avewright/proofwiki-math huggingface dataset:
Theorem: Characterization of Unconditional Convergence of Series in Banach Space

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 $X$.
{{TFAE}}
:$(1): \quad$ $\ds \sum_{n \mathop = 1}^\infty x_n$ converges unconditionally
:$(2): \quad$ for each strictly increasing sequence $\sequence {n_k}_{k \mathop \in \N}$ in $\N$, $\ds \sum_{k \mathop = 1}^\infty x_{n_k}$ converges
:$(3): \quad$ $\ds \sum_{n \mathop = 1}^\infty \epsilon_n x_n$ converges for each $\sequence {\epsilon_n}_{n \mathop \in \N} \in \set {-1, 1}^\N$
:$(4): \quad$ for each $\epsilon > 0$, there exists $n \in \N$ such that for all finite sets $F \subseteq \set {j : j \ge n + 1}$, we have $\ds \norm {\sum_{j \mathop \in F} x_j}_X < \epsilon$

Proof:
=== $(4)$ implies $(1)$ ===
Suppose that:
:for each $\epsilon > 0$, there exists $n \in \N$ such that for all finite sets $F \subseteq \set {j : j \ge n + 1}$ we have $\ds \norm {\sum_{j \mathop \in F} x_j}_X < \epsilon$
Let $\pi : \N \to \N$ be a permutation.
Let $\epsilon > 0$.
Take $M \in \N$ such that for every finite set $F \subseteq \set {j : j \ge M + 1}$, we have:
:$\ds \norm {\sum_{j \mathop \in F} x_j}_X < \epsilon$
Since $\pi$ is an injection, there exists $N \in \N$ such that $\map \pi n \ge M + 1$ for each $n \ge N$.
Then for $n > m \ge N$, we have that $\set {\map \pi j : m + 1 \le j \le n}$ is a finite subset of $\set {j : j \ge M + 1}$. 
Hence:
:$\ds \norm {\sum_{k \mathop = 1}^n x_{\map \pi k} - \sum_{k \mathop = 1}^m x_{\map \pi k} } = \norm {\sum_{k \mathop = m + 1}^n x_{\map \pi k} }_X < \epsilon$ for $n > m \ge N$.
Hence the sequence $\sequence {\sum_{k \mathop = 1}^n x_{\map \pi k} }_{n \mathop \in \N}$ is Cauchy.
Since $X$ is a Banach space, we have that:
:$\ds \sum_{n \mathop = 1}^\infty x_{\map \pi n}$ is convergent. 
Since $\pi : \N \to \N$ is an arbitrary permutation, we have that:
:$\ds \sum_{n \mathop = 1}^\infty x_n$ converges unconditionally
{{qed|lemma}}
Theorem: Weakly Compact Set in Banach Space is Angelic

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\struct {K, w}$ be compact.
Then $\struct {K, w}$ is angelic.

Proof:
Let $\cl_w$ be the closure taken in $\struct {X, w}$. 
Let $\cl$ be the closure taken in $\struct {X, \norm {\, \cdot \,}_X}$.
Let $A \subseteq K$.
Let $x \in \map {\cl_w} A$.
From Banach Space with Weak Topology has Countable Tightness, $\struct {X, w}$ has countable tightness.
Hence there exists a countable set $S \subseteq A$ such that $x \in \map {\cl_w} S$.
Let:
:$L = \map \cl {\map \span S}$
From Characterization of Separable Normed Vector Space, $L$ is separable.
From Topological Closure is Closed, $L$ is closed in $\struct {X, \norm {\, \cdot \,}_X}$.
From Closed Linear Span is Closed Vector Subspace, $L$ is a norm-closed linear subspace of $X$.
From Closed Subspace of Banach Space forms Banach Space, $L$ is a Banach space.
From Closed Vector Subspace in Normed Vector Space is Weakly Closed, we have that $L$ is also closed in $\struct {X, w}$.
Hence we have $\map {\cl_w} S \subseteq L$.
Further, since $S \subseteq A$, we have $\map {\cl_w} S \subseteq \map {\cl_w} A \subseteq K$.
Hence $\map {\cl_w} S$ is closed in $\struct {K, w}$.
Since $\struct {K, w}$ is compact, we have that $\struct {\map {\cl_w} S, w}$ is compact from Closed Subspace of Compact Space is Compact.
Since $L$ is separable, we obtain that $\struct {\map {\cl_w} S, w}$ is metrizable from Weakly Compact Set in Separable Banach Space is Weakly Metrizable.
Recall that $x \in \map {\cl_w} S$.
Hence, from Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists a sequence in $S$ such that $x_n \to x$ in $\struct {\map {\cl_w} S, w}$.
Since $S \subseteq A$, we have obtained a sequence in $A$ converging weakly to $x$.
Since $x \in \map {\cl_w} A$ was arbitrary, we have that $\struct {K, w}$ is angelic.
{{qed}}
Theorem: Weakly Compact Set in Banach Space whose Normed Dual is Weak-* Separable is Weakly Metrizable

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space.
Let $w$ be the weak topology on $X$.
Let $\struct {K, w}$ be compact with $K \subseteq X$.
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 $X^\ast$.
Suppose that $\struct {X^\ast, w^\ast}$ is separable.
Then $\struct {K, w}$ is metrizable.

Proof:
Let $\SS_\ast$ be a countable everywhere dense subset of $\struct {X^\ast, w^\ast}$.
From Weak-* Dense Subset of Normed Dual Space Separates Points, $\SS_\ast$ separates points.
Let $\SS = \set {f \restriction_K : f \in \SS_\ast}$. 
Let $\tau$ be the initial topology on $K$ generated by $\SS$.
From Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable, $\tau$ is metrizable.
Let $\iota : \struct {K, w} \to \struct {K, \tau}$ be the inclusion mapping.
First, $\iota$ is clearly a bijection.
We want to show that it is a homeomorphism.
From Subspace Topology on Initial Topology is Initial Topology on Restrictions, the weak topology on $K$ is generated by $\set {f \restriction_K : f \in X^\ast}$. 
Hence $\tau \subseteq w$.
Hence $\iota$ is continuous.
Further, since $\tau$ is metrizable, it is Hausdorff from Metric Space is Hausdorff.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota$ is a homeomorphism.
Hence $w = \tau$ and $\struct {K, w}$ is metrizable.
{{qed}}
Category:Weakly Compact Sets
Category:Weak-* Topologies
Category:Metrizable Spaces
lfcgkpw7qukspvy88recndpb758yfw0
Theorem: Weakly Compact Set in Separable Banach Space is Weakly Metrizable

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a separable Banach space.
Let $w$ be the weak topology on $X$.
Let $\struct {K, w}$ be compact with $K \subseteq X$.
Let $\struct {K, w}$ is metrizable.

Proof:
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 $X^\ast$.
From Normed Dual Space of Separable Normed Vector Space is Weak-* Separable, $\struct {X^\ast, w^\ast}$ is separable.
From Weakly Compact Set in Banach Space whose Normed Dual is Weak-* Separable is Weakly Metrizable, $\struct {K, w}$ is metrizable.
{{qed}}
Category:Weakly Compact Sets
Category:Metrizable Spaces
Category:Separable Spaces
Category:Banach Spaces
1em48i5cziqrf06sp7ocmqq11k3fic3
Theorem: Closed Vector Subspace in Normed Vector Space is Weakly Closed

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $Y$ be a closed vector subspace of $X$.
Then $Y$ is weakly closed.

Proof:
From Linear Subspace is Convex Set, $Y$ is convex. 
Further, $Y$ is norm closed in $\struct {X, \norm {\, \cdot \,}_X}$.
From Mazur's Theorem: Corollary, $Y$ is weakly closed.
{{qed}}
Category:Normed Vector Spaces
d800cw8h3af1ldz56zg7syvq6rfhglr
Theorem: Bounded Subset of Banach Space is Relatively Weakly Compact iff Weak-* Closure in Second Normed Dual is Contained in Embedding of Original Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $S \subseteq X$ be bounded.
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $w^\ast$ be the weak-$\ast$ topology on $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$.
Let $\cl_w$ be closure taken in $\struct {X, w}$.
Let $\cl_{w^\ast}$ be closure taken in $\struct {X^{\ast \ast}, w^\ast}$.
Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.
Then $S$ is relatively compact in $\struct {X, w}$ {{iff}} $\map {\cl_{w^\ast} } {\iota S} \subseteq \iota X$. 
That is, $\struct {\map {\cl_w} S, w}$ is compact {{iff}} $\map {\cl_{w^\ast} } {\iota S} \subseteq \iota X$.

Proof:
=== Necessary Condition ===
Suppose that $\struct {\map {\cl_w} S, w}$ is compact.
Let $\phi \in \map {\cl_{w^\ast} } {\iota S}$.
Then by Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {\phi_\lambda}_{\lambda \mathop \in \Lambda}$ in $\iota S$ converging weak-$\ast$ to $\phi$.
Write $\phi_\lambda = \iota x_\lambda$ for $x_\lambda \in S$.
Then $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ is a net in $S$.
Since $\struct {\map {\cl_w} S, w}$ is compact, there exists a directed set $\struct {A, \sqsubseteq}$ and an increasing cofinal mapping $\mu : A \to \Lambda$ such that:
:$\family {x_{\map \mu \alpha} }_{\alpha \mathop \in A}$ is a subnet of $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$
:$\family {x_{\map \mu \alpha} }_{\alpha \mathop \in A}$ converges weakly to $x \in \map {\cl_w} S$.
From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual, $\iota : \struct {X, w} \to \struct {\iota X, w^\ast}$ is a homeomorphism.
Hence $\family {\phi_{\map \mu \alpha} }_{\alpha \mathop \in A} = \family {\iota x_{\map \mu \alpha} }_{\alpha \mathop \in A}$ converges to $\iota x$.
From Subnet of Convergent Net Converges to Same Limit, we have that $\family {\phi_{\map \mu \alpha} }_{\alpha \mathop \in A}$ converges to $\phi$. 
From Weak-* Topology is Hausdorff, $\struct {X^{\ast \ast}, w^\ast}$ is a Hausdorff space.
From Characterization of Hausdorff Property in terms of Nets, $\family {\phi_{\map \mu \alpha} }_{\alpha \mathop \in A}$ has a unique limit. 
Hence $\phi = \iota X$.
That is, $\phi \in \iota X$.
Since $\phi \in \map {\cl_{w^\ast} } {\iota S}$ was arbitrary, we have $\map {\cl_{w^\ast} } {\iota S} \subseteq \iota X$.
{{qed|lemma}}
Theorem: Bounded Subset of Normed Dual Space is Relatively Weak-* Compact

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach 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 the weak-$\ast$ topology on $X^\ast$. 
Let $\cl_{w^\ast}$ be the closure taken in $\struct {X^\ast, w^\ast}$.
Let $S \subseteq X^\ast$ be bounded.
Then $S$ is relatively compact in $\struct {X^\ast, w^\ast}$.
That is, $\struct {\map {\cl_{w^\ast} } S, w^\ast}$ is compact.

Proof:
Let $B_{X^\ast}^-$ be the closed unit ball in $X^\ast$. 
Let:
:$\ds \delta = \sup_{f \mathop \in S} \norm f_{X^\ast}$
Then we have:
:$S \subseteq \delta B_{X^\ast}^-$
From the Banach-Alaoglu Theorem, $\struct {B_{X^\ast}^-, w^\ast}$ is compact.
Hence from Dilation of Compact Set in Topological Vector Space is Compact, $\struct {\delta B_{X^\ast}^-, w^\ast}$ is compact.
From Weak-* Topology is Hausdorff, $\struct {X^\ast, w^\ast}$ is Hausdorff.
From Compact Subspace of Hausdorff Space is Closed, $\delta B_{X^\ast}$ is closed in $\struct {X^\ast, w^\ast}$.
From Set is Closed iff Equals Topological Closure and Topological Closure of Subset is Subset of Topological Closure, we have that:
:$\map {\cl_{w^\ast} } S \subseteq  \delta B_{X^\ast}^-$
From Closed Subspace of Compact Space is Compact, $\struct {\map {\cl_{w^\ast} } S, w^\ast}$ is compact.
{{qed}}
Category:Weak-* Topologies
1i7jt9639cdvejnqonijcq5uo1yk3jo
Theorem: Tail of Directed Set is Directed

Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\lambda_0 \in \Lambda$.
Let $\Lambda_0 = \set {\lambda \in \Lambda : \lambda \succeq \lambda_0}$.
Let $\preceq_0$ be the restriction of $\preceq$ to $\Lambda_0$.
Then $\struct {\Lambda_0, \preceq_0}$ is directed.

Proof:
Let $\mu_1, \mu_2 \in \Lambda_0$. 
We can find $\lambda \in \Lambda$ such that $\mu_1 \preceq \lambda$ and $\mu_2 \preceq \lambda$.
Since $\lambda_0 \preceq \mu_1$ and $\lambda_0 \preceq \mu_2$, we have that $\lambda_0 \preceq \lambda$ by transitivity.
Hence $\lambda \in \Lambda_0$, with $\mu_1 \preceq \lambda$ and $\mu_2 \preceq \lambda$.
Hence $\struct {\Lambda_0, \preceq_0}$ is directed.
{{qed}}
Category:Directed Preorderings
2mf7ka6c7omifye0txh8e05ugo6bqsk
Theorem: Translation of Bounded Subset of Normed Vector Space is Bounded

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space.
Let $A \subseteq X$ be a bounded set.
Let $x \in X$.
Then $A + x$ is bounded.

Proof:
Since $A$ is bounded, there exists $M > 0$ such that:
:$\norm y_X \le M$ for all $y \in A$.
Hence from {{NormAxiomVector|3}}, we have:
:$\norm {y + x}_X \le M + \norm x_X < \infty$ for all $y \in A$.
Hence $\norm u_X \le M + \norm x_X$ for all $u \in A + x$.
Hence $A + x$ is bounded.
{{qed}}
Category:Normed Vector Spaces
jm4kauyvybi0vonrdrtt4j55uakyyaq
Theorem: Topological Closure of Translation in Topological Vector Space is Translation of Topological Closure

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $x \in X$. 
Let $A \subseteq X$ be non-empty.
Let $\cl$ be closure taken in $\struct {X, \tau}$.
Then $\map \cl {A + x} = \map \cl A + x$.

Proof:
From Characterization of Closedness in terms of Nets, we have $y \in \map \cl {A + x}$ {{iff}} there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {u_\lambda + x}_{\lambda \mathop \in \Lambda}$ with $u_\lambda \in A$ for each $\lambda \in \Lambda$ such that:
:$\family {u_\lambda + x}_{\lambda \mathop \in \Lambda}$ converges to $y$.
From Constant Sequence in Topological Space Converges and Linear Combination of Convergent Sequences in Topological Vector Space is Convergent, this is equivalent to:
:$\family {u_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $y - x$.
From Characterization of Closedness in terms of Nets, there exists such a $\family {u_\lambda}_{\lambda \mathop \in \Lambda}$ {{iff}} $y - x \in \map \cl A$.
That is, {{iff}} $y \in \map \cl A + x$. 
{{qed}}
Category:Topological Vector Spaces
17jg5tv690181n214od5jal7ukthf6u
Theorem: Complex-Valued Function is Continuous iff Real Part and Imaginary Part are Continuous

Let $\struct {X, \tau}$ be a topological space.
Let $f : X \to \C$ be a function.
Then $f$ is continuous {{iff}} both the real part $\map \Re f$ and the imaginary part $\map \Im f$ are continuous.

Proof:
=== Necessary Condition ===
Suppose that $f$ is continuous.
From Real and Imaginary Part Projections are Continuous, $z \mapsto \map \Re z$ and $z \mapsto \map \Im z$ are continuous.
From Composite of Continuous Mappings is Continuous, $\map \Re f$ and $\map \Im f$ are continuous.
{{qed|lemma}}
Theorem: Eberlein-Šmulian Characterization of Reflexive Banach Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Then $\struct {X, \norm {\, \cdot \,}_X}$ is reflexive {{iff}}:
:every bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ has a weakly convergent subsequence.

Proof:
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$.
Theorem: Separable Banach Space is Linearly Isomorphic to Quotient of 1-Sequence Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$ be the $1$-sequence space.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a separable Banach space over $\GF$.
Then there exists a closed vector subspace $M$ of $\map {\ell_1} \GF$ such that:
:$X$ is linearly isomorphic to the normed quotient vector space $\map {\ell_1} \GF/M$.

Proof:
From 1-Sequence Space Continuously Surjects into Separable Banach Space, there exists a bounded linear surjection $Q : \map {\ell_1} \GF \to X$.
From Kernel of Bounded Linear Transformation is Closed Linear Subspace, $\ker Q$ is a closed vector subspace of $X$.
Let $M = \ker Q$.
Then the normed quotient vector space $\map {\ell_1} \GF/M$ is well-defined.
From Continuous Surjection Induces Continuous Bijection from Quotient Space and Continuous Linear Transformation between Topological Vector Spaces is Bounded, there exists a bounded bijection $\widetilde Q : \map {\ell_1} \GF/M \to X$.
From the Banach Isomorphism Theorem, $\widetilde Q$ is a linear isomorphism.
{{qed}}
Theorem: Weak Compactness is Equivalent to Norm Compactness in Schur Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space that is a Schur space.
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$.
Let $K \subseteq X$.
Then $\struct {K, w}$ is compact {{iff}} $\struct {K, \norm {\, \cdot \,}_X}$ is compact.

Proof:
=== Necessary Condition ===
Suppose that $\struct {K, w}$ is compact.
From the Eberlein-Šmulian Theorem, $\struct {K, w}$ is sequentially compact.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.
Since $\struct {K, w}$ is sequentially compact, there exists a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ converging weakly to $x \in K$.
Since $\struct {X, \norm {\, \cdot \,}_X}$ is a Schur space, we also have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ converges in norm to $x \in K$.
Hence every sequence in $K$ has a convergent subsequence with limit in $K$.
Hence $\struct {K, \norm {\, \cdot \,}_X}$ is sequentially compact.
From Sequentially Compact Metric Space is Compact, $\struct {K, \norm {\, \cdot \,}_X}$ is compact.
{{qed|lemma}}
Theorem: Weak Topology and Norm Topology on Normed Vector Space Coincide iff Finite Dimensional

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $d$ be the metric induced on $X$ by $\norm {\, \cdot \,}$.
Let $\tau$ be the topology on $X$ induced by $d$. 
Let $w$ be the weak topology on $X$.
Then $\tau = w$ {{iff}} $X$ is finite-dimensional.

Proof:
=== Necessary Condition ===
Suppose that $X$ is not finite-dimensional.
Since $\tau$ is induced by a metric, it is metrizable.
From Weak Topology on Infinite Dimensional Normed Vector Space is not Metrizable, $w$ is not metrizable.
Hence $\tau \ne w$. 
{{qed|lemma}}
Theorem: Topologies with Same Convergent Nets are Equal

Let $X$ be a set.
Let $\tau_1$ and $\tau_2$ be topologies on $X$ such that:
:a net converges to $x$ in $\struct {X, \tau_1}$ {{iff}} it converges to $x$ in $\struct {X, \tau_2}$.
Then $\tau_1 = \tau_2$.

Proof:
Let $U \subseteq X$.
By Characterization of Openness in terms of Nets, we have that $U \in \tau_1$ {{iff}} for each:
:$x \in U$
:directed set $\tuple {\Lambda, \preceq}$
:net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$ in $\struct {X, \tau_1}$
there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.
By Characterization of Openness in terms of Nets, we have that $U \in \tau_2$ {{iff}} for each:
:$x \in U$
:directed set $\tuple {\Lambda, \preceq}$
:net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$ in $\struct {X, \tau_2}$
there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.
Since:
:a net converges to $x$ in $\struct {X, \tau_1}$ {{iff}} it converges to $x$ in $\struct {X, \tau_2}$
we have that $U \in \tau_1$ {{iff}} $U \in \tau_2$.
Hence $\tau_1 = \tau_2$.
{{qed}}
Category:Nets (Set Theory)
p85vj7akt494au609khwfftgnq2dwfi
Theorem: Reflexive Space is Banach Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a reflexive normed vector space.
Then $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space.

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.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_X}$.
From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have that:
:$\norm {x_n - x_m}_X = \norm {\iota x_n - \iota x_m}_{X^{\ast \ast} }$
Hence $\sequence {\iota x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$.
From Normed Dual Space is Banach Space, $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ is a Banach space.
Hence there exists $\phi \in X^{\ast \ast}$ such that $\iota x_n \to \phi$ in $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$.
Since $\iota$ is an isometric isomorphism, we have $X^{\ast \ast} = \iota X$.
Hence we have $\phi \in \iota X$.
So there exists $x \in X$ such that $\phi = \iota x$.
Then we have:
:$\norm {x_n - x}_X = \norm {\iota x_n - \phi}_{X^{\ast \ast} } \to 0$
Hence we have that $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$. 
{{qed}}
Category:Reflexive Spaces
Category:Banach Spaces
sv9wqk9x2m1l46bvk3zodcydtl09f39
Theorem: Reflexive Banach Space is Schur Space iff Finite Dimensional

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a reflexive Banach space over $\GF$.
Then $\struct {X, \norm {\, \cdot \,}_X}$ is a Schur space {{iff}} it is finite-dimensional.

Proof:
=== Necessary Condition ===
Suppose that $\struct {X, \norm {\, \cdot \,}_X}$ is a Schur space.
Let $B_X^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,}_X}$.
Let:
:$S_X = \set {x \in X : \norm x_X = 1}$
Since $\struct {X, \norm {\, \cdot \,}_X}$ is reflexive, $B_X^-$ is weakly compact from Kakutani's Theorem.
From Weak Compactness is Equivalent to Norm Compactness in Schur Space, $B_X^-$ is compact.
From Closed Subspace of Compact Space is Compact, $S_X$ is compact.
From Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact, $X$ is finite-dimensional.
{{qed|lemma}}
Theorem: Subsequence Characterization of Cauchy Sequence in Topological Vector Space

Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence {{iff}}:
:for all strictly increasing sequences $\sequence {n_k}_{k \mathop \in \N}$ and $\sequence {m_k}_{k \mathop \in \N}$ in $\N$, we have that:
::$x_{n_k} - x_{m_k} \to {\mathbf 0}_X$ in $\struct {X, \tau}$.

Proof:
=== Necessary Condition ===
Suppose that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence.
Let $\sequence {n_k}_{k \mathop \in \N}$ and $\sequence {m_k}_{k \mathop \in \N}$ in $\N$ be strictly increasing sequences.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$ in $\struct {X, \tau}$. 
There exists $N \in \N$ such that for $n, m \ge N$, we have:
:$x_n - x_m \in U$ for $n, m \ge N$.
Taking $k \ge N$, we have that $n_k \ge N$ and $m_k \ge N$, and hence:
:$x_{n_k} - x_{m_k} \in U$ for $k \ge N$. 
So $x_{n_k} - x_{m_k} \to {\mathbf 0}_X$ in $\struct {X, \tau}$.
{{qed|lemma}}
Theorem: Weakly Cauchy Sequence in Normed Vector Space is Bounded

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weak Cauchy sequence in $X$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is bounded.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
For each $f \in X^\ast$, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is a Cauchy sequence.
From Cauchy's Convergence Criterion, we have that $\sequence {\map f {x_n} }_{n \mathop \in \N}$ converges for each $f \in X^\ast$.
From Convergent Sequence in Metric Space is Bounded, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is bounded for each $f \in X^\ast$.
That is:
:$\ds \sup_{n \mathop \in \N} \cmod {\map f {x_n} } < \infty$ for each $f \in X^\ast$.
From Subset of Normed Vector Space providing Bounded Point Evaluations is Norm Bounded, we therefore have:
:$\ds \sup_{n \mathop \in \N} \norm {x_n}_X < \infty$
{{qed}}