avewright/proofwiki-math huggingface dataset:
Theorem: Bijective Identification Mapping is Homeomorphism from Original Topology to Identification Topology

Let $\struct {X, \tau_X}$ be a topological space.
Let $f : X \to Y$ be a bijection.
Let $\tau_Y$ be the identification topology on $Y$ induced by $T$. 
Then $f : \struct {X, \tau_X} \to \struct {Y, \tau_Y}$ is a homeomorphism.

Proof:
From the definition of the identification topology, we have:
:$\tau_Y = \set {V \in \map \PP Y : f^{-1} \sqbrk V \in \tau_X}$
From Identification Mapping is Continuous, $f$ is continuous.
Further, let $V \in \tau_X$.
We show that $f \sqbrk V \in \tau_Y$.
From Image of Preimage under Mapping, we have that $f^{-1} \sqbrk {f \sqbrk V} = V \in \tau_X$ since $f$ is a bijection.
That is, $f^{-1} \sqbrk {f \sqbrk V} \in \tau_X$.
That is, $f \sqbrk V \in \tau_Y$. 
Hence $f$ is an open mapping.
Since $f$ is also a continuous bijection, we have that $f$ is a homeomorphism.
{{qed}}
Category:Identification Topology
ood2masdm7voe4shg84f8rl6o31kbri
Theorem: Unconditionally Convergent Complex Series is Absolutely Convergent

Let $\sequence {a_n}_{n \mathop \in \N}$ be a complex sequence such that:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges unconditionally.
Then:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.

Proof:
Let $\pi : \N \to \N$ be a permutation.
Then:
:$\ds \sum_{n \mathop = 1}^\infty a_{\map \pi n} = \sum_{n \mathop = 1}^\infty \map \Re {a_{\map \pi n} } + i \sum_{n \mathop = 1}^\infty \map \Im {a_{\map \pi n} }$ converges.
From Convergence of Series of Complex Numbers by Real and Imaginary Part:
:$\ds \sum_{n \mathop = 1}^\infty \map \Re {a_{\map \pi n} }$ and $\ds \sum_{n \mathop = 1}^\infty \map \Im {a_{\map \pi n} }$
converge.
Since $\pi$ was arbitrary, we have that: 
:$\ds \sum_{n \mathop = 1}^\infty \map \Re {a_n}$ and $\ds \sum_{n \mathop = 1}^\infty \map \Im {a_n}$ converge unconditionally.
From Real Series is Unconditionally Convergent iff Absolutely Convergent:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map \Re {a_n} } < \infty$ and $\ds \sum_{n \mathop = 1}^\infty \cmod {\map \Im {a_n} } < \infty$
Writing:
:$a_n = \map \Re {a_n} + i \map \Im {a_n}$
we obtain:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} \le \sum_{n \mathop = 1}^\infty \cmod {\map \Re {a_n} } + \sum_{n \mathop = 1}^\infty \cmod {\map \Im {a_n} } < \infty$
by the Triangle Inequality: Complex Numbers.
Hence:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
{{qed}}
Category:Unconditionally Convergent Series
Category:Absolutely Convergent Series
aui589mfsq1tjn69jyh8rbvh6pkyqv4
Theorem: Sum of Schauder Basis of Zero-Limit Sequences is Weakly Unconditionally Cauchy

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis for $\map {c_0} \GF$ given by Space of Zero-Limit Sequences admits Schauder Basis.
Then:
:$\ds \sum_{n \mathop = 1}^\infty e_n$ is weakly unconditionally Cauchy.

Proof:
Let $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$ be the $1$-sequence space.
Let $\struct {\map {c_0^\ast} \GF, \norm {\, \cdot \,}_{c_0^\ast} }$ be the normed dual space of $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
We aim to show that:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map f {x_n} } < \infty$ for each $f \in \map {c_0^\ast} \GF$.
For each $x \in \map {\ell_1} \GF$, define $f_x : \map {c_0} \GF \to \GF$ by:
:$\ds \map {f_x} y = \sum_{i \mathop = 1}^n x_i y_i$
By Normed Dual Space of Space of Zero-Limit Sequences is Isometrically Isomorphic to 1-Sequence Space, $f_x \in \map {c_0^\ast} \GF$ for each $x \in \map {\ell_1} \GF$.
Define $T : \map {\ell_1} \GF \to \map {c_0^\ast} \GF$ by:
:$T x = f_x$ for each $x \in \map {\ell_1} \GF$
From Normed Dual Space of Space of Zero-Limit Sequences is Isometrically Isomorphic to 1-Sequence Space, $T$ is an isometric isomorphism.
Let $f \in \map {c_0^\ast} \GF$. 
Then since $T$ is a bijection, there exists $x \in \map {\ell_1} \GF$ such that $f = f_x$. 
We then have:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map f {e_n} } = \sum_{n \mathop = 1}^\infty \cmod {\map {f_x} {e_n} } = \sum_{n \mathop = 1}^\infty \cmod {x_n}$
Since $x \in \map {\ell_1} \GF$, we have that:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map f {e_n} } < \infty$
as required.
{{qed}}
Category:Space of Zero-Limit Sequences
Category:Weakly Unconditionally Cauchy Series
kxqf09ptjjk1sigg46el1lvx5gae5kr
Theorem: Series is Weakly Unconditionally Cauchy iff Linear Transformation sending Schauder Basis of Space of Zero-Limit Sequences to Terms of Sequence is Bounded

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis for $\map {c_0} \GF$ given by Space of Zero-Limit Sequences admits Schauder Basis.
Then:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ is weakly unconditionally Cauchy
{{iff}}:
:there exists a bounded linear transformation $T : \map {c_0} \GF \to X$ such that $T e_n = x_n$.

Proof:
=== Necessary Condition ===
Suppose that:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ is weakly unconditionally Cauchy
Let $\struct {\map {c_{00} } \GF, \norm {\, \cdot \,}_\infty}$ be the space of almost-zero sequences.
Define a mapping $T : \map {c_{00} } \GF \to X$ by:
:$\ds T \xi = \sum_{n \mathop = 1}^\infty \map \xi n x_n$ for each $\xi \in \map {c_{00} } \GF$.
This map is well-defined since this sum is finite.
Further, for $\xi_1, \xi_2 \in \map {c_{00} } \GF$ and $\lambda \in \GF$, we have:
{{begin-eqn}}
{{eqn	| l = \map T {\xi_1 + \lambda \xi_2}
		| r = \sum_{n \mathop = 1}^\infty \map {\paren {\xi_1 + \lambda \xi_2} } n x_n
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map {\xi_1} n x_n + \lambda \sum_{n \mathop = 1}^\infty \map {\xi_2} n x_n 
}}
{{eqn	| r = T \xi_1 + \lambda T \xi_2 
}}
{{end-eqn}}
Hence $T$ is linear.
From Characterization of Weak Unconditional Cauchyness of Series in Banach Space, there exists $C > 0$ such that:
:$\ds \norm {\sum_{n \mathop = 1}^\infty \map \xi n x_n}_X \le C \norm \xi_\infty$ for all $\xi \in \map {c_{00} } \GF$. 
That is:
:$\norm {T \xi}_X \le C \norm \xi_\infty$ for all $\xi \in \map {c_{00} } \GF$.
Hence $T : \map {c_{00} } \GF \to X$ is a bounded linear transformation.
From Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain, there exists an extension $\widetilde T : \map {c_0} \GF \to X$ that is still a bounded linear transformation.
Further, we have:
{{begin-eqn}}
{{eqn	| l = \map {\widetilde T} {e_n}
		| r = \map T {e_n}
}}
{{eqn	| r = \sum_{j \mathop = 1}^\infty \map {e_n} j x_j
}}
{{eqn	| r = \sum_{j \mathop = 1}^\infty \delta_{n j} x_j 
}}
{{eqn	| r = x_n
}}
{{end-eqn}}
Hence $\widetilde T$ is our desired bounded linear transformation.
{{qed|lemma}}
Theorem: Rearrangement of Convergent Sequence Converges to Same Limit

Let $\struct {X, \tau}$ be a topological space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a convergent sequence in $X$ with limit $x$.
Let $\pi : \N \to \N$ be a permutation.
Then $\sequence {x_{\map \pi n} }_{n \mathop \in \N}$ converges to $x$ in $\struct {X, \tau}$.

Proof:
Let $U$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$, there exists $N \in \N$ such that:
:$x_n \in U$ for $n \ge N$.
Since $\pi$ is a bijection, there exists $M \in \N$ such that:
:$\set {1, 2, \ldots, N} \subseteq \pi \sqbrk {\set {1, 2, \ldots, M} }$ 
Then, since $\pi$ is a bijection: 
:for $m \ge M + 1$, we have $\map \pi m > N$. 
Hence, $x_{\map \pi m} \in U$ for $m \ge M + 1$. 
Since $U$ was an arbitrary open neighborhood of $x$, we have that $\sequence {x_{\map \pi n} }_{n \mathop \in \N}$ converges to $x$ in $\struct {X, \tau}$.
{{qed}}
Category:Topological Spaces
62kaz4vpz40rke87m2scg91oug7d6jd
Theorem: Countably Infinite Set has Uncountable Family of Subsets with Finite Intersection

Let $\CC$ be a countably infinite set.
Then there exists an uncountable family $\SS \subseteq \powerset \CC$ such that for each $A, B \in \SS$, we have that:
:$A \cap B$ is finite.

Proof:
Since $\CC$ is countably infinite, there exists a bijection $f : \CC \to \N$. 
From Rational Numbers are Countably Infinite, the set of rational numbers $\Q$ is countably infinite. 
Hence, there exists a bijection $g : \Q \to \N$.
Then $h = f^{-1} \circ g : \Q \to \CC$ is a bijection.
We will now prove the theorem for $\Q$, with the existence of $h$ allowing us to deduce the theorem for $\CC$. 
Let $\theta \in \R \setminus \Q$. 
From Rationals are Everywhere Dense in Topological Space of Reals, there exists a sequence $\sequence {\map {q_n} \theta}_{n \mathop \in \N}$ in $\Q$ with $\map {q_n} \theta \to \theta$ as $n \to \infty$.
Let $A_\theta = \set {\map {q_n} \theta : n \in \N}$.
Note that $A_\theta$ must be countably infinite, since $\sequence {\map {q_n} \theta}_{n \mathop \in \N} \subseteq \Q$ is not eventually $\theta \in \R \setminus \Q$. 
Now take $\theta_1 \ne \theta_2$
{{WLOG}} suppose that $\theta_1 < \theta_2$. 
Take $\epsilon > 0$ such that:
:$\ds \epsilon < \frac {\theta_2 - \theta_1} 2$
so that:
:$\theta_1 + \epsilon < \theta_2 - \epsilon$
Since $\map {q_n} {\theta_1} \to \theta_1$, there exists $N_1 \in \N$ such that $\map {q_n} {\theta_1} \in \openint {\theta_1 - \epsilon} {\theta_1 + \epsilon}$ for $n \ge N_1$.
Since $\map {q_n} {\theta_2} \to \theta_2$, there exists $N_2 \in \N$ such that $\map {q_n} {\theta_2} \in \openint {\theta_2 - \epsilon} {\theta_2 + \epsilon}$ for $n \ge N_2$.
These two sets are disjoint sets.
Hence we have:
:$A_{\theta_1} \cap A_{\theta_2} \subseteq \set {\map {q_n} {\theta_1} : 1 \le N < N_1} \cup \set {\map {q_n} {\theta_2} : 1 \le N < N_2}$
Hence $A_{\theta_1} \cap A_{\theta_2}$ is finite. 
Since each $A_\theta$ is countably infinite, this establishes that each $A_\theta$ is distinct.
We set $S = \set {A_\theta : \R \setminus \Q}$.
From Irrational Numbers are Uncountably Infinite, since each $A_\theta$ is distinct, $S$ is uncountable.
Hence we have satisfied the requirement for $\Q$. 
Now let $\AA_\theta = h \sqbrk {A_\theta} \subseteq \CC$ for each $\theta \in \R$.
From Image of Intersection under Injection, we have $\AA_{\theta_1} \cap \AA_{\theta_2} = h \sqbrk {A_{\theta_1} \cap A_{\theta_2} }$ for $\theta_1 \ne \theta_2$.
From Image of Mapping from Finite Set is Finite, $\AA_{\theta_1} \cap \AA_{\theta_2}$ is finite.
Since $h$ is a bijection, we have that each $\AA_\theta$ is distinct. 
Set $\SS = \set {\AA_\theta : \theta \in \R}$.
From Irrational Numbers are Uncountably Infinite, since each $A_\theta$ is distinct, $\SS$ is uncountable.
Hence we are done.
{{qed}}
Theorem: Composition Mapping of Restriction Mappings

Let $S$, $T$, and $U$ be sets.
Let $f: S \to T$ and $g: T \to U$ be mappings.
Let $A \subseteq S$, $B \subseteq T$ and $C \subseteq U$.
Let $f \sqbrk S \subseteq B$ and $g \sqbrk T \subseteq C$.
Then:
{{begin-eqn}}
{{eqn| l = \paren{g \circ f} \restriction_{A \times C}
     | r = \paren{g \restriction_{B \times C} } \circ \paren{f \restriction_{A \times B} }
}}
{{end-eqn}}
where:
:$\paren{g \circ f} \restriction_{A \times C}$ is the restriction of $g \circ f$ to $A \times C$
:$f \restriction_{A \times B}$ is the restriction of $f$ to $A \times B$
:$g \restriction_{B \times C}$ is the restriction of $g$ to $B \times C$

Proof:
We have:
{{begin-eqn}}
{{eqn| q = \forall a \in A
     | l = \map {\paren{g \circ f} \restriction_{A \times C} } a
     | r = \map {g \circ f} a
     | c = {{Defof|Restricted Mapping}}
}}
{{eqn| r = \map g {\map f a}
     | c = {{Defof|Composite Mapping}}
}}
{{eqn| r = \map {g \restriction_{B \times C} } {\map f a}
     | c = {{Defof|Restricted Mapping}} with $\map f a \in B$ and $g \sqbrk T \subseteq C$
}}
{{eqn| r = \map {g \restriction_{B \times C} } {\map {f \restriction_{A \times B} } a}
     | c = {{Defof|Restricted Mapping}} with $a \in A$ and $f \sqbrk S \subseteq B$   
}}
{{eqn| r = \map{\paren{\paren{g \restriction_{B \times C} } \circ \paren{f \restriction_{A \times B} } } } a
     | c = {{Defof|Composite Mapping}}
}}
{{end-eqn}}
From Equality of Mappings:
{{begin-eqn}}
{{eqn| l = \paren{g \circ f} \restriction_{A \times C}
     | r = \paren{g \restriction_{B \times C} } \circ \paren{f \restriction_{A \times B} }
}}
{{end-eqn}}
{{qed}}
Category:Composite Mappings
Category:Restrictions
9w8yu6wchyhzez4x34eg79t4abzrubw
Theorem: Closure of Totally Bounded Subspace is Totally Bounded

Let $\struct {M, d}$ be a metric space.
Let $A \subseteq M$ be totally bounded.
Let $\cl$ be the topological closure.
Then $\map \cl A$ is totally bounded.

Proof:
Let $\epsilon > 0$.
Since $A$ is totally bounded, there exists $x_1, \ldots, x_n \in A$ such that:
:$\ds A \subseteq \bigcup_{i \mathop = 1}^n \map {B_{\epsilon/2} } {x_i}$
where $\map {B_\epsilon} {x_i}$ is the open ball of radius $\epsilon$, centered at $x_i$.
From Set Closure Preserves Set Inclusion, we have:
:$\ds \map \cl A \subseteq \map \cl {\bigcup_{i \mathop = 1}^n \map {B_{\epsilon/2} } {x_i} }$
From Closure of Finite Union equals Union of Closures, we have:
:$\ds \map \cl {\bigcup_{i \mathop = 1}^n \map {B_{\epsilon/2} } {x_i} } = \bigcup_{i \mathop = 1}^n \map \cl {\map {B_{\epsilon/2} } {x_i} }$
From Closure of Open Ball in Metric Space, we have $\map \cl {\map {B_{\epsilon/2} } {x_i} } \subseteq \map {B_{\epsilon/2}^-} {x_i}$ where $\map {B_{\epsilon/2}^-} {x_i}$ denotes the closed ball.
Note that $\map d {x, x_i} \le \epsilon/2$ implies that $\map d {x, x_i} < \epsilon$.
Hence $\map {B_{\epsilon/2}^-} {x_i} \subseteq \map {B_\epsilon} {x_i}$.
We therefore have:
:$\ds \map \cl A \subseteq \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i}$
Since $\epsilon$ was arbitrary, we have that $\map \cl A$ is totally bounded.
{{qed}}
Category:Totally Bounded Metric Spaces
ob4frt90034yjz1bkhtwxe9o1s64slr
Theorem: Subspace of Complete Metric Space is Relatively Compact iff Totally Bounded

Let $\struct {M, d}$ be a metric space.
Let $A \subseteq M$.
We have that $A$ is relatively compact in $\struct {M, d}$ {{iff}} $A$ is totally bounded.

Proof:
Let $\cl$ be the topological closure.
Theorem: Unconditionally Convergent Series in Banach Space is Weakly Unconditionally Cauchy

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ is unconditionally convergent.
Then:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ is weakly unconditionally Cauchy.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $f \in X^\ast$.
Let $\pi : \N \to \N$ be a permutation. 
Then:
:$\ds \sum_{n \mathop = 1}^\infty x_{\map \pi n}$ converges.
We then have:
{{begin-eqn}}
{{eqn	| l = \map f {\sum_{n \mathop = 1}^\infty x_{\map \pi n} }
		| r = \map f {\lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N x_{\map \pi n} }
}}
{{eqn	| r = \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \map f {x_{\map \pi n} }
		| c = Continuous Mappings preserve Convergent Sequences, {{Defof|Linear Functional}}
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map f {x_{\map \pi n} }
}}
{{end-eqn}}
Since $\pi$ was arbitrary, we have that:
:$\ds \sum_{n \mathop = 1}^\infty \map f {x_n}$ is unconditionally convergent.
From Unconditionally Convergent Complex Series is Absolutely Convergent, we have that:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {\map f {x_n} } < \infty$
Since $f \in X^\ast$ was arbitrary, we have that:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ is weakly unconditionally Cauchy.
{{qed}}
Theorem: Weakly Open Set in Hausdorff Topological Vector Space is Open

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$ with weak topology $w$. 
Let $U$ be an open set in $\struct {X, w}$.
Then $U$ is open in $\struct {X, \tau}$.

Proof:
Let $X^\ast$ be the topological dual of $X$.
From the definition of the weak topology, $w$ is the initial topology on $X$ generated by $X^\ast$. 
Since every function in $X^\ast$ is continuous, we have that $w \subseteq \tau$ from Domain Topology Contains Initial Topology iff Mappings are Continuous.
{{qed}}
Category:Topological Vector Spaces
Category:Weakly Open Sets
cnoakn2oc7pob1gp7f8xu8lavbyzxqp
Theorem: Weak-* Open Set in Normed Dual Space is Open

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $w^\ast$ be a weak-$\ast$ topology on $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
Let $U$ be an open set in $\struct {X^\ast, w^\ast}$. 
Then $U$ is open in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.

Proof:
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
From the definition of the weak-$\ast$ topology, $w^\ast$ is the initial topology on $X^{\ast \ast}$ generated by $\iota X$.
By the definition of the second normed dual, $\iota x : X^\ast \to \GF$ is continuous for each $x \in X$. 
Hence by Domain Topology Contains Initial Topology iff Mappings are Continuous, we have that the topology on $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ is finer than $w^\ast$.
Hence it follows that any open set in $\struct {X^\ast, w^\ast}$ is open in $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
{{qed}}
Category:Weak-* Topologies
Category:Normed Dual Spaces
l3w1f7u58nedxhj52c99tm06a48jv4x
Theorem: Weak-* Topology and Norm Topology Coincide on Norm Compact Subsets of Normed Dual Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $w^\ast$ be the weak-$\ast$ topology on $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$.
Let $\struct {K, \norm {\, \cdot \,}_{X^\ast} }$ with $K \subseteq X^\ast$ be compact.
Then the open sets of $\struct {K, w^\ast}$ and $\struct {K, \norm {\, \cdot \,}_{X^\ast} }$ coincide.

Proof:
We show that the identity mapping $\iota : \struct {K, \norm {\, \cdot \,}_{X^\ast} } \to \struct {K, w^\ast}$ is a homeomorphism.
From Weak-* Open Set in Normed Dual Space is Open, every open set in $\struct {K, w^\ast}$ is open in $\struct {K, \norm {\, \cdot \,}_{X^\ast} }$.
Hence $\iota : \struct {K, \norm {\, \cdot \,}_{X^\ast} } \to \struct {K, w^\ast}$ is continuous.
It is also clearly a bijection.
From Weak-* Topology is Hausdorff, $\struct {K, w^\ast}$ is Hausdorff.
Further, $\struct {K, \norm {\, \cdot \,}_{X^\ast} }$ is compact.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota : \struct {K, \norm {\, \cdot \,}_{X^\ast} } \to \struct {K, w^\ast}$ is a homeomorphism.
Hence $U$ is open in $\struct {K, \norm {\, \cdot \,}_{X^\ast} }$ {{iff}} $U$ is open in $\struct {K, w^\ast}$.
{{qed}}
Category:Weak-* Topologies
he9to5c9t6q888ql79ltv2ssb0lqnvw
Theorem: Weak Topology and Original Topology Coincide on Originally Compact Subsets of Hausdorff Topological Vector Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a Hausdorff topological vector space over $\GF$ with weak topology $w$.
Let $\struct {K, \tau}$ be compact. 
Let $U \subseteq X$.
Then $U$ is open in $\struct {K, \tau}$ {{iff}} $U$ is open in $\struct {K, w}$.

Proof:
We show that the identity mapping $\iota : \struct {K, \tau} \to \struct {K, w}$ is a homeomorphism.
From Weakly Open Set in Hausdorff Topological Vector Space is Open, every open set in $\struct {K, w}$ is open in $\struct {K, \tau}$.
Hence $\iota : \struct {K, \norm {\, \cdot \,}_{X^\ast} } \to \struct {K, w^\ast}$ is continuous.
It is also clearly a bijection.
From Weak Topology on Topological Vector Space over Hausdorff Topological Field is Hausdorff, $\struct {K, w^\ast}$ is Hausdorff.
Further, $\struct {K, \tau}$ is compact.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota : \struct {K, \tau} \to \struct {K, w}$ is a homeomorphism.
Hence $U$ is open in in $\struct {K, \tau}$ {{iff}} $U$ is open in $\struct {K, w}$. 
{{qed}}
Category:Weak Topologies on Topological Vector Spaces
ir2hqjparp7tk0118wthsl4nec07h0h
Theorem: Compact Linear Transformation is Weakly Compact

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces over $\GF$.
Let $T : X \to Y$ be compact linear transformation.
Then $T$ is weakly compact.

Proof:
Let $B_X^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\cl$ be the topological closure taken in $\struct {X, \norm {\, \cdot \,}_X}$.
Since $T$ is compact, $\map \cl {T \sqbrk {B_X^-} }$ is compact.
From Weak Topology and Original Topology Coincide on Originally Compact Subsets of Hausdorff Topological Vector Space, we have that $\map \cl {T \sqbrk {B_X^-} }$ is weakly compact.
Hence $T$ is weakly compact.
{{qed}}
Category:Weakly Compact Linear Transformations
Category:Compact Linear Transformations
her84xeobwaudjavucj48991yr2ggh6
Theorem: Subset of Totally Bounded Metric Space is Totally Bounded

Let $\struct {M, d}$ be a totally bounded metric space.
Let $A \subseteq M$.
Then $A$ is totally bounded.

Proof:
Let $\epsilon > 0$.
Since $M$ is totally bounded, there exists $x_1, \ldots, x_n \in M$ such that:
:$\ds M = \bigcup_{j \mathop = 1}^n \map {B_{\epsilon/2} } {x_j}$
where $\map {B_{\epsilon/2} } {x_j}$ is the open ball with radius $\epsilon/2$ and center $x_i$. 
We have:
:$\ds A \subseteq \bigcup_{j \mathop = 1}^n \map {B_{\epsilon/2} } {x_j}$
{{WLOG}} suppose that $A \cap \map {B_{\epsilon/2} } {x_j} \ne \O$ for each $1 \le j \le n$, else we can remove these balls from the union.
For each $1 \le j \le n$, select $a_j \in A \cap \map {B_{\epsilon/2} } {x_j}$.
We argue that $\map {B_{\epsilon/2} } {x_j} \subseteq \map {B_\epsilon} {a_j}$. 
Let $x \in \map {B_{\epsilon/2} } {x_j}$. 
Then $\map d {x, x_j} < \epsilon/2$.
Hence by {{Metric-space-axiom|2}}, we have $\map d {x, a_j} \le \map d {x, x_j} + \map d {x_j, a_j} < \epsilon$.
Hence $x \in \map {B_\epsilon} {a_j}$.
So we have $\map {B_{\epsilon/2} } {x_j} \subseteq \map {B_\epsilon} {a_j}$ for each $1 \le j \le n$.
We therefore have:
:$\ds A \subseteq \bigcup_{j \mathop = 1}^n \map {B_\epsilon} {a_j}$ 
with $a_1, \ldots, a_n \in A$.
Hence $A$ is totally bounded.
{{qed}}
Category:Totally Bounded Metric Spaces
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