avewright/proofwiki-math huggingface dataset:
Theorem: Reflexive Banach Space is Weakly Sequentially Complete

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 weakly sequentially complete.

Proof:
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weak Cauchy sequence in $X$.
From Weakly Cauchy Sequence in Normed Vector Space is Bounded, $\sequence {x_n}_{n \mathop \in \N}$ is bounded.
Then for each $f \in X^\ast$, $\sequence {\map f {x_n} }_{n \mathop \in \N}$ is a Cauchy sequence.
From the Eberlein-Šmulian Characterization of Reflexive Banach Space, $\sequence {x_n}_{n \mathop \in \N}$ has a weakly convergent subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ with limit $x$.
Hence:
:$\map f {x_{n_k} } \to \map f x$ as $k \to \infty$ for each $f \in X^\ast$.
From Convergent Subsequence of Cauchy Sequence: Normed Vector Space, we have:
:$\map f {x_n} \to \map f x$ as $n \to \infty$ for each $f \in X^\ast$.
Hence $\sequence {x_n}_{n \mathop \in \N}$ is weakly convergent.
Hence every weak Cauchy sequence in $X$ is weakly convergent.
Hence $\struct {X, \norm {\, \cdot \,}_X}$ is weakly sequentially complete.
{{qed}}
Theorem: Banach Space with Schur Property is Weakly Sequentially Complete

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space with the Schur property.
Then $\struct {X, \norm {\, \cdot \,}_X}$ is weakly sequentially complete.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a weak Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_X}$.
From Subsequence Characterization of Cauchy Sequence in Topological Vector Space, we therefore have that:
: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:
::$x_{n_k} - x_{m_k} \weakconv {\mathbf 0}_X$
Since $\struct {X, \norm {\, \cdot \,}_X}$ is a Schur space, we have:
:$\norm {x_{n_k} - x_{m_k} }_X \to 0$
Then, from Subsequence Characterization of Cauchy Sequence in Normed Vector Space, $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_X}$.
Since $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space, $\sequence {x_n}_{n \mathop \in \N}$ converges.
{{qed}}
Theorem: Rearrangement of Unconditionally Convergent Series in Banach Space Converges to Same Value

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.
Let:
:$\ds x = \sum_{n \mathop = 1}^\infty x_n$
Let $\pi : \N \to \N$ be a permutation.
We have:
:$\ds \sum_{n \mathop = 1}^\infty x_{\map \pi n} = x$

Proof:
Let $\epsilon > 0$.
Take $N_1 \in \N$ such that:
:$\ds \norm {\sum_{n \mathop = 1}^N x_n - x}_X < \frac \epsilon 2$
Take $N_2 \ge N_1$ such that:
:$\ds \norm {\sum_{n \mathop \in F} x_n}_X < \frac \epsilon 4$ for every finite subset $F$ of $\set {N_2 + 1, \ldots}$.
Finally, since $\pi : \N \to \N$ is a permutation, there exists $N \in \N$ with $N \ge N_2$ such that:
:$\set {1, 2, \ldots, N_2} \subseteq \pi \sqbrk {\set {1, 2, \ldots, N} }$
Let $n \ge N$.
Consider the difference:
:$\ds \sum_{n \mathop = 1}^N x_n - \sum_{n \mathop = 1}^N x_{\map \pi n}$
The $x_1, \ldots, x_{N_2}$ from the first sum will cancel with a subset of $\set {x_{\map \pi 1}, \ldots, x_{\map \pi N} }$
Left over will be some $x_j$ from the first sum for which there does not exist $1 \le n \le N$ with $j = \map \pi n$, and $x_n$ for $1 \le n \le N$ such that $\map \pi n > N$.
In both cases, the index will be in $\set {N_2 + 1, \ldots}$. 
Hence we can write:
:$\ds \sum_{n \mathop = 1}^N x_n - \sum_{n \mathop = 1}^N x_{\map \pi n} = \sum_{n \in F_1} x_n - \sum_{n \in F_2} x_n$
where $F_1, F_2$ are finite subsets of $\set {N_2 + 1, \ldots}$.
Hence:
:$\ds \norm {\sum_{n \in F_1} x_n} < \frac \epsilon 4$
and:
:$\ds \norm {\sum_{n \in F_2} x_n} < \frac \epsilon 4$
By {{NormAxiomVector|3}}, we then have:
:$\ds \norm {\sum_{n \mathop = 1}^N x_n - \sum_{n \mathop = 1}^N x_{\map \pi n} }_X < \frac \epsilon 2$ for $n \ge N$.
Hence for $n \ge N$ we have:
{{begin-eqn}}
{{eqn	| l = \norm {\sum_{n \mathop = 1}^N x_{\map \pi n} - x}_X
		| o = \le
		| r = \norm {\sum_{n \mathop = 1}^N x_{\map \pi n} - \sum_{n \mathop = 1}^N x_n}_X + \norm {\sum_{n \mathop = 1}^N x_n - x}_X 
}}
{{eqn	| o = < 
		| r = \frac \epsilon 2 + \frac \epsilon 2 
}}
{{eqn	| r = \epsilon 
}}
{{end-eqn}}
{{qed}}
Theorem: Characterization of Weak Unconditional Cauchyness of Series in Banach Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $\map {c_{00} } \GF$ be the space of almost-zero sequences on $\GF$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
{{TFAE}}
:$(1): \quad$ $\ds \sum_{n \mathop = 1}^\infty x_n$ is weakly unconditionally Cauchy
:$(2): \quad$ there exists $C > 0$ such that for all  $\xi \in \map {c_{00} } \GF$ we have:
::::$\ds \norm {\sum_{n \mathop = 1}^\infty \map \xi n x_n}_X \le C \max_{n \mathop \in \N} \cmod {\map \xi n}$
:$(3): \quad$ there exists $C' > 0$ such that for every finite subset of $\N$ and each $\sequence {\epsilon_n}_{n \mathop \in \N} \in \set {z \in \GF : \cmod z = 1}^\N$:
::::$\ds \norm {\sum_{n \mathop \in F} \epsilon_n x_n}_X \le C'$

Proof:
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map {c_{00} } \GF$.
Theorem: Space of Almost-Zero Sequences is Everywhere Dense in Space of Zero-Limit Sequences

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $\struct {\map {c_{00} } \GF, \norm {\, \cdot \,}_\infty}$ be the space of almost-zero sequences.
Then $\struct {\map {c_{00} } \GF, \norm {\, \cdot \,}_\infty}$ is everywhere dense in $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.

Proof:
Let $\phi \in \map {c_0} \GF$.
Let $\epsilon > 0$.
Then there exists $N \in \N$ such that:
:$\ds \sup_{n \mathop \ge N} \cmod {\map \phi n} < \epsilon$
Define $\phi_\epsilon \in \map {c_{00} } \GF$ by:
:$\ds \map {\phi_\epsilon} k = \begin{cases}\map \phi k & n \le N \\ 0 & n > N\end{cases}$
Take $n \ge N$.
We have:
:$\ds \sup_{k \mathop \in \N} \cmod {\map \phi k - \map {\phi_\epsilon} k} = \max \set {\sup_{k \mathop \le N} \cmod {\map \phi k - \map {\phi_\epsilon} k}, \sup_{k \mathop > N} \cmod {\map \phi k - \map {\phi_\epsilon} k} }$
For $\map \phi k = \map {\phi_\epsilon} k$ for $k \le N$, while $\map {\phi_\epsilon} k = 0$ for $k > N$ hence we have:
:$\ds \sup_{k \mathop \in \N} \cmod {\map \phi k - \map {\phi_\epsilon} k} < \epsilon$
That is:
:$\norm {\phi - \phi_\epsilon}_\infty < \epsilon$
with $\phi_\epsilon \in \map {c_{00} } \GF$. 
Since $\epsilon > 0$ and $\phi \in \map {c_0} \GF$ were arbitrary, we have that:
:$\struct {\map {c_{00} } \GF, \norm {\, \cdot \,}_\infty}$ is everywhere dense in $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
{{qed}}
Category:Space of Almost-Zero Sequences
Category:Space of Zero-Limit Sequences
gjq5a8gh1fpfr726oefyzcxr5w38zl6
Theorem: Space of Zero-Limit Sequences admits Schauder Basis

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Define $e_n \in \map {c_0} \GF$ by:
:$\ds \map {e_n} k = \begin{cases}1 & n = k \\ 0 & n \ne k\end{cases}$ 
Then $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis for $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.

Proof:
For each $n \in \N$, define $e_n^\ast : \map {c_0} \GF \to \GF$ by:
:$\map {e_n^\ast} \phi = \map \phi n$
for each $\phi \in \map {c_0} \GF$.
We argue that $e_n^\ast$ is a bounded linear functional.
Let $\phi, \psi \in \map {c_0} \GF$ and $\lambda \in \GF$.
We have:
{{begin-eqn}}
{{eqn	| l = \map {e_n^\ast} {\phi + \lambda \psi}
		| r = \map {\paren {\phi + \lambda \psi} } n
}}
{{eqn	| r = \map \phi n + \lambda \map \psi n
}}
{{eqn	| r = \map {e_n^\ast} \phi + \lambda \map {e_n^\ast} \psi 
}}
{{end-eqn}}
Hence $e_n^\ast$ is a linear functional.
We have:
:$\ds \cmod {\map {e_n^\ast} \phi} = \cmod {\map \phi n} \le \sup_{n \mathop \in \N} \cmod {\map \phi n} = \norm \phi_\infty$
Hence $e_n^\ast$ is a bounded linear functional.
Further, by the definition of $e_i$ we have:
:$\map {e_j^\ast} {e_i} = \delta_{i j}$
where $\delta$ is the Kronecker $\delta$.
From Condition for Sequence to be Schauder Basis in terms of Coordinate Functionals, it is enough to show that:
:$\ds x = \sum_{n \mathop = 1}^\infty \map {e_n^\ast} x e_n$ in $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$
for each $x \in \map {c_0} \GF$.
Let $\epsilon > 0$. 
Let $x \in \map {c_0} \GF$.
Pick $N \in \N$ such that:
:$\ds \sup_{k > M} \cmod {\map x k} < \epsilon$
for $M \ge N$. 
Let $M \ge N$.  
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {x - \sum_{n \mathop = 1}^M \map {e_n^\ast} x e_n}_\infty 
		| r = \max \set {\sup_{k \le M} \cmod {\map x k - \sum_{n \mathop = 1}^M \map {e_n^\ast} x \map {e_n} k}, \sup_{k > M} \cmod {\map x k - \sum_{n \mathop = 1}^M \map {e_n^\ast} x \map {e_n} k} }
}}
{{eqn	| r = \max \set {\sup_{k \le M} \cmod {\map x k - \sum_{n \mathop = 1}^M \map x n \delta_{n k} }, \sup_{k > M} \cmod {\map x k} }
}}
{{eqn	| r = \max \set {\sup_{k \le M} \cmod {\map x k - \map x k}, \sup_{k > M} \cmod {\map x k} }
		| c = {{Defof|Kronecker Delta}}
}}
{{eqn	| r = \sup_{k > M} \cmod {\map x k}
}}
{{eqn	| o = <
		| r = \epsilon 
}}
{{end-eqn}}
Hence we have that:
:$\ds \lim_{M \mathop \to \infty} \sum_{n \mathop = 1}^M \map {e_n^\ast} x e_n = x$ in $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
Hence $\sequence {e_n^\ast}_{n \mathop \in \N}$ is a sequence of bounded linear functionals on $\map {c_0} \GF$ such that:
:$(1): \quad$ $\map {e_j^\ast} {e_i} = \delta_{i j}$ for $i, j \in \N$
:$(2): \quad$ $\ds x = \sum_{n \mathop = 1}^\infty \map {e_n^\ast} x e_n$ in $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
From Condition for Sequence to be Schauder Basis in terms of Coordinate Functionals, we have that $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis for $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$.
{{qed}}
Category:Space of Zero-Limit Sequences
Category:Schauder Bases
tothsjmlu4bunxphzujvpks42mytbi5
Theorem: Infinite Subseries of Unconditionally Convergent Series is Unconditionally Convergent

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.
Let $A \subseteq \N$ be infinite. 
Then:
:$\ds \sum_{n \mathop \in A} x_n = \sum_{n \mathop = 1}^\infty \map {\chi_A} n x_n$ is unconditionally convergent
where $\chi_A$ is the indicator function of $A$.

Proof:
From Characterization of Unconditional Convergence of Series in Banach Space:
: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 $\ds \norm {\sum_{j \mathop \in F} x_j}_X < \epsilon$
From Subset of Finite Set is Finite, we can swap $F$ for $F \cap A$, where we interpret the sum over the empty set as ${\mathbf 0}_X$.
Hence we obtain:
:for each $\epsilon > 0$, there exists $n \in \N$ such that for all finite sets $F \cap A \subseteq \set {j : j \ge n + 1}$, we have $\ds \norm {\sum_{j \mathop \in F \cap A} x_j}_X < \epsilon$
That is:
: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 $\ds \norm {\sum_{j \mathop \in F} \map {\chi_A} n x_j}_X < \epsilon$
{{qed}}
Theorem: Limit Superior in terms of Partitioning Subsequences

Let $\sequence {a_n}_{n \mathop \in \N}$ be a real sequence that is bounded above.
Let $\SS = \set {S_1, \ldots, S_N}$ be a finite partition of $\N$.
Let $S_j = \set {\map {\phi_j} k : k \in \N}$ where $\sequence {\map {\phi_j} k}_{k \mathop \in \N}$ is an increasing sequence.
We therefore have:
:$\ds \limsup_{k \mathop \to \infty} x_k = \max_{1 \le j \le N} \limsup_{k \mathop \to \infty} x_{\map {\phi_j} k}$
where $\limsup$ denotes the limit superior.

Proof:
Let $M > 0$.
We first write:
:$\ds \set {x_k : k \ge M} = \bigcup_{j \mathop = 1}^N \set {x_{\map {\phi_j} k} : \map {\phi_j} k \ge M}$
We then have, from Supremum of Union of Bounded Above Sets of Real Numbers:
:$\ds \sup_{k \ge M} x_k = \max_{1 \le j \le N} \sup_{\map {\phi_j} k \ge M} x_{\map {\phi_j} k}$
From Limit of Maximum of Convergent Sequences is Maximum of Limits, we then have:
:$\ds \limsup_{k \mathop \to \infty} x_k = \max_{1 \le j \le N} \limsup_{k \mathop \to \infty} x_{\map {\phi_j} k}$
{{qed}}
Category:Limits Superior
izqsc8di8mr35z5yais01kbdz8osg8f
Theorem: Negative of Limit Superior is Limit Inferior of Negative

Let $\sequence {a_n}_{n \mathop \in \N}$ be a real sequence that is bounded above.
Then we have:
:$\ds -\limsup_{n \mathop \to \infty} a_n = \liminf_{n \mathop \to \infty} \paren {-a_n}$

Proof:
For each $N \in \N$, we have:
:$\ds -\sup_{n \ge N} a_n = \inf_{n \ge N} \paren {-a_n}$
from Negative of Supremum is Infimum of Negatives.
From Multiple Rule for Real Sequences, we have:
:$\ds -\sup_{n \ge N} a_n \to -\limsup_{n \mathop \to \infty} a_n$ as $N \to \infty$.
Also:
:$\ds \inf_{n \ge N} \paren {-a_n} \to \liminf_{n \mathop \to \infty} a_n$ as $N \to \infty$.
Hence we have, from Convergent Real Sequence has Unique Limit:
:$\ds -\limsup_{n \mathop \to \infty} a_n = \liminf_{n \mathop \to \infty} \paren {-a_n}$
{{qed}}
Category:Limits Superior
Category:Limits Inferior
nqej7z7kob0s8f0konah9wdyqvpaxjv
Theorem: Negative of Limit Inferior is Limit Superior of Negative

Let $\sequence {a_n}_{n \mathop \in \N}$ be a real sequence that is bounded below .
Then we have:
:$\ds -\liminf_{n \mathop \to \infty} a_n = \limsup_{n \mathop \to \infty} \paren {-a_n}$

Proof:
For each $N \in \N$, we have:
:$\ds -\inf_{n \ge N} a_n = \sup_{n \ge N} \paren {-a_n}$
from Negative of Infimum is Supremum of Negatives.
From Multiple Rule for Real Sequences, we have:
:$\ds -\inf_{n \ge N} a_n \to -\liminf_{n \mathop \to \infty} a_n$ as $N \to \infty$.
Also:
:$\ds \sup_{n \ge N} \paren {-a_n} \to \limsup_{n \mathop \to \infty} a_n$ as $N \to \infty$.
Hence we have, from Convergent Real Sequence has Unique Limit:
:$\ds -\liminf_{n \mathop \to \infty} a_n = \limsup_{n \mathop \to \infty} \paren {-a_n}$
{{qed}}
Category:Limits Superior
Category:Limits Inferior
b4g253m1u0wt61y14sz5wxvfnxbm5hc
Theorem: Limit Inferior in terms of Partitioning Subsequences

Let $\sequence {a_n}_{n \mathop \in \N}$ be a real sequence that is bounded below.
Let $\SS = \set {S_1, \ldots, S_N}$ be a finite partition of $\N$.
Let $S_j = \set {\map {\phi_j} k : k \in \N}$ where $\sequence {\map {\phi_j} k}_{k \mathop \in \N}$ is an increasing sequence.
We therefore have:
:$\ds \liminf_{k \mathop \to \infty} x_k = \min_{1 \le j \le N} \liminf_{k \mathop \to \infty} x_{\map {\phi_j} k}$
where $\liminf$ denotes the limit inferior.

Proof:
We have:
{{begin-eqn}}
{{eqn	| l = \liminf_{k \mathop \to \infty} x_k
		| r = -\limsup_{k \mathop \to \infty} \paren {-x_k}
		| c = Negative of Limit Inferior is Limit Superior of Negative
}}
{{eqn	| r = -\max_{1 \le j \le N} \limsup_{k \mathop \to \infty} \paren {-x_{\map {\phi_j} k} }
}}
{{eqn	| r = -\max_{1 \le j \le N} -\liminf_{k \mathop \to \infty} x_{\map {\phi_j} k}
		| c = Negative of Limit Inferior is Limit Superior of Negative
}}
{{eqn	| r = \min_{1 \le j \le N} \liminf_{k \mathop \to \infty} x_{\map {\phi_j} k}
		| c = Negative of Supremum is Infimum of Negatives
}}
{{end-eqn}}
{{qed}}
Category:Limits Inferior
Category:Limits Superior
429lj0j258v2bt99iuc1iy8ryvokeuy
Theorem: Real Series is Unconditionally Convergent iff Absolutely Convergent

Let $\sequence {a_n}_{n \mathop \in \N}$ be a real sequence.
Then $\ds \sum_{n \mathop = 1}^\infty a_n$ is unconditionally convergent {{iff}} it is absolutely convergent.

Proof:
=== Necessary Condition ===
Suppose that:
:$\ds \sum_{n \mathop = 1}^\infty a_n$ is unconditionally convergent
{{AimForCont}}:
:$\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} = \infty$
From Riemann's Rearrangement Theorem, there exists a permutation $\pi : \N \to \N$ such that:
:$\ds \sum_{n \mathop = 1}^\infty a_{\map \pi n} = \infty$
This contradicts that $\ds \sum_{n \mathop = 1}^\infty a_n$ is unconditionally convergent.
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent.
{{qed|lemma}}
Theorem: Gantmacher's Theorem

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces over $\GF$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ and $\struct {Y^\ast, \norm {\, \cdot \,}_{Y^\ast} }$ be the respective normed dual spaces.
Let $w$ denote the weak topology.
Let $w^\ast$ denote the weak-$\ast$ topology.
Let $\struct {X^{\ast \ast}, \norm {\, \cdot \,}_{X^{\ast \ast} } }$ and $\struct {Y^{\ast \ast}, \norm {\, \cdot \,}_{Y^{\ast \ast} } }$ be the respective second normed duals.
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation on $X$.
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation on $Y$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator for $T$.
Let $T^{\ast \ast} : X^{\ast \ast} \to Y^{\ast \ast}$ be the dual operator for $T^{\ast \ast}$.
{{TFAE}}
:$(1): \quad$ $T$ is weakly compact
:$(2): \quad$ $T^{\ast \ast} \sqbrk {X^{\ast \ast} } \subseteq \iota_Y Y$
:$(3): \quad$ $T^\ast : \struct {Y^\ast, w^\ast} \to \struct {X^\ast, w}$ is continuous
:$(4): \quad$ $T^\ast$ is weakly compact.

Proof:
Let $w$ and $w$ denote the weak topology of the relevant space.
Theorem: Characterization of von Neumann-Bounded Set in Weak Topology

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 $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,} }$. 
Let $B \subseteq X$.
Then $B$ is von Neumann-bounded in $\struct {X, w}$ {{iff}}:
:$\ds \sup_{x \mathop \in B} \cmod {\map f x} < \infty$ for all $f \in X^\ast$.

Proof:
=== Necessary Condition ===
Suppose that $B$ is von Neumann-bounded in $\struct {X, w}$.
Let $f \in X^\ast$.
From Open Sets in Weak Topology of Topological Vector Space:
:$V = \set {y \in X : \cmod {\map f y} < 1}$ is an open neighborhood of $\mathbf 0_X$ in $\struct {X, w}$.
Hence there exists $r > 0$ such that:
:$B \subseteq r V$
That is, if $x \in B$ then there exists $v \in V$ such that $x = r y$.
Since $f$ is a linear functional, we have $\map f x = r \map f y$. 
Then we have $\cmod {\map f x} < r$ for each $x \in B$.
In other words:
:$\ds \sup_{x \mathop \in B} \cmod {\map f x} < \infty$
{{qed|lemma}}
Theorem: Eberlein-Šmulian Characterization of Weakly Compact Linear Transformation

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 a bounded linear transformation.
Then $T$ is weakly compact {{iff}}:
:for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$, there exists a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:
::$\sequence {T x_{n_k} }_{k \mathop \in \N}$ converges weakly.

Proof:
Let $w$ be the weak topology on $\struct {X, \norm {\, \cdot \,}_X}$. 
Let $\cl$ be the closure taken in $\struct {X, \norm {\, \cdot \,}_X}$. 
Let $\cl_w$ be the closure taken in the weak topology.
Let $B_X^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,}_X}$.
Theorem: Composition of Weakly Compact Linear Transformation and Bounded Linear Transformation is Weakly Compact

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Z, \norm {\, \cdot \,}_Z}$ be Banach spaces over $\GF$.
Let $T : X \to Y$ be a weakly compact linear transformation.
Let $A : Y \to Z$ be a bounded linear transformation.
Let $B : Z \to X$ be a bounded linear transformation.
Then:
:$(1): \quad$ $T B : Z \to Y$ is weakly compact
:$(2): \quad$ $A T : X \to Z$ is weakly compact.

Proof:
=== $T B$ is weakly compact ===
By the Eberlein-Šmulian Characterization of Weakly Compact Linear Transformation, it is enough to show that:
:for each bounded sequence $\sequence {z_n}_{n \mathop \in \N}$ in $Z$, there exists a subsequence $\sequence {z_{n_k} }_{k \mathop \in \N}$ such that:
::$\sequence {T B z_{n_k} }_{k \mathop \in \N}$ converges weakly in $Y$.
Let $\sequence {z_n}_{n \mathop \in \N}$ be a bounded sequence in $Z$.
Since $B$ is bounded, we have:
:$\norm {B z_n}_X \le M \norm {z_n}_Z$
for some $M > 0$.
Hence $\sequence {B z_n}_{n \mathop \in \N}$ is a bounded sequence in $X$.
Since $T$ is weakly compact:
:there exists a subsequence $\sequence {B z_{n_k} }_{k \mathop \in \N}$ such that:
::$\sequence {T B z_{n_k} }_{k \mathop \in \N}$ converges weakly in $Y$.
by the Eberlein-Šmulian Characterization of Weakly Compact Linear Transformation.
Hence by the Eberlein-Šmulian Characterization of Weakly Compact Linear Transformation, $T B$ is weakly compact.
{{qed|lemma}}
Theorem: Complement of Subgroup is not necessarily Unique

Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $K$ be a complement of $H$ in $G$.
Then it is not necessarily the case that $K$ is the only complement of $H$ in $G$.

Proof:
Proof by Counterexample:
Let $\Z_2 = \set {e, a}$ be the cyclic group of order 2, where $a$ denotes the non-identity element.
Let $G = \set {e, a, b, c}$ be the Klein Four-Group
Let $H = \set {e, a}$ be a subgroup of $G$.
Let $K_1 = \set {e, b}$ be a subgroup of $G$.
Let $K_2 = \set {e, c}$ be a subgroup of $G$.
Claim: $K_1$ and $K_2$ are different complements of $H$ in $G$.
{{Recall|Complement of Subgroup}}
{{:Definition:Complement of Subgroup/Definition 1}}
We check that $K_1$ is a complement of $H$:
{{begin-eqn}}
{{eqn | l = H K_1
      | r = \set {e, a} \set {e, b}
}}
{{eqn | r = \set {e, a, b, a b}
      | c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {e, a, b, c} 
      | c = {{Defof|Klein Four-Group}}
}}
{{eqn | r = G
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = H \cap K_1
      | r = \set {e, a} \cap \set {e, b}
}}
{{eqn | r = \set e
}}
{{end-eqn}}
Hence $K_1$ is a complement of $H$.
Similarly we check that $K_2$ is a complement of $H$:
{{begin-eqn}}
{{eqn | l = H K_2
      | r = \set {e, a} \set {e, c}
}}
{{eqn | r = \set {e, a, c, a c}
      | c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {e, a, b, c} 
      | c = {{Defof|Klein Four-Group}}
}}
{{eqn | r = G
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = H \cap K_1
      | r = \set {e, a} \cap \set {e, c}
}}
{{eqn | r = \set e
}}
{{end-eqn}}
Hence $K_2$ is a complement of $H$.
The result follows.
{{qed}}
Theorem: Intersection of Event with Complement Can't Happen

Let the probability space of an experiment $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
Let $A \in \Sigma$ be an events of $\EE$, so that $A \subseteq \Omega$.
Then:
:$A \cap \overline A = \O$
where $\overline A$ is the complementary event to $A$.
That is:
:$A \cap \overline A$ is an impossibility
or:
:$\map \Pr {A \cap \overline A} = 0$
That is, $A$ and $\overline A$ are disjoint.

Proof:
By definition:
:$A \subseteq \Omega$
and:
:$\overline A = \relcomp \Omega A$
From Intersection with Relative Complement is Empty:
:$A \cap \overline A = \O$
We then have from Probability of Empty Event is Zero that:
:$\map \Pr \Omega = 0$
The result follows by definition of impossible event.
{{qed}}