in Cartesian Plane
}}
{{eqn | r = \frac y r \frac r x
| c =
}}
{{eqn | r = \frac y x
| c =
}}
{{eqn | r = \tan \theta
| c = Tangent of Angle in Cartesian Plane
}}
{{end-eqn}}
When $\cos \theta = 0$ the expression $\dfrac {\sin \theta} {\cos \theta}$ is not defined.
{{qed}}
\end{proof}
|
22569
|
\section{Tangent of 22.5 Degrees}
Tags: Tangent Function
\begin{theorem}
:$\tan 22.5^\circ = \tan \dfrac \pi 8 = \sqrt 2 - 1$
where $\tan$ denotes tangent.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan 22.5^\circ
| r = \tan \frac {45^\circ} 2
| c =
}}
{{eqn | r = \frac {1 - \cos 45^\circ} {\sin 45^\circ}
| c = Half Angle Formulas/Tangent/Corollary 2
}}
{{eqn | r = \frac {1 - \frac {\sqrt 2} 2} {\frac {\sqrt 2} 2}
| c = Cosine of 45 Degrees and Sine of 45 Degrees
}}
{{eqn | r = \sqrt 2 - 1
| c = multiplying top and bottom by $\sqrt 2$
}}
{{end-eqn}}
{{qed}}
Category:Tangent Function
\end{proof}
|
22570
|
\section{Tangent of 67.5 Degrees}
Tags: Tangent Function
\begin{theorem}
:$\tan 67.5 \degrees = \tan \dfrac {3 \pi} 8 = \sqrt 2 + 1$
where $\tan$ denotes tangent.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan 67.5 \degrees
| r = \map \tan {45 \degrees + 22.5 \degrees}
| c =
}}
{{eqn | r = \frac {\tan 45 \degrees + \tan 22.5 \degrees} {1 - \tan 45 \degrees \tan 22.5 \degrees}
| c = Tangent of Sum
}}
{{eqn | r = \frac {1 + \paren {\sqrt 2 - 1} } {1 - 1 \times \paren {\sqrt 2 - 1} }
| c = Tangent of $45 \degrees$ and Tangent of $22.5 \degrees$
}}
{{eqn | r = \frac {\sqrt 2} {2 - \sqrt 2}
| c = simplifying
}}
{{eqn | r = \frac {\sqrt 2 \paren {2 + \sqrt 2} } {\paren {2 - \sqrt 2} \paren {2 + \sqrt 2} }
| c = multiplying top and bottom by $2 + \sqrt 2$
}}
{{eqn | r = \frac {2 \sqrt 2 + 2} {4 - 2}
| c = Difference of Two Squares
}}
{{eqn | r = \sqrt 2 + 1
| c = simplifying
}}
{{end-eqn}}
{{qed}}
Category:Tangent Function
\end{proof}
|
22571
|
\section{Tangent of Angle in Cartesian Plane}
Tags: Trigonometry, Tangent Function, Analytic Geometry
\begin{theorem}
Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\tan \theta = \dfrac y x$
where $\tan$ denotes the tangent of $\theta$.
\end{theorem}
\begin{proof}
:500px
Let a unit circle $C$ be drawn with its center at the origin $O$.
Let a tangent line be drawn to $C$ parallel to $PS$ meeting $C$ at $R$.
Let $Q$ be the point on $OP$ which intersects this tangent line.
$\angle OSP = \angle ORQ$, as both are right angles.
Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$.
By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar.
Thus:
Then:
{{begin-eqn}}
{{eqn | l = \frac y x
| r = \frac {SP} {OS}
| c =
}}
{{eqn | r = \frac {RQ} {OR}
| c = {{Defof|Similar Triangles}}
}}
{{eqn | r = RQ
| c = $OP$ is Radius of the Unit Circle
}}
{{eqn | r = \tan \theta
| c = {{Defof|Tangent Function|subdef = Definition from Circle}}
}}
{{end-eqn}}
When $\theta$ is obtuse, the same argument holds, but both $x$ and $\tan \theta$ are negative.
When $\theta = \dfrac \pi 2$ we have that $x = 0$.
Then $OP$ is parallel to the tangent line at $R$ which it therefore does not meet.
Thus when $\theta = \dfrac \pi 2$, it follows that $\tan \theta$ is not defined.
Likewise $\dfrac y x$ is not defined when $x = 0$.
Thus the relation holds for $\theta = \dfrac \pi 2$.
When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis.
In this case, $y$ is negative.
Thus the relation continues to hold.
When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\tan \theta = 0 = \dfrac y x$.
Hence the result.
{{qed}}
\end{proof}
|
22572
|
\section{Tangent of Complement equals Cotangent}
Tags: Cotangent Function, Tangent Function
\begin{theorem}
:$\map \tan {\dfrac \pi 2 - \theta} = \cot \theta$ for $\theta \ne n \pi$
where $\tan$ and $\cot$ are tangent and cotangent respectively.
That is, the cotangent of an angle is the tangent of its complement.
This relation is defined wherever $\sin \theta \ne 0$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map \tan {\frac \pi 2 - \theta}
| r = \frac {\map \sin {\frac \pi 2 - \theta} } {\map \cos {\frac \pi 2 - \theta} }
| c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\cos \theta} {\sin \theta}
| c = Sine and Cosine of Complementary Angles
}}
{{eqn | r = \cot \theta
| c = Cotangent is Cosine divided by Sine
}}
{{end-eqn}}
The above is valid only where $\sin \theta \ne 0$, as otherwise $\dfrac {\cos \theta} {\sin \theta}$ is undefined.
From Sine of Multiple of Pi it follows that this happens when $\theta \ne n \pi$.
{{qed}}
\end{proof}
|
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