avewright/proofwiki-math huggingface dataset:
Theorem: Closed Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable

Let $X$ be a separable normed vector space.
Let $\sequence {x_n}_{n \in \N}$ be a countable everywhere dense subset of $X$.
Let $X^\ast$ be the normed dual space of $X$. 
Let $\lambda > 0$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$. 
Let $w^\ast$ be the weak-$\ast$ topology.
Define $d : X^\ast \times X^\ast \to \R$ by:
:$\ds \map d {f, g} = \sum_{n \mathop = 1}^\infty \frac 1 {2^n} \min \set {\norm {\map f {x_n} - \map g {x_n} }_X, 1}$ for each $f, g \in X^\ast$. 
Then $\struct {\lambda B^-_{X^\ast}, w^\ast}$ is metrized by the restriction $d'$ of $d$ to $\lambda B^-_{X^\ast}$. 
Further, we have:
:$\map {d'} {f, g} = \map {d'} {f - g, {\mathbf 0}_X}$ for each $f, g \in X^\ast$.

Proof:
From the Banach-Alaoglu Theorem, we have that $\struct {B^-_{X^\ast}, w^\ast}$ is compact.
From Dilation of Compact Set in Topological Vector Space is Compact, $\struct {\lambda B^-_{X^\ast}, w^\ast}$ is compact.
From Weak-* Compact Set in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {\lambda B^-_{X^\ast}, w^\ast}$ is metrized by the metric $d$, and:
:$\map {d'} {f, g} = \map {d'} {f - g, {\mathbf 0}_X}$ for each $f, g \in X^\ast$. 
{{qed}}
Theorem: Topological Closure in Coarser Topology is Larger

Let $X$ be a set. 
Let $\tau_1$ and $\tau_2$ be topologies on $X$ such that:
:$\tau_1 \subseteq \tau_2$
That is, such that $\tau_1$ is coarser than $\tau_2$. 
Let $S \subseteq X$.
Then we have: 
:$\map {\cl_2} S \subseteq \map {\cl_1} S$
where $\cl_1$ and $\cl_2$ denote topological closure in $\struct {X, \tau_1}$ and $\struct {X, \tau_2}$ respectively.

Proof:
Let $\CC_1$ be the set of closed sets $C$ in the topological space $\struct {X, \tau_1}$ such that $S \subseteq C$.
Let $\CC_2$ be the set of closed sets $C$ in the topological space $\struct {X, \tau_2}$ such that $S \subseteq C$.
Let $C \in \CC_1$.
Then from Closed Set in Coarser Topology is Closed in Finer Topology:
:$C \in \CC_2$
Hence by definition of subset:
:$\CC_1 \subseteq \CC_2$
From Intersection is Decreasing: 
:$\bigcap \CC_2 \subseteq \bigcap \CC_1$
From the definition of topological closure:
:$\cl_1 := \bigcap \CC_1$
and:
:$\cl_2 := \bigcap \CC_2$
Hence the result:
:$\map {\cl_2} S \subseteq \map {\cl_1} S$
{{qed}}
Category:Set Closures
1xneanqfqxk520td2j2zwi7liji8cfh
Theorem: Closed Unit Ball in Normed Dual Space is Weak-* Closed

Let $X$ be a normed vector space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball in $X^\ast$. 
Then we have that $B^-_{X^\ast}$ is weak-$\ast$ closed.

Proof:
From Weak-* Topology is Hausdorff, $\struct {X^\ast, w^\ast}$ is Hausdorff.
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Compact Subspace of Hausdorff Space is Closed, it follows that $B^-_{X^\ast}$ is weak-$\ast$ closed.
{{qed}}
Category:Weak-* Topologies
Category:Normed Dual Spaces
ru6rxncnsohq5g7cc6f1f8h21na7wcy
Theorem: Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Separable

Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.
Then $\struct {B^-_{X^\ast}, w^\ast}$ is separable.

Proof:
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Closed Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable. 
From Compact Metric Space is Separable, we have that $\struct {B^-_{X^\ast}, w^\ast}$ is separable.
{{qed}}
Category:Weak-* Topologies
Category:Separable Spaces
s76484go3uyj3bpijjiapyeqnixhjy8
Theorem: Compactness and Sequential Compactness are Equivalent in Metric Spaces

Let $\struct {X, d}$ be a metric space. 
Then $\struct {X, d}$ is compact {{iff}} it is sequentially compact.

Proof:
=== Necessary Condition ===
This is precisely Compact Subspace of Metric Space is Sequentially Compact in Itself.
{{qed|lemma}}
Theorem: Compact Metric Space is Separable

Let $\struct {X, d}$ be a compact metric space.
Then $\struct {X, d}$ is separable.

Proof:
From Compactness and Sequential Compactness are Equivalent in Metric Spaces, $\struct {X, d}$ is sequentially compact. 
From Sequentially Compact Metric Space is Separable, $\struct {X, d}$ is separable.
{{qed}}
Category:Separable Spaces
Category:Compact Metric Spaces
Category:Metric Spaces
0c0b8ghlibyso00ajtn701d817kswic
Theorem: Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Sequentially Compact

Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.
Then $\struct {B^-_{X^\ast}, w^\ast}$ is sequentially compact.

Proof:
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Closed Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable. 
From Compactness and Sequential Compactness are Equivalent in Metric Spaces, $\struct {B^-_{X^\ast}, w^\ast}$ is sequentially compact.
{{qed}}
Category:Sequentially Compact Spaces
Category:Weak-* Topologies
fb0ul7irudndpkmpyp6gr7wogapveeo
Theorem: Normed Dual Space of Separable Normed Vector Space is Weak-* Separable

Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Then $\struct {X^\ast, w^\ast}$ is separable.

Proof:
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
From Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Separable, $\struct {B^-_{X^\ast}, w^\ast}$ is separable space.
Let $S$ be a countable dense subset of $B^-_{X^\ast}$. 
For $n \in \N$, we have: 
:$\map {\cl_{w^\ast} } {n S} = n B^-_{X^\ast}$ 
from Dilation of Closure of Set in Topological Vector Space is Closure of Dilation.
Now, note that: 
:$\ds \bigcup_{n \mathop = 1}^\infty n S$ is countable
from Countable Union of Countable Sets is Countable.
From Closure of Union contains Union of Closures, we therefore have: 
{{begin-eqn}}
{{eqn	| l = \map {\cl_{w^\ast} } {\bigcup_{n \mathop = 1}^\infty n S}
		| o = \supseteq
		| r = \bigcup_{n \mathop = 1}^\infty \map {\cl_{w^\ast} } {n S}
}}
{{eqn	| r = \bigcup_{n \mathop = 1}^\infty B^-_{X^\ast}
}}
{{eqn	| r = X^\ast
}}
{{end-eqn}}
So:
:$\ds \map {\cl_{w^\ast} } {\bigcup_{n \mathop = 1}^\infty n S} = X^\ast$
So $\struct {X^\ast, w^\ast}$ is separable. 
{{qed}}
Theorem: Infinite-Dimensional Banach Space has Uncountable Dimension

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be an infinite-dimensional Banach space over $\Bbb F$. 
Let $B$ be a Hamel basis for $X$.
Then $B$ is uncountable.
That is, $\dim X > \aleph_0$.
{{explain|While the title talks about having an uncountable dimension, the result itself talks about the Hamel basis being uncountable. Can the two concepts be linked such that the result matches the title?}}

Proof:
Suppose that $B$ is countable. 
Let: 
:$B = \set {e_n : n \in \N}$
For each $n \in \N$, let: 
:$F_n = \span \set {e_k : k \le n}$
Since $\dim F_n = n$, $F_n$ clearly has finite dimension. 
We show that: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$ 
Let $x \in X$. 
Then there exists $n_1, \ldots, n_k \in \N$ and $\alpha_1, \ldots, \alpha_n \in \Bbb F$ such that: 
:$\ds x = \sum_{j \mathop = 1}^k \alpha_j e_{n_j}$
{{explain|Why is the above true?}}
Setting:
:$\ds N = \max_{1 \le j \le k} n_j$ 
we have that $x \in F_N$, hence:
:$\ds X = \bigcup_{n \mathop = 1}^\infty F_n$ 
From Finite Dimensional Subspace of Normed Vector Space is Closed, $F_n$ is closed for each $n \in \N$. 
From Interior of Proper Subspace of Normed Vector Space is Empty, $F_n$ has empty interior for each $n \in \N$.
So each $F_n$ is nowhere dense.
Since $X$ is a complete metric space, it is a Baire space.
However, we have written $X$ as a countable union of nowhere dense sets.
This contradicts the Baire Category Theorem.
So $B$ is uncountable.
{{qed}}
Category:Banach Spaces
bvow4314dm92c8ixf4hskjiorfw76t9
Theorem: Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional normed vector space.
Let $X^\ast$ be the normed dual space of $X$.
Then $X^\ast$ is infinite dimensional.

Proof:
{{MissingLinks}}
{{AimForCont}} $X^\ast$ is finite dimensional.
Then from Normed Dual Space of Finite-Dimensional Vector Space is Topologically Isomorphic to Original Space:
:$X^{\ast \ast} \cong X^{\ast}$
From Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space: Corollary:
:$\dim X^{\ast \ast} = \dim X^\ast$
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $\iota$ is a linear isometry from Evaluation Linear Transformation on Normed Vector Space is Linear Isometry.
Hence $\iota : X \to \iota X$ is an isometric isomorphism.
From Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space: Corollary:
:$\dim X = \dim \iota X$
From Image of Vector Subspace under Linear Transformation is Vector Subspace:
:$\iota X$ is a linear subspace of $X^{\ast \ast}$.
From Dimension of Proper Subspace is Less Than its Superspace:
:$\dim \iota X \le \dim X^{\ast \ast}$
So:
:$\dim X \le \dim X^{\ast \ast} = \dim X^\ast$
But $\dim X$ is not finite.
Hence $\dim X^\ast$ is not finite. 
This contradicts our assumption that $X^\ast$ is finite dimensional.
Hence by Proof by Contradiction $X^\ast$ is infinite dimensional.
{{qed}}
Category:Normed Vector Spaces
Category:Normed Dual Spaces
8vb2pb9s2854q6joqzfhag5n1ntp4vm
Theorem: Subset of Normed Vector Space is von Neumann-Bounded in Weak Topology iff Norm Bounded

Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space. 
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$. 
Let $B \subseteq X$. 
Then $B$ is bounded in $X$ {{iff}} it is von Neumann-bounded in $\struct {X, w}$.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Theorem: Union of Image of Convergent Sequence and Limit in Topological Space is Compact

Let $\struct {X, \tau}$ be a topological space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence such that: 
:$x_n \to x$
Then: 
:$\ds \set {x_n : n \in \N} \cup \set x$ is compact.

Proof:
Let $\family {U_\alpha}_{\alpha \in A}$ be an open cover for $\ds \set {x_n : n \in \N} \cup \set x$.
Take $\beta \in A$ be such that $x \in U_\beta$. 
Since $x_n \to x$, there exists $N \in \N$ such that $x_n \in U_\beta$ for $n \ge N$. 
For each $n < N$, we can select $U_{\alpha_1}, \ldots, U_{\alpha_{N - 1} }$ such that $x_j \in U_{\alpha_j}$ for each $1 \le j \le N - 1$.
Then we have: 
:$\ds \set {x_n : n \in \N} \cup \set x \subseteq U_\beta \cup \bigcup_{j \mathop = 1}^{N - 1} U_{\alpha_j}$ 
So every open cover has a finite subcover.
So $\ds \set {x_n : n \in \N} \cup \set x$ is compact.
{{qed}}
Category:Compact Topological Spaces
qqhxb463cdgsg0pqm08r4oyyypzmd02
Theorem: Characterization of von Neumann-Boundedness in terms of Local Basis

Let $\struct {X, \tau}$ be a topological vector space.
Let $\mathcal B$ be a local basis for $\mathbf 0_X$ in $\struct {X, \tau}$.
Let $E \subseteq X$. 
Then $E$ is von Neumann-bounded {{iff}}:
:for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$.

Proof:
=== Necessary Condition ===
Suppose that $E$ is von Neumann-bounded.
Then: 
:for each open neighbourhood $V$ of ${\mathbf 0}_X$, there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$
Since $\mathcal B$ consists of open neighbourhoods of ${\mathbf 0}_X$, we in particular have: 
:for each $V \in \mathcal B$ there exists $s > 0$ such that $E \subseteq t V$ for each $t > s$. 
{{qed|lemma}}
Theorem: Characterization of Cauchy Sequence in Topological Vector Space in terms of Local Basis

Let $\struct {X, \tau}$ be a topological vector space.
Let $\BB$ be a local basis for $\mathbf 0_X$ in $\struct {X, \tau}$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence. 
Then $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy {{iff}}:
:for each $V \in \BB$ there exists $N \in \N$ such that $x_n - x_m \in V$ for $n, m \ge N$.

Proof:
=== Necessary Condition ===
Suppose that $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Then: 
:for each open neighborhood $V$ of $\mathbf 0_X$ there exists $N \in \N$ such that: 
::$x_n - x_m \in V$ for each $n, m \ge N$. 
Since $\BB$ consists of open neighbourhoods of $\mathbf 0_X$, we in particular have:
:for each $V \in \BB$ there exists $N \in \N$ such that $x_n - x_m \in V$ for $n, m \ge N$. 
{{qed|lemma}}
Theorem: Convergent Sequence in Topological Vector Space is Cauchy

Let $\struct {X, \tau}$ be a topological vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a convergent sequence with $x_n \to x$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.

Proof:
Let $V$ be an open neighborhood of ${\mathbf 0}_X$. 
From {{Corollary|Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods|1}}, there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq V$
Since $x_n \to x$, there exists $N \in \N$ such that: 
:$x_n \in x + U$ 
for $n \ge N$. 
Since $U$ is symmetric, we have $U = -U$ and so: 
:$-x_m \in -x + U$ 
for $m \ge N$.
Then we have: 
:$x_n - x_m \in U + U \subseteq V$ 
for $n, m \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
{{qed}}
Category:Topological Vector Spaces
Category:Cauchy Sequences in Topological Vector Spaces
185lq8ljjr2ux4h0wkl2ktjdhlkneow
Theorem: Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded

Let $\struct {X, \tau}$ be a topological vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence. 
Let:
:$E = \set {x_n : n \in \N}$
Then $E$ is von Neumann-bounded.

Proof:
Let $W$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, take a balanced open neighborhood of ${\mathbf 0_X}$, $V \subseteq W$. 
From {{Corollary|Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods|1}}, there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq V$
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, we can take $U$ to be balanced.
Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that: 
:$x_n \in x_N + U$
for $n \ge N$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, there exists $M \in \N$ such that: 
:$x_N \in k U$
Then we have: 
:$x_n \in x_N + U \subseteq k U + U \subseteq k \paren {U + U} \subseteq k V$
for $n \ge N$. 
Applying Topological Vector Space as Union of Dilations of Open Neighborhood of Origin again, for each $1 \le j \le N - 1$ we have $r_j \in \N$ such that $x_j \in r_j V$. 
Taking $K = \max \set {r_1, \ldots, r_{N - 1}, k}$, we have: 
:$x_n \in K V$ for all $n \in \N$.
Since $V$ is balanced, we have that: 
:$x_n \in s V$ for all $s > K$. 
That is: 
:$x_n \in s W$ for all $s > K$. 
{{qed}}
Category:Cauchy Sequences in Topological Vector Spaces
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
osuxvm8nwfdlyhprkc9ssw4dn6qd2d3
Theorem: Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded

Let $\struct {X, \tau}$ be a topological vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence with $x_n \to x$. 
Let: 
:$E = \set {x_n : n \in \N}$
Then $E$ is von Neumann-bounded.

Proof:
From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {x_n}_{n \in \N}$ is a Cauchy sequence.
From Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded, $E$ is von Neumann-bounded.
{{qed}}
Theorem: Image of Convergent Sequence in Topological Vector Space is von Neumann-Bounded

Let $\struct {X, \tau}$ be a topological vector space. 
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence with $x_n \to x$. 
Let: 
:$E = \set {x_n : n \in \N}$
Then $E$ is von Neumann-bounded.

Proof:
From Union of Image of Convergent Sequence and Limit in Topological Space is Compact, $E \cup \set x$ is compact.
From Compact Subspace of Topological Vector Space is von Neumann-Bounded, $E \cup \set x$ is von Neumann bounded.
From Subset of von Neumann-Bounded Set is von Neumann-Bounded, $E$ is von Neumann bounded.
{{qed}}
Theorem: Von Neumann-Bounded Set in Weak-* Topology of Normed Dual of Banach Space is Norm Bounded

Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Let $E$ be a von Neumann-bounded subset of $\struct {X^\ast, w^\ast}$.
Then $E$ is a von Neumann-bounded subset of $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.

Proof:
Let $x \in X$. 
Then from Open Sets in Weak-* Topology of Topological Vector Space:
:$V = \set {f \in X^\ast : \cmod {\map f x} < 1}$ is an open neighborhood of $\mathbf 0_{X^\ast}$ in $\struct {X^\ast, w^\ast}$.
Then there exists $r > 0$ such that: 
:$E \subseteq r V$ 
Then for each $f \in E$, we have $f = r g$ for some $g \in V$. 
Then $\cmod {\map f x} = r \cmod {\map g x} < r$.
So:
:$\ds \sup_{f \in E} \cmod {\map f x} < \infty$
for each $x \in X$.
Since $X$ is a Banach space, we may apply the Banach-Steinhaus Theorem to obtain: 
:$\ds \sup_{f \in E} \norm f_{X^\ast} < \infty$
That is, there exists $M > 0$ such that: 
:$\norm f_{X^\ast} \le M$ for all $f \in E$. 
From Characterization of von Neumann-Boundedness in Normed Vector Space, we may conclude that $E$ is von Neumann-bounded in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$. 
{{qed}}
Category:Weak-* Topologies
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
8m3zmabomqdo6fcy4wayuangb3e05s7
Theorem: Point Evaluations are Continuous Linear Functionals on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

Let $K$ be a compact Hausdorff space. 
Let $\map \CC K$ be the space of complex-valued continuous functions of compact Hausdorff space.
For $k \in K$, define $\delta_k : \map \CC K \to \C$ by: 
:$\map {\delta_k} f = \map f k$ 
for each $f \in \map \CC K$.
Then $\delta_k$ is a continuous linear functional on $\map \CC K$ and further: 
:$\norm {\delta_k}_{\map \CC {K}^\ast} = 1$

Proof:
Let $k \in K$. 
$\delta_k$ is clearly linear, so we need to show that it is continuous.
From Continuity of Linear Functionals, it is enough to show that $\delta_k$ is bounded.
For each $f \in \map \CC K$ we have: 
:$\ds \cmod {\map f k} \le \sup_{x \in K} \cmod {\map f x}$ 
That is: 
:$\cmod {\map {\delta_k} f} \le \norm f_\infty$
If $K$ is a singleton, then we have $\cmod {\map {\delta_k} f} = \norm f_\infty$ for each $f \in \map \CC K$ and hence $\norm {\delta_k}_{\map \CC {K}^\ast} = 1$. 
If $K$ is not a singleton, pick distinct points $k, \ell \in K$.
From Urysohn's Lemma, there exists a continuous function $f : K \to \closedint 0 1 \subseteq \C$ with $\map f \ell = 0$ and $\map f k = 1$.
For this $f \in \map \CC K$, we have: 
:$\cmod {\map {\delta_k} f} = \cmod {\map f k} = 1 = \sup_{x \in K} \cmod {\map f x} = \norm f_\infty$ 
Hence $\norm {\delta_k}_{\map \CC K^\ast} = 1$ in this case as well.
{{qed}}
Category:Space of Continuous Functions on Compact Hausdorff Space
6hxnsen5ix66agyr8m4dl641stiyx81
Theorem: Space of Complex-Valued Continuous Functions on Compact Hausdorff Space is Separable iff Space is Metrizable

Let $K$ be a compact Hausdorff space. 
Let $\map \CC K$ be the space of complex-valued continuous functions of compact Hausdorff space.
Then $\map \CC K$ is separable {{iff}} the topology on $K$ is metrizable.

Proof:
=== Necessary Condition ===
Suppose that $\map \CC K$ is separable.
Then by Closed Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{\map \CC K^\ast}, w^\ast}$ is metrizable.
For each $k \in K$, define $\delta_k : \map \CC K \to \C$ by: 
:$\map {\delta_k} f = \map f k$ 
for each $f \in \map \CC K$.
In Point Evaluations are Continuous Linear Functionals on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space, it is established that $\delta_k$ is a continuous linear functional with $\norm {\delta_k}_{\map \CC K^\ast} = 1$.
So we can define $\delta : K \to \struct {B^-_{\map \CC K^\ast}, w^\ast}$ by: 
:$\map \delta k = \delta_k$
for each $k \in K$.
We show that $\delta$ is injective and continuous.
If $K$ is a singleton, there is nothing to show, so suppose that $K$ is not a singleton.
Let $k, \ell \in K$ have $k \ne \ell$.
By Urysohn's Lemma, there exists a continuous mapping $f : K \to \closedint 0 1$ with $\map f k = 0$ and $\map f \ell = 0$. 
So $\map {\delta_k} f \ne \map {\delta_\ell} f$ for this $f \in \map \CC K$. 
So $\delta_k \ne \delta_\ell$ if $k \ne \ell$.
We now want to show that $\delta$ is continuous.
By Continuity in Initial Topology, it is enough to show that for each $f \in \map \CC K^\ast$, $f^\wedge \circ \delta \in \map \CC K$ is continuous.
For $k \in K$, we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {f^\wedge \circ \delta} } k
		| r = \map {\delta_k} f
}}
{{eqn	| r = \map f k
}}
{{end-eqn}}
hence $f^\wedge \circ \delta = f \in \map \CC K$. 
So $\delta$ is continuous mapping.
So $\delta : K \to \struct {\map \delta K, w^\ast}$ is a continuous bijection.
From Weak-* Topology is Hausdorff, we have that $\struct {\map \delta K, w^\ast}$ is Hausdorff.
So $\delta$ is a continuous bijection between a compact space $K$ and a Hausdorff space $\struct {\map \delta K, w^\ast}$.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, this map is a homeomorphism.
So $K$ is homeomorphic to $\struct {\map \delta K, w^\ast}$.
Since $\struct {B^-_{\map \CC K^\ast}, w^\ast}$ is metrizable, so is $\struct {\map \delta K, w^\ast}$ from Subspace of Metric Space is Metric Space.
So $K$ has a metrizable topology.
{{qed|lemma}}
Theorem: Characterization of Paracompactness in T3 Space/Lemma 16

:$\forall n \in \N_{> 0}, \forall y, z \in X : y \ne z \leadsto \map {A_n} z \cap V_{n+1} \sqbrk {\map {A_n} y} = \O$

Proof:
Let $n \in \N_{> 0}$.
Let $y, z \in X : y \ne z$.
Theorem: Weak-* Metrizability of Closed Unit Ball in Normed Dual of Normed Vector Space implies Original Space is Separable

Let $X$ be a normed vector space such that: 
:$\struct {B_{X^\ast}^-, w^\ast}$ is metrizable
where $X^\ast$ is the normed dual of $X$ and $B_{X^\ast}^-$ is the closed unit ball of $X^\ast$. 
Then $X$ is separable.

Proof:
Let: 
:$K = \struct {B_{X^\ast}^-, w^\ast}$
From the Banach-Alaoglu Theorem, $K$ is compact.
From Weak-* Topology is Hausdorff, $K$ is Hausdorff.
Since $K$ is metrizable, $\map \CC K$ is separable from Space of Complex-Valued Continuous Functions on Compact Hausdorff Space is Separable iff Space is Metrizable.
Define $T : X \to \map \CC K$ by: 
:$T x = x^\wedge \restriction_{B_{X^\ast}^-}$
Clearly $T$ is a linear transformation.
Further, we have for each $f \in X^\ast$ with $\norm f_{X^\ast} = 1$:
:$\ds \cmod {\map {\paren {T x} } f} = \cmod {\map f x} \le \norm x$
from Fundamental Property of Norm on Bounded Linear Functional.
From Existence of Support Functional, for each $x \in X$ there exists $f \in X^\ast$ with $\norm f_{X^\ast} = 1$ and $\map f x = \norm x$.
Hence we have:
:$\norm {T x}_{\map \CC K} = \norm x$ 
So $T$ is a linear isometry, hence so is $T^{-1} : T \sqbrk X \to X$.
Since $\map \CC K$ is separable, $T \sqbrk X$ is separable.
From Separability of Normed Vector Space preserved under Isometric Isomorphism, $X$ is separable.
{{qed}}