avewright/proofwiki-math huggingface dataset:
Theorem: Intersections of Slices form Neighborhood Basis for Standard Topology of Compact Convex Subset of Hausdorff Locally Convex Space

Let $\struct {X, \PP}$ be a Hausdorff locally convex space with its standard topology.
Let $X^\ast$ be the topological dual space of $X$.
Let $K$ be a compact convex subset of $X$.
Let $x_0 \in K$.
Let $V$ be a open neighborhood of $x_0$ in $K$. 
Let $\FF$ be the set of non-empty finite subsets of $X^\ast$. 
Let:
:$V_{f, \alpha} = \set {x \in K : \map \Re {\map f x} < \alpha}$
Then there exists $F \in \FF$ and $\alpha_f \in \R$ for each $f \in F$ such that:
:$\ds x_0 \in \bigcap_{f \in F} V_{f, \alpha_f} \subseteq V$
That is, if we set:
:$\ds \BB = \set {\bigcap_{f \in F} V_{f, \alpha_f} : F \in \FF, \, \alpha_f \in \R}$
then $\BB$ is a neighborhood basis for $x_0$ in the subspace topology of $K$.

Proof:
Since $V$ is open, $K \setminus V$ is closed.
Let $x \in K \setminus V$.
Since $\struct {X, \PP}$ is Hausdorff, $\set {x_0}$ is compact.
From:
:Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Compact Convex Set and Closed Convex Set (Real Case) if $\GF = \R$
:Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Compact Convex Set and Closed Convex Set (Complex Case) if $\GF = \C$
there exists $f \in X^\ast$ such that:
applied to $K$ and $\set {x_0}$, there exists $g_x \in X^\ast$ and $\beta_x \in \R$ such that:
:$\ds \map \Re {\map {g_x} {x_0} } < \beta_x < \inf_{y \mathop \in K} \map \Re {\map {g_x} y} \le \map \Re {\map {g_x} x}$
Hence for each $x \in K \setminus V$:
:$x \in \set {y \in X : \map \Re {\map {g_x} y} > \beta_x}$
Hence we have:
:$\ds K \setminus V \subseteq \bigcup_{x \in K \setminus V} \set {y \in X : \map \Re {\map {g_x} y} > \beta_x}$
From Closed Subspace of Compact Space is Compact, $K \setminus V$ is compact.
Hence there exists $x_1, \ldots, x_n \in K \setminus V$ such that:
:$\ds K \setminus V \subseteq \bigcup_{i \mathop = 1}^n \set {y \in X : \map \Re {\map {g_{x_i} } y} > \beta_{x_i} }$
That is:
:$\ds K \setminus V \subseteq \bigcup_{i \mathop = 1}^n \set {y \in K : \map \Re {\map {g_{x_i} } y} > \beta_{x_i} }$
Now set $f_i = g_{x_i}$ and $\alpha_i = \beta_{x_i}$ for each $i \in \set {1, \ldots, n}$.
From Set Complement inverts Subsets, applied in $K$, we have:
:$\ds \bigcap_{i \mathop = 1}^n \set {y \in K : \map \Re {\map {f_i} y} \le \alpha_i} \subseteq V$
In particular:
:$\ds \bigcap_{i \mathop = 1}^n \set {y \in K : \map \Re {\map {f_i} y} < \alpha_i} \subseteq V$
Further, $\map \Re {\map {f_i} {x_0} } < \alpha_i$ for each $i \in \set {1, \ldots, n}$, so we have:
:$\ds x_0 \in \bigcap_{i \mathop = 1}^n \set {y \in K : \map \Re {\map {f_i} y} < \alpha_i}$
Taking $F = \set {f_1, \ldots, f_n}$ and $\alpha_{f_i} = \alpha_i$ for each $i \in \set {1, \ldots, n}$ completes the proof.
{{qed}}
Category:Locally Convex Spaces
Category:Convex Sets (Vector Spaces)
iudvsdi0xygc7ipg25mjmgp2sfqh6kt
Theorem: Convex Combination contained in Convex Set

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $K$ be a convex subset of $X$.
Let $\lambda_1, \ldots, \lambda_n \in \openint 0 1$ be such that:
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$ 
Let $u_1, \ldots, u_n \in K$.
Then we have:
:$\ds \sum_{i \mathop = 1}^n \lambda_i u_i \in K$

Proof:
For each $n \in \N$, let $\map P n$ be the proposition:
:for each $\lambda_1, \lambda_2, \ldots, \lambda_n \in \openint 0 1$ with:
::$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
:and $u_1, u_2, \ldots, u_n \in U$, we have: 
::$\ds \sum_{i \mathop = 1}^n \lambda_i u_i \in K$
We proceed by induction.
Theorem: Extreme Point of Convex Set not contained in Convex Hull of Subset not containing Point

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $K$ be a convex subset of $X$. 
Let $x \in K$ be an extreme point of $K$. 
Let $C \subseteq K$ be such that $x \not \in C$. 
Let $\map {\operatorname {conv} } C$ be the convex hull of $C$.
Then $x \not \in \map {\operatorname {conv} } C$.

Proof:
Suppose that $x \in \map {\operatorname {conv} } C$.
Let $\lambda_0, \ldots, \lambda_n \in \openint 0 1$ and $u_0, \ldots, u_n \in C$ be such that:
:$\ds \sum_{i \mathop = 0}^n \lambda_i = 1$
and:
:$\ds x = \sum_{i \mathop = 0}^n \lambda_i u_i$
Write:
:$\ds x = \lambda_0 u_0 + \sum_{i \mathop = 1}^n \lambda_i u_i$
and further:
:$\ds x = \lambda_0 u_0 + \paren {1 - \lambda_0} \sum_{i \mathop = 1}^n \paren {\frac {\lambda_i} {1 - \lambda_0} } u_i$
We have:
{{begin-eqn}}
{{eqn	| l = \sum_{i \mathop = 1}^n \frac {\lambda_i} {1 - \lambda_0}
		| r = \frac {\sum_{i \mathop = 1}^n \lambda_i} {1 - \lambda_0}
}}
{{eqn	| r = \frac {\sum_{i \mathop = 0}^n \lambda_i - \lambda_0} {1 - \lambda_0}
}}
{{eqn	| r = \frac {1 - \lambda_0} {1 - \lambda_0}
}}
{{eqn	| r = 1
}}
{{end-eqn}}
From Convex Combination contained in Convex Set, we have:
:$\ds \sum_{i \mathop = 1}^n \paren {\frac {\lambda_i} {1 - \lambda_0} } u_i \in K$ 
Hence since $x$ is an extreme point of $K$, we have:
:$\ds x = \sum_{i \mathop = 1}^n \paren {\frac {\lambda_i} {1 - \lambda_0} } u_i = u_0 \in C$
contradicting that $x \not \in C$. 
{{qed}}
Category:Convex Hulls
Category:Extreme Points of Convex Sets
pp3zl16arrmiipjywz8zyuz547u887n
Theorem: Convex Hull of Finite Union of Convex Sets

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $K_1, \ldots, K_n \subseteq X$ be convex.
Let:
:$\ds K = \bigcup_{i \mathop = 1}^n K_i$
Let $\map {\operatorname {conv} } K$ be the convex hull of $K$.
Then:
:$\ds \map {\operatorname {conv} } K = \set {\sum_{i \mathop = 1}^n \lambda_i x_i : \lambda_i \in \R_{> 0}, \, x_i \in K_i, \, \sum_{i \mathop = 1}^n t_i = 1}$

Proof:
We first show that:
:$\ds \set {\sum_{i \mathop = 1}^n \lambda_i x_i : \lambda_i \in \R_{> 0}, \, x_i \in K_i, \, \sum_{i \mathop = 1}^n t_i = 1} \subseteq \map {\operatorname {conv} } K$
Let $x_i \in K_i$ for $i \in \set {1, \ldots, n}$ and $t_1, \ldots, t_n \in \R_{> 0}$ be such that:
:$\ds \sum_{i \mathop = 1}^n t_i = 1$
In particular, we have $x_1, \ldots, x_n \in K \subseteq \map {\operatorname {conv} } K$ for each $i \in \set {1, \ldots, n}$. 
Hence from Convex Combination contained in Convex Set:
:$\ds \sum_{i \mathop = 1}^n t_i x_i \in \map {\operatorname {conv} } K$
as required.
We now take $x \in \map {\operatorname {conv} } K$. 
Then there exists $u_1, \ldots, u_N \in K$ and $\lambda_1, \ldots, \lambda_N \in \R_{> 0}$ such that:
:$\ds \sum_{i \mathop = 1}^N \lambda_i = 1$
and:
:$\ds x = \sum_{i \mathop = 1}^N \lambda_i u_i$
By the definition of $K$, for each $i \in \set {1, \ldots, N}$ there exists $j \in \set {1, \ldots, n}$ such that $u_i \in K_j$.
For each $j \in \set {1, \ldots, n}$, let $I_j$ be the set of $i \in \set {1, \ldots, N}$ such that $j$ is least with $u_i \in K_j$.
Clearly the $I_j$s are then pairwise disjoint.
{{WLOG}} that $I_j \ne \O$ for each $j \in \set {1, \ldots, n}$.
If $I_j = \O$ for any $j$, simply remove $K_j$ from the union.
We can rearrange the sum as:
:$\ds x = \sum_{j \mathop = 1}^n \sum_{i \mathop \in I_j} \lambda_i u_i$
That is:
:$\ds x = \sum_{j \mathop = 1}^n \paren {\sum_{i \mathop \in I_j} \lambda_i} \sum_{i \mathop \in I_j} \frac {\lambda_i} {\sum_{i \mathop \in I_j} \lambda_i} u_i$
We prove that $x$ is now in the desired form.
We have:
:$\ds \sum_{i \mathop \in I_j} \frac {\lambda_i} {\sum_{i \mathop \in I_j} \lambda_i} = \frac {\sum_{i \mathop \in I_j} \lambda_i} {\sum_{i \mathop \in I_j} \lambda_i} = 1$ 
Since $u_i \in K_j$ for each $i \in I_j$ and each $K_j$ is convex, we have:
:$\ds \sum_{i \mathop \in I_j} \frac {\lambda_i} {\sum_{i \mathop \in I_j} \lambda_i} u_i \in K_j$
from Convex Combination contained in Convex Set.
Hence set:
:$\ds x_j = \sum_{i \mathop \in I_j} \frac {\lambda_i} {\sum_{i \mathop \in I_j} \lambda_i} u_i \in K_j$
and:
:$\ds t_j = \sum_{i \mathop \in I_j} \lambda_i$
We then have:
:$\ds \sum_{j \mathop = 1}^n t_j = \sum_{j \mathop = 1}^n \sum_{i \mathop \in I_j} \lambda_i = \sum_{i \mathop = 1}^N \lambda_i = 1$
with:
:$\ds x = \sum_{j \mathop = 1}^n t_j x_j$ 
Hence $x$ has the desired form and:
:$\ds \map {\operatorname {conv} } K \subseteq \set {\sum_{i \mathop = 1}^n t_i x_i : t_i \in \R_{> 0}, \, x_i \in K_i, \, \sum_{i \mathop = 1}^n t_i = 1}$
Hence the result.
{{qed}}
Category:Convex Sets (Vector Spaces)
Category:Convex Hulls
0joxa7plt6ry1sesv2d61m8r2tlifp6
Theorem: Closure of Set with Compact Closed Convex Hull in Hausdorff Locally Convex Space contains Extreme Points of Convex Hull

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ with its standard topology.
Let $S \subseteq X$ be such that:
:$K = \map \cl {\map {\operatorname {conv} } S}$ is compact
where $\operatorname {conv}$ denotes convex hull.
Let $\map E K$ be the set of extreme points of $K$.
Then $\map E K \subseteq \map \cl S$.

Proof:
Let $X^\ast$ be the topological dual space of $X$.
{{AimForCont}} suppose that:
:$\map E K \not \subseteq \map \cl S$
Take $x_0 \in \map E K \setminus \map \cl S$.
Then $X \setminus \map \cl S$ is an open neighborhood of $x_0$ in $X$.
By Slices form Neighborhood Basis of Extreme Point of Compact Convex Subset of Hausdorff Locally Convex Space, there exists $f \in X^\ast$ and $\alpha \in \R$ such that:
:$x_0 \in \set {x \in K : \map \Re {\map f x} < \alpha} \subseteq X \setminus \map \cl S$
Hence we have:
:$\map \Re {\map f {x_0} } < \alpha$
and:
:$\map \Re {\map f x} \ge \alpha$ for $x \in \map \cl S$
Now, since $f$ is continuous, $\map \Re f$ is continuous from Real Part of Continuous Function is Continuous.
Hence:
:$\set {x \in K : \map \Re {\map f x} \ge \alpha}$ is closed as the preimage of the closed set $\hointr \alpha \infty$ under the continuous map $\map \Re f$.
From Real Part of Linear Functional is Linear Functional, $\map \Re f$ is $\R$-linear.
Now, for $x, y \in \set {x \in K : \map \Re {\map f x} \ge \alpha}$ and $t \in \openint 0 1$, we have:
:$\map \Re {\map f {t x + \paren {1 - t} y} } = t \map \Re {\map f x} + \paren {1 - t} \map \Re {\map f y} \ge t \alpha + \paren {1 - t} \alpha = \alpha$
Hence:
:$\set {x \in K : \map \Re {\map f x} \ge \alpha}$ is convex.
From:
:$S \subseteq \set {x \in K : \map \Re {\map f x} \ge \alpha}$
we obtain:
:$\map {\operatorname {conv} } S \subseteq \set {x \in K : \map \Re {\map f x} \ge \alpha}$
from Convex Hull preserves Subsets and Convex Hull is Smallest Convex Set containing Set: Corollary.
Since $\set {x \in K : \map \Re {\map f x} \ge \alpha}$ is closed, we have:
:$\map \cl {\set {x \in K : \map \Re {\map f x} \ge \alpha} } = \set {x \in K : \map \Re {\map f x} \ge \alpha}$
from Set is Closed iff Equals Topological Closure.
From Set Closure Preserves Set Inclusion, we then have:
:$\map \cl {\map {\operatorname {conv} } S} \subseteq \set {x \in K : \map \Re {\map f x} \ge \alpha}$
Since $\map \Re {\map f {x_0} } > \alpha$, it follows that $x_0 \not \in \map \cl {\map {\operatorname {conv} } S}$.
However, $\map \cl {\map {\operatorname {conv} } S} = K$, so we have that $x_0 \not \in K$.
This is a contradiction, since $x_0 \in \map E K \subseteq K$.
Hence we have:
:$\map E K \subseteq \map \cl S$
{{qed}}
Category:Convex Hulls
Category:Extreme Points of Convex Sets
Category:Locally Convex Spaces
6mpjdwm1xbms9gxegphadlv9zh6b6b0
Theorem: Product of Cardinalities of Centralizer and Conjugacy Class

Let $\struct {G, \circ}$ be a group and let $a \in G$.
Let $\map {C_G} a$ denote the centralizer of $a$ in $G$.
Let $\conjclass a$ denote the conjugacy class of $a$ in $G$.
Then:
:$\order {\map {C_G} a} \times \order {\conjclass a} = \order G$
where $\order G$ denotes the order of $G$.

Proof:
From Number of Conjugates is Number of Cosets of Centralizer:
:$\order {\conjclass a} = \index G {\map {C_G} a}$
where $\index G {\map {C_G} a}$ is the index of $\map {C_G} a$ in $G$.
From Lagrange's Theorem (Group Theory):
:$\index G {\map {C_G} a} = \dfrac {\order G} {\order {\map {C_G} a} }$
Hence the result:
:$\order {\map {C_G} a} \times \order {\conjclass a} = \order G$
{{qed}}
Theorem: Degree of Characteristic Polynomial of Matrix equals Order of Matrix

Let $R$ be a commutative ring with unity.
Let $\mathbf A$ be a square matrix over $R$ of order $n > 0$.
Let $\mathbf I_n$ be the $n \times n$ identity matrix.
Let $R \sqbrk x$ be the polynomial ring in one variable over $R$.
Let $\map {p_{\mathbf A} } x$ be the characteristic polynomial of $\mathbf A$.
Then the degree of $\map {p_{\mathbf A} } x$ is equal to the order of $\mathbf A$.

Proof:
From Existence of Schur Decomposition for Square Matrix, matrix $\mathbf A$ is similar to an upper triangular matrix $\mathbf T$.
Hence:
{{begin-eqn}}
{{eqn | l = \map {p_{\mathbf A} } x
      | r = \map \det {x \mathbf I_n - \mathbf A}
      | c = {{Defof|Characteristic Polynomial of Matrix}} (alternate definition)
}}
{{eqn | r = \map \det {x \mathbf I_n - \mathbf T}
      | c = Similar Matrices have Same Characteristic Polynomial
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {x - T_{i i} }
      | c = Determinant of Upper Triangular Matrix
}}
{{end-eqn}}
where $T_{i i}$ is the $i^{th}$ diagonal element of $T$.
Each multiplicand $x - T_{i i}$ of the product is a polynomial of degree $1$ because the highest power of the variable $x$ is $1$.
Furthermore, each multiplicand $x - T_{i i}$ of the product is a monic polynomial because the coefficient of the highest power of $x$ is $1$.
From Product of Monic Polynomials, the degree of the product of $n$ polynomials of degree $1$ is $n$.
{{qed}}
Theorem: Character is Constant on Conjugacy Class

Let $G$ be a finite group.
Let $\conjclass g$ be the conjugacy class for some $g \in G$.
Let $\chi_\rho$ be the character associated with an irreducible linear representation $\rho$ of $G$.
Then:
:$\forall x \in \conjclass g: \map {\chi_\rho} x = \map {\chi_\rho} g$
That is, character is constant across all elements of a given conjugacy class.

Proof:
{{Recall|Character (Representation Theory)|character}}
{{:Definition:Character (Representation Theory)}}
Let $x = h^{-1} g h$ for some $h \in G$.
{{begin-eqn}}
{{eqn | l = \map {\chi_\rho} x
      | r = \map \tr {\map \rho x}
}}
{{eqn | r = \map \tr {\map \rho {h^{-1} g h} }
}}
{{eqn | r = \map \tr {\paren {\map \rho h}^{-1} \map \rho g \map \rho h}
      | c = $\rho$ is a group homomorphism
}}
{{eqn | r = \map \tr {\map \rho g}
      | c = Similar Matrices have same Traces
}}
{{eqn | r = \map {\chi_\rho} g
}}
{{end-eqn}}
Since the choice of $x$ is arbitrary in $C_g$, the result follows.
{{qed}}
Theorem: Zero Sets are Exactly the Closed G-Delta Sets in T4 Space

Let $T = \struct {S, \tau}$ be a $T_4$ topological space.
Then the zero sets in $T$ are precisely the closed $G_\delta$ subsets of $S$.

Proof:
Let $T = \struct {S, \tau}$ be a $T_4$ space.
Theorem: Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology

Let $\GF \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Schauder basis for $X$.
Let $\sequence {S_n}_{n \mathop \in \N}$ be the projections associated with $\sequence {x_n}_{n \mathop \in \N}$. 
Then $\sequence {S_n}_{n \mathop \in \N}$ converges to the identity mapping in the strong operator topology.

Proof:
From Characterization of Convergence in Strong Operator Topology, it suffices to show:
:$S_n x \to x$ for each $x \in X$.
Let $\sequence {x_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {x_n}_{n \mathop \in \N}$.
By the definition of a Schauder basis, we have:
:$\ds x = \sum_{k \mathop = 1}^\infty \map {x_k^\ast} x x_k$ for each $x \in X$.
That is:
:$\ds x = \lim_{n \mathop \to \infty} \sum_{k \mathop = 1}^n \map {x_k^\ast} x x_k = \lim_{n \mathop \to \infty} \map {S_n} x$ for each $x \in X$.
This is what we aimed to show.
{{qed}}
Category:Projections Associated with Schauder Bases
Category:Strong Operator Topology
Category:Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology
ebrj1pwcp3kjolc6mx6tizuant215tf
Theorem: Schwarz-Pick Theorem

Let $\Bbb D$ be the unit disk in the complex plane centered at $0$.
Let $f: \Bbb D \to \Bbb D$ be a holomorphic function.
Then:
:$\forall z \in \Bbb D: \dfrac {\cmod {\map {f'} z} } {1 - \cmod {\map f z}^2} \le \dfrac 1 {1 - \cmod z^2}$
where $\cmod z$ denotes the modulus of $z$.
Moreover, if there exists a $z_0 \in \Bbb D$ such that:
:$\dfrac {\cmod {\map {f'} {z_0} } } {1 - \cmod {\map {f'} {z_0} }^2} = \dfrac 1 {1 - \cmod {z_0}^2}$
then $f$ is a Möbius transformation.

Proof:
=== Lemma ===
Let $z \in \Bbb D$ be arbitrary.
Let $\varphi_z$ be the Möbius transformation defined as:
:$\map {\varphi_z} w := \dfrac {w - z} {1 - \overline z w}$
where $\overline z$ denotes the complex conjugate of $z$.
Then $\varphi_z$ is a holomorphic bijection from $\Bbb D$ to $\Bbb D$ such that:
:its inverse is $\varphi_{-z}$
:its derivative $\map {\varphi_z'} w$ is $\dfrac {1 - \overline z z} {\paren {1 - \overline z w}^2}$
:$\map {\varphi_z} z = 0$
:$\map {\varphi_z} 0 = z$
Theorem: Parity Check Detects Single Error

Let $\LL$ be a '''linear $\tuple {n, 2}$-code''' with a parity check.
Then $\LL$ is such that one transmission error will be detected.

Proof:
Let $w$ be a transmitted codeword from $\LL$.
Let $s$ be the sum of the bits of $w$.
Let $w'$ be the received word corresponding to $w$.
Let $s'$ be the sum of the bits of $w'$.
Suppose that $w'$ has one transmission error.
Then either $w'$ has a $1$ instead of $0$, or $0$ instead of $1$.
In the first case, $s' = s + 1$.
In the second case, $s' = s - 1$.
In both cases, the parity of $s'$ is different from the parity of $s$.
That means the parity of $s'$ does not match the parity of the codewords of $\LL$.
Hence the result.
{{qed}}
Theorem: Space in which Every Closed Set is a Zero Set is T4

Let $T = \struct{S, \tau}$ be a topological space such that every closed set is a zero set.
Then $T$ is a $T_4$ space.

Proof:
Let $C, D \subseteq S$ be disjoint closed sets.
By hypothesis, there exist continuous functions:
:$f: S \to \closedint 0 1$ such that $C = f^{-1} (\set 0)$
:$g: S \to \closedint 0 1$ such that $D = g^{-1} (\set 0)$
Define $h: S \to \closedint 0 1$ by:
:$\map h x := \dfrac{\map f x}{\map f x + \map g x}$
Since $f(x), g(x) \ge 0$ and $C \cap D = \O$, it follows that for all $x \in S$ we have $\map f x + \map g x > 0$. So $h$ is well-defined and continuous as a quotient of continuous functions with non-vanishing denominator.
Now observe:
* If $x \in C$, then $\map f x = 0$ and $\map g x > 0$, so $\map h x = 0$.
* If $x \in D$, then $\map g x = 0$ and $\map f x > 0$, so $\map h x = 1$.
Hence, $h$ is a Urysohn function separating $C$ and $D$.
By the Urysohn's Lemma Converse, $T$ is $T_4$.
{{qed}}
Theorem: Seminorm on Vector Space induced by Linear Transformation is Seminorm

Let $\GF \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\GF$.
Let $p$ be a seminorm on $Y$.
Define:
:$\map q x = \map p {T x}$
for each $x \in X$.
Then $q$ is a seminorm on $X$.

Proof:
=== Proof of $(\text N 2)$ ===
Let $x \in X$ and $\lambda \in \GF$.
We have:
{{begin-eqn}}
{{eqn	| l = \map q {\lambda x}
		| r = \map p {\map T {\lambda x} }
}}
{{eqn	| r = \map p {\lambda T x}
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \cmod \lambda \map p {T x}
		| c = {{NormAxiomVector|2}} for $p$
}}
{{eqn	| r = \cmod \lambda \map q x
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Finite Supremum of Family of Seminorms is Seminorm

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $\PP$ be a set of seminorms on $X$ such that:
:$\ds \sup_{p \in \PP} \map p x < \infty$ for each $x \in X$.
Define $q : X \to \R$ by:
:$\ds \map q x = \sup_{p \in \PP} \map p x$ for each $x \in X$.
Then $q$ is a seminorm on $X$.

Proof:
=== Proof of $(\text N 2)$ ===
Let $x \in X$ and $\lambda \in \GF$.
We have:
{{begin-eqn}}
{{eqn	| l = \map q {\lambda x}
		| r = \sup_{p \in \PP} \map p {\lambda x}
}}
{{eqn	| o = \le
		| r = \sup_{p \in \PP} \cmod \lambda \map p x
		| c = {{NormAxiomVector|2}} for $p$
}}
{{eqn	| r = \cmod \lambda \sup_{p \in \PP} \map p x
		| c = Multiple of Supremum
}}
{{eqn	| r = \cmod \lambda \map q x
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Complete Norms on Vector Space Equivalent iff Identity Mapping Bounded

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be norms on $X$ such that:
:$\struct {X, \norm {\, \cdot \,}_i}$ is a Banach space for $i \in \set {1, 2}$.
Let $I : \struct {X, \norm {\, \cdot \,}_1} \to \struct {X, \norm {\, \cdot \,}_2}$ be the identity mapping.
Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent {{iff}} $I$ is bounded.

Proof:
=== Necessary Condition ===
Suppose that:
:$\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent.
Then there exists $K > 0$ such that:
:$\norm {I x}_2 = \norm x_2 \le K \norm x_1$ 
That is:
:$I : \struct {X, \norm {\, \cdot \,}_1} \to \struct {X, \norm {\, \cdot \,}_2}$ is bounded.
{{qed|lemma}}
Theorem: Characterization of Perfectly T4 Space

A topological space $T = \struct {S, \tau}$ is perfectly $T_4$ {{iff}} every closed set is a zero set.

Proof:
=== Sufficient Condition ===
Let $T$ be a perfectly $T_4$ space.
By definition, every closed set in $T$ is a $G_\delta$ set.
Moreover, since $T$ is also a $T_4$ space, by Zero Sets are Exactly the Closed G-Delta Sets in T4 Space, every closed $G_\delta$ set is a zero set.
Hence, every closed set in $T$ is a zero set.
Theorem: Norm Induced by Projections Associated with Schauder Basis is Norm

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.
Let $E = \sequence {e_n}_{n \mathop \in \N}$ be a Schauder basis for $X$.
Let $\sequence {S_n}_{n \mathop \in \N}$ be the projections associated with $\sequence {e_n}_{n \mathop \in \N}$.
Define $\norm {\, \cdot \,}_E : X \to \R$ by:
:$\ds \norm x_E = \sup_{n \mathop \in \N} \norm {\map {S_n} x}$ for each $x \in X$.
Then $\norm {\, \cdot \,}_E$ is well-defined and a norm on $X$.

Proof:
=== $\norm {\, \cdot \,}_E$ is well-defined ===
We need to show that:
:$\ds \sup_{n \mathop \in \N} \norm {\map {S_n} x} < \infty$ for each $x \in X$.
From Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology and Characterization of Convergence in Strong Operator Topology we have:
:$\map {S_n} x \to x$ for each $x \in X$.
From Modulus of Limit: Normed Vector Space, we have:
:$\norm {\map {S_n} x} \to \norm x$ for each $x \in X$. 
Then, from the definition of a convergent sequence, there exists $N \in \N$ such that:
:$\norm {\map {S_n} x} \le \norm x + 1$ for $n \ge N$.
Then we have:
:$\ds \sup_{n \mathop \in \N} \norm {\map {S_n} x} \le \max \set {\norm {\map {S_1} x}, \norm {\map {S_2} x}, \ldots, \norm {\map {S_{N - 1} } x}, \norm x + 1} < \infty$
Hence $\norm {\, \cdot \,}_E$ is well-defined.
{{qed|lemma}}
Theorem: Perfectly T4 Property is Hereditary

Let $T = \struct{S, \tau}$ be a perfectly $T_4$ space.
Then every subspace of $T$ is also perfectly $T_4$.
That is, the property of being perfectly $T_4$ is hereditary.

Proof:
{{Recall|Perfectly T4 Space|perfectly $T_4$ space}}
{{:Definition:Perfectly T4 Space}}
Let $H \subseteq S$ be a subspace of $T$, with the subspace topology $\tau_H$.
Let $F \subseteq H$ be a closed set in $H$.
Then, by Closed Set in Topological Subspace there exists a closed set $F_0$ of $T$ such that $F = F_0 \cap H$.
Since $T$ is perfectly $T_4$, using Characterization of Perfectly T4 Space there exists a continuous mapping $f: S \to \closedint 0 1$ such that $f^{-1} \sqbrk {\set 0} = F_0$.
Let $g := f \restriction_H: H \to \closedint 0 1$ be the restriction of $f$ to $H$.
From Restriction of Continuous Mapping is Continuous, it follows that $g$ is continuous.
Hence:
:$g^{-1} \sqbrk {\set 0} = f^{-1} \sqbrk {\set 0} \cap H = F_0 \cap H = F$
Thus, $F$ is a zero set in $H$.
Since $F \subseteq H$ was arbitrary, it follows from the Characterization of Perfectly T4 Space that $H$ is perfectly $T_4$.
{{qed}}