avewright/proofwiki-math huggingface dataset:
Theorem: Modified Fort Space is not T4

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_4$ space.

Proof:
We have:
:Modified Fort Space is not Normal
:Modified Fort Space is $T_1$
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
But $T$ is a $T_1$ space which is not a normal space.
So it follows that $T$ cannot be a $T_4$ space.
{{qed}}
Theorem: Modified Fort Space is not T5

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_5$ space.

Proof:
We have:
:Modified Fort Space is not $T_4$
:$T_5$ Space is $T_4$
Hence the result.
{{qed}}
Theorem: Modified Fort Space is not T2.5

Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_{2 \frac 1 2}$ space.

Proof:
We have:
:Modified Fort Space is not $T_2$
:$T_{2 \frac 1 2}$ Space is $T_2$
Hence the result.
{{qed}}
Category:Modified Fort Spaces
Category:Examples of T2.5 Spaces
jjc0pl9sbihif40wgfx3djy50x19jeb
Theorem: Discrete Space is Perfectly Normal

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a perfectly normal space.

Proof:
{{Recall|Perfectly Normal Space|perfectly normal space|index = 1}}
{{:Definition:Perfectly Normal Space/Definition 1}}
From Discrete Space is Perfectly $T_4$:
:$T$ is a perfectly $T_4$ space.
From Discrete Space is $T_1$:
:$T$ is a $T_1$ space.
The result follows.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Perfectly Normal Spaces
jsk08tciuzdu2dtfv4czgrla75pk79x
Theorem: Discrete Space is Perfectly T4

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a perfectly $T_4$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Perfectly $T_4$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Perfectly T4 Spaces
7fu0d78os74btwrgj1hwdt94zq47532
Theorem: Discrete Space is T0

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_0$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_0$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of T0 Spaces
14lps3o67ylt0umgz3afxs5f5j223mm
Theorem: Discrete Space is T1

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_1$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_1$.
{{qed}}
Theorem: Discrete Space is T2

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_2$ (Hausdorff) space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_2$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Hausdorff Spaces
jiee21lnl9kzptjqb1mbehkxt4jokmp
Theorem: Discrete Space is Semiregular

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a semiregular space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Semiregular.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Semiregular Spaces
tihq69w1fk6xl4xxroxjpqmubr05f3q
Theorem: Discrete Space is T2.5

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_{2 \frac 1 2}$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_{2 \frac 1 2}$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of T2.5 Spaces
it6qzgs6rsmrvde7adi0cekobkui12b
Theorem: Discrete Space is T3.5

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_{3 \frac 1 2}$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_{3 \frac 1 2}$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of T3.5 Spaces
0vt4ev7pelth3quz5vs2zoxvxikn2z5
Theorem: Discrete Space is T4

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_4$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_4$ Space.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of T4 Spaces
logo1ub1kwnajkbpqr88dxbitzu320p
Theorem: Discrete Space is T5

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a $T_5$ space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is $T_5$.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of T5 Spaces
l3e4gsx8vrfl9e89wdjwty10vzoesqp
Theorem: Discrete Space is Regular

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a regular space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Regular Space.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Regular Spaces
tvmv0jwemimffmkira0r5kyqifub5hb
Theorem: Discrete Space is Urysohn

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is an Urysohn space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Urysohn.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Urysohn Spaces
3jtw0ar6aduz0skw5pt0ntkiysxtxkc
Theorem: Discrete Space is Completely Regular

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a completely regular space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Completely Regular.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Completely Regular Spaces
32yy8u8qtlpi8ywibyhm99r1c21sonl
Theorem: Discrete Space is Completely Normal

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a completely normal space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Completely Normal.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Completely Normal Spaces
qarjdg84medbnfxk9hetz0nirigy5md
Theorem: Discrete Space is Normal

Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ is a normal space.

Proof:
We have that a Standard Discrete Metric induces Discrete Topology.
The result follows from Metric Space is Normal Space.
{{qed}}
Category:Discrete Space satisfies all Separation Properties
Category:Discrete Topologies
Category:Examples of Normal Spaces
gf5e76vtgowo0p4ozgr9c8y4ritnoce
Theorem: Reflection Formula (Geometry)

Let $H$ be a Hilbert space.
Let $v, a, \in H$ be arbitrary elements of $H$.
Let $A \subset H$ denote the hyperplane orthogonal to $a$.
The reflection of $v$ through $A$ is given by:
:$\map {\operatorname {Ref}_a} v = v - 2 \dfrac {\langle v, a \rangle} {\langle a, a \rangle} a$

Proof:
By Orthogonal Projection onto Orthocomplement, we may write:
:$v = v^\parallel + v^\perp$
such that $v^\perp \in A$ and $v^\parallel \in A^\perp$.
$v^\parallel$ is the projection of $v$ onto $a$ given by:
:$v^\parallel = \dfrac {\langle v, a\rangle} {\langle a, a \rangle} a$
{{explain|What is $v^\perp$?}}
Then:
:$\map {\operatorname {Ref}_a} v := v^\perp - v^\parallel = (v - v^\parallel) - v^\parallel = v - 2 \dfrac {\langle v, a\rangle} {\langle a, a \rangle} a$
{{Handwaving|Why is $\map {\operatorname {proj}_a} v {{=}} \map {\operatorname {proj}_{A^\perp} } v$ and why does this formula aligns with the geometric description of reflection?}}
Category:Orthogonal Projections
Category:Orthocomplements
Category:Geometric Reflections
Category:Isometries (Euclidean Geometry)
Category:Euclidean Geometry
7ykwhnktxnrnpb6vnz3c4cbt4atk3lt
Theorem: Cartan Subalgebra as Maximal Abelian Subalgebra

Let $\mathfrak g$ be a finite-dimensional, semisimple Lie algebra over an algebraically closed field.
Let $\mathfrak h \subset \mathfrak g$ be a Lie subalgebra.
$\mathfrak h$ is a Cartan subalgebra of $\mathfrak{g}$ {{iff}}:
{{begin-axiom}}
{{axiom | n = 1 | t = $\mathfrak h$ is a maximal Abelian subalgebra of $\mathfrak g$}} 
{{axiom | n = 2 | t = $\forall H \in \mathfrak h$, the adjoint representation $\operatorname {ad}_H: \mathfrak g \to \mathfrak g$ is diagonalizable}}
{{end-axiom}}

Proof:
{{ProofWanted}}
Category:Cartan Subalgebras
Category:Lie Subalgebras
ftb08i9k1jptgxmlc56s5kje7zkedim
Theorem: Weyl Group Permutes Root System

Let $\struct {E, \Phi}$ be a root system with Weyl group $\mathscr W$.
Then $\mathscr W$ permutes $\Phi$.

Proof:
{{tidy}}
Let $\sigma$ be an element of $\mathscr W$.
$\map \sigma \Phi \subset \Phi$ by Axiom (3) of a Root System.
Since $\sigma \in \mathscr W \subset \map {GL} E$, its elements are invertible by definition of the general linear group over $E$.
The result follows by Inverse Mapping is Bijection.
{{qed}}
Category:Weyl Groups
Category:Root Systems
mxwfr5twt3a46q1mmc5wzig0bhrqf0s
Theorem: Root System has Base

Let $\struct {E, \Phi}$ be a root system with base $\Delta \subset \Phi$.
Let $P_\alpha$ denote the hyperplane orthogonal to $\alpha\in\Phi$.
{{Disambiguate|Definition:Orthogonal}}
For $\gamma \in E$, let $\map {\Phi^+} \gamma : = \set {\alpha \in \Phi: \innerprod \alpha \gamma > 0}$.
Call $\gamma$ '''regular''' {{iff}} $\gamma \notin E \setminus \ds \bigcup_{\alpha \mathop \in \Phi}P_\alpha$.
Call $\alpha \in \map {\Phi^+} \gamma$ '''decomposable''' {{iff}} $\alpha = \beta_1 + \beta_2$ for some $\beta_1, \beta_2 \in \map {\Phi^+} \gamma$, otherwise call it '''indecomposable'''.
Then:
:for $\gamma$ regular in $\Phi$ the set $\map \Delta \gamma$ of all indecomposable roots in $\map {\Phi^+} \gamma$ is a base for $\Phi$
:every base of $\Phi$ is obtainable in this manner.

Proof:
By Vector Space over an Infinite Field is not equal to the Union of Proper Subspaces, regular elements of $E$ exist.
{{ProofWanted}}
Theorem: Complete Metric Space is Non-Meager

Let $M = \struct {A, d}$ be a complete metric space.
Then $M$ is non-meager.

Proof:
From the Baire Category Theorem for Complete Metric Spaces, $\struct {S, d}$ is a Baire space.
By Baire Space is Non-Meager, it follows that $T$ is non-meager.
{{qed}}
{{ADC|Baire Space is Non-Meager}}
Theorem: Hilbert Cube is Locally Injectively Path-Connected

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a locally injectively path-connected space.

Proof:
{{Recall|Locally Injectively Path-Connected Space|locally injectively path-connected space}}
{{:Definition:Locally Injectively Path-Connected Space}}
Let $\alpha \in I^\omega$ be arbitrary.
Let $\map {B_\epsilon} \alpha$ be an open ball of $\alpha$ in $M$.
Let $y \in \map {B_\epsilon} \alpha$ be arbitrary.
We have:
:$\forall t \in \closedint 0 1: \map d {x, t x + \paren {1 - t} y} \le \map d {x, y}$
Hence the line segment from $\alpha$ to $y$ lies entirely inside $\map {B_\epsilon} \alpha$
Hence from Open Balls form Basis for Open Sets of Metric Space, $M$ is a locally injectively path-connected space by definition.
{{handwaving|that's cutting the corner}}
Theorem: Linearly Ordered Space is Regular

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $\struct {S, \tau}$ is regular.

Proof:
From Linearly Ordered Space is Normal, $\struct {S, \tau}$ is a normal space.
The result follows from Normal Space is Regular.
{{qed}}
Category:Linearly Ordered Spaces
Category:Examples of Regular Spaces
eka6nca65nyxkdr4pb4bh8ig3qdecuz
Theorem: Ordinal Space is Regular

Let $\Gamma$ denote a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both regular spaces.

Proof:
By definition, $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both linearly ordered spaces.
The result follows from Linearly Ordered Space is Regular.
{{qed}}
Category:Ordinal Spaces
Category:Examples of Regular Spaces
cyxxvsr3hccqkeepy2w93wym3tjmpnt
Theorem: Set with no Limit Point is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be such that $H$ has no limit point.
Then $H$ is a closed set of $T$.

Proof:
{{Recall|Closed Set (Topology)|closed set}}
{{:Definition:Closed Set (Topology)/Definition 2}}
But {{hypothesis}} $H$ has no limit point.
Hence vacuously $H$ contains all its limit points.
Hence the result.
{{qed}}
Category:Limit Points of Sets
Category:Closed Sets
n2ueqsosjp8s7sq3r6lp395vm1eqj0k
Theorem: Equivalence of Definitions of T3.5 Space

{{TFAE|def = T3.5 Space|view = a $T_{3 \frac 1 2}$ space}}
Let $T = \struct {S, \tau}$ be a topological space.

Proof:
=== Definition 1 implies Definition 2 ===
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space by definition $1$.
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space|index = 1}}
{{:Definition:T3.5 Space/Definition 1}}
{{ProofWanted}}
{{Recall|T3.5 Space|$T_{3 \frac 1 2}$ space|index = 2}}
{{:Definition:T3.5 Space/Definition 2}}
Thus $T = \struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space by definition $2$.
{{qed|lemma}}
Theorem: Topological Space is Compact iff Every Ultrafilter Converges

A topological space $T = \struct {S, \tau}$ is '''compact''' {{iff}} every ultrafilter on $S$ converges to a point in $S$.

Proof:
From Topological Space is Compact iff Every Filter has Adherent Point, every filter on $S$ has an adherent point in $S$.
It remains to be proved that:
:every filter on $S$ has an adherent point in $S$ {{iff}} every ultrafilter on $S$ converges to a point in $S$.
Theorem: Hermitian Matrix is Hermitian Operator

A Hermitian matrix is a Hermitian operator.

Proof:
Let $\mathbf A \in \mathbb C^{n \times n}$ be a Hermitian matrix.
Let $\mathbf v, \mathbf w \in \mathbb C^n$ be vectors.
Then:
{{begin-eqn}}
{{eqn | l = \innerprod {\mathbf A \mathbf v} {\mathbf w}
      | r = \mathbf w^\dagger \paren {\mathbf A \mathbf v}
      | c = {{Defof|Complex Vector Inner Product}}
}}
{{eqn | r = \paren {\mathbf w^\dagger \mathbf A} \mathbf v
      | c = Matrix Multiplication is Associative
}}
{{eqn | r = \paren {\mathbf A^\dagger \mathbf w}^\dagger \mathbf v
      | c = Hermitian Conjugate of Matrix Product and Hermitian Conjugate is Involution
}}
{{eqn | r = \paren {\mathbf A \mathbf w}^\dagger \mathbf v
      | c = {{Defof|Hermitian Matrix}}
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf A \mathbf w}
      | c = {{Defof|Complex Vector Inner Product}}
}}
{{end-eqn}}
Thus, $\mathbf A = \mathbf A^*$, where $\mathbf A^*$ is the adjoint of $\mathbf A$.
Thus, $\mathbf A$ is a Hermitian operator.
{{qed}}
Category:Hermitian Operators
Category:Hermitian Matrices
Category:Complex Vector Spaces
0ir5swm0ngwh7y0nu8yro8tgx75m524
Theorem: Hermitian Conjugate of Matrix Product

Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined.
Then $\mathbf B^\dagger \mathbf A^\dagger$ is defined, and:
:$\paren {\mathbf A \mathbf B}^\dagger = \mathbf B^\dagger \mathbf A^\dagger$
where $\mathbf X^\dagger$ is the Hermitian conjugate of $\mathbf X$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {\mathbf A \mathbf B}^\dagger
      | r = \overline {\paren {\mathbf A \mathbf B}^\intercal}
      | c = {{Defof|Hermitian Conjugate}}
}}
{{eqn | r = \overline {\mathbf B^\intercal \mathbf A^\intercal}
      | c = Transpose of Matrix Product
}}
{{eqn | r = \overline {\mathbf B}^\intercal \overline {\mathbf A}^\intercal
      | c = Product of Complex Conjugates
}}
{{eqn | r = \mathbf B^\dagger \mathbf A^\dagger
      | c = {{Defof|Hermitian Conjugate}}
}}
{{end-eqn}}
{{qed}}
Category:Hermitian Conjugates
Category:Conventional Matrix Multiplication
1w9xbkehu2y7w9nozw4anbnhrhvsgf6
Theorem: Arc-Connected Space is Path-Connected

Let $T = \struct {S, \tau}$ be a topological space which is arc-connected.
Then $T$ is path-connected.

Proof:
Let $T = \struct {S, \tau}$ be arc-connected.
Then $\forall x, y \in S$, there exists an arc $f: \closedint 0 1 \to S$ such that $\map f 0 = x$ and $\map f 1 = y$.
From Arc is Injective Path, $f$ is {{afortiori}} an injective path.
The result follows from Injectively Path-Connected Space is Path-Connected.
{{qed}}
Theorem: Joining Arcs makes Another Arc

Let $T$ be a topological space.
Let $\mathbb I \subseteq \R$ be the closed unit interval $\closedint 0 1$.
Let $a, b, c$ be three distinct points of $T$.
Let $f, g: \mathbb I \to T$ be arcs in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
Let $h: \mathbb I \to T$ be the mapping given by:
:<nowiki>$\map h x = \begin {cases}
\map f {2 x} & : x \in \closedint 0 {\dfrac 1 2} \\
\map g {2 x - 1} & : x \in \closedint {\dfrac 1 2} 1
\end {cases}$</nowiki>
Then either:
:$h$ is an arc in $T$
or
:There exists some restriction of $h$ which, possibly after reparametrisation, is an arc in $T$.

Proof:
{{ProofWanted}}
Category:Arcs (Topology)
n5bxcpqirj8m3kvlv7vuib2tptlvmqe
Theorem: Arc-Connectedness is Equivalence Relation

Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is arc-connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.

Proof:
{{ProofWanted|While it is expected that this is similar to Injective Path-Connectedness is Equivalence Relation, no assumptions made until it has been thought about}}
Category:Examples of Equivalence Relations
Category:Arc-Connectedness
gs3kxsjw86hh21zqpy8iqc8h8sd3ji7
Theorem: Complex Vector Inner Product is Complex Inner Product

The complex vector inner product is an example of a complex inner product.

Proof:
{{Recall|Complex Vector Inner Product|complex vector inner product}}
{{:Definition:Complex Vector Inner Product}}
Let $\mathbf x, \mathbf y, \mathbf z \in \mathbb C^n$ be vectors.
Theorem: Number of Roots of Polynomial With Complex Coefficients Equals Degree of Polynomial

Every non-constant polynomial of degree $n > 0$ with coefficients in $\C$ has $n$ roots in $\C$ including multiple roots.

Proof:
Proof by induction:
For all $n \in \mathbb N_{>0}$, let $\map P n$ be the proposition:
:$\map \deg p = n \implies p$ has $n$ roots
where $p$ is some polynomial with coefficients in $\C$.
Theorem: Linear Combination of Eigenvectors Corresponding to Same Eigenvalue of Linear Operator is Eigenvector

Let $K$ be a field. 
Let $V$ be a vector space over $K$. 
Let $\mathbf T : V \to V$ be a linear operator.
Let $\lambda \in K$ be an eigenvalue of $\mathbf T$ with geometric multiplicity $n$.
Let $\mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n$ be eigenvectors corresponding to $\lambda$. 
Let $\mathbf v$ be a non-zero linear combination of $\mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n$:
:$\mathbf v = \ds \sum_{i \mathop = 1}^n c_i \mathbf v_i \neq \mathbf 0$
where $c_i \in K$.
Then, $\mathbf v$ is an eigenvector of $\mathbf T$ with an eigenvalue $\lambda$.

Proof:
Applying $\mathbf T$ to $\mathbf v$ results in:
{{begin-eqn}}
{{eqn | l = \mathbf T \mathbf v
      | r = \mathbf T \paren {\sum_{i \mathop = 1}^n c_i \mathbf v_i}
}}
{{eqn | r = \sum_{i \mathop = 1}^n c_i \mathbf T \mathbf v_i
      | c = $\mathbf T$ is a linear operator
}}
{{eqn | r = \sum_{i \mathop = 1}^n c_i \lambda \mathbf v_i
      | c = All $\mathbf v_i$ have same eigenvalue $\lambda$
}}
{{eqn | r = \lambda \paren {\sum_{i \mathop = 1}^n c_i \mathbf v_i}
      | c = Indexed Summation of Multiple of Mapping
}}
{{eqn | r = \lambda \mathbf v
}}
{{end-eqn}}
Thus, $\mathbf v$ is an eigenvector of $\mathbf T$ with an eigenvalue $\lambda$.
{{qed}}
Theorem: Weakly Sigma-Locally Compact Space is Hemicompact

Let $T = \struct {S, \tau}$ be a weakly $\sigma$-locally compact space.
Then $T$ is a hemicompact space.

Proof:
{{Recall|Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space}}
{{:Definition:Weakly Sigma-Locally Compact Space}}
{{Recall|Hemicompact Space|hemicompact space}}
{{:Definition:Hemicompact Space}}
{{ProofWanted}}
Theorem: Extremally Disconnected and Locally Hausdorff Space is Sequentially Discrete

Let $T = \struct {S, \tau}$ be an extremally disconnected space which is also locally Hausdorff.
Then $T$ is sequentially discrete.

Proof:
{{Recall|Sequentially Discrete Space|sequentially discrete topological space}}
{{:Definition:Sequentially Discrete Space}}
Let $T = \struct {S, \tau}$ be an extremally disconnected space which is also locally Hausdorff.
{{Recall|Extremally Disconnected Space|extremally disconnected space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
{{Recall|Locally Hausdorff Space|locally Hausdorff space}}
{{:Definition:Locally Hausdorff Space}}
{{ProofWanted}}
Theorem: Hermitian Operator is Normal

All Hermitian operators are normal operators.

Proof:
Let $\mathbf T$ be a Hermitian operator.
Then,
{{begin-eqn}}
{{eqn | l = \mathbf T^* \mathbf T
      | r = \mathbf T \mathbf T
      | c = {{Defof|Hermitian Operator}}
}}
{{eqn | r = \mathbf T \mathbf T^*
      | c = {{Defof|Hermitian Operator}}
}}
{{end-eqn}}
Thus, $\mathbf T$ is a normal operator.
{{qed}}
Category:Hermitian Operators
Category:Normal Operators
09epgttrco7wtfkcefk8krvjspwj5yw
Theorem: Topological Space with Countable K-Network has Countable Network

Let $T = \struct {S, \tau}$ be a topological space which has a countable $k$-network.
Then $T$ has a countable network.

Proof:
Follows directly from K-Network is Network.
{{qed}}
Theorem: K-Network is Network

Let $\NN$ be a $k$-network on a topological space $T = \struct {S, \tau}$.
Then $\NN$ is also a network on $T$.

Proof:
{{Recall|K-Network|$k$-network}}
{{:Definition:K-Network/K-Network}}
{{Recall|Network (Topology)|network}}
{{:Definition:Network (Topology)/Definition 1}}
Let $T = \struct {S, \tau}$ have a $k$-network $\NN$.
Let $U \in \tau$ be arbitrary.
From Empty Set is Compact, $\O$ is a compact space.
From Empty Set is Subset of All Sets, $\O \subseteq U$.
Hence by definition of $k$-network, there exists a subset $\NN^*$ of $\NN$ such that:
:$\O \subseteq \bigcup \NN^* \subseteq U$
As $U$ is arbitrary, $\NN$ is a network on $T$ by definition.
{{qed}}
Theorem: Fully T4 Space is Paracompact

Let $T = \struct {S, \tau}$ be a fully $T_4$ space.
Then $T$ is paracompact.

Proof:
{{Recall|Paracompact Space|paracompact}}
{{:Definition:Paracompact Space}}
Let $T = \struct {S, \tau}$ be a fully $T_4$ space.
{{Recall|Fully T4 Space|fully $T_4$}}
{{:Definition:Fully T4 Space}}
Let $\UU$ be an open cover for $T$.
Then, from the definition of barycentric refinement, there exists a cover $\VV$ for $T$ such that:
:$\ds \forall x \in S: \exists U \in \UU: \paren {\bigcup \set {V \in \VV: x \in V} } \subseteq U$
{{proof wanted|This is clearly a stronger statement than Fully Normal Space is Paracompact, and developed recently}}