avewright/proofwiki-math huggingface dataset:
Theorem: Sum of Closures is Subset of Closure of Sum in Topological Vector Space

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A, B \subseteq X$. 
Then: 
:$A^- + B^- \subseteq \paren {A + B}^-$
where $A^-$, $B^-$ and $\paren {A + B}^-$ denote the closures of $A$, $B$ and $A + B$.

Proof:
Let $x \in A^-$ and $y \in B^-$.
From Point in Set Closure iff Limit of Net, there exists directed sets $\struct {\Lambda_1, \preceq_1}$ and $\struct {\Lambda_2, \preceq_2}$ and nets $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ valued in $A$ and $\sequence {y_\lambda}_{\lambda \mathop \in \Lambda_2}$ valued in $B$ such that:
:$\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$ converges to $x$
:$\sequence {y_\lambda}_{\lambda \mathop \in \Lambda_2}$ converges to $y$.
Define a relation $\sqsubseteq$ on $\Lambda_1 \times \Lambda_2$ by: 
:$\tuple {\lambda_1, \lambda_2} \sqsubseteq \tuple {\mu_1, \mu_2}$
for $\tuple {\lambda_1, \lambda_2}, \tuple {\mu_1, \mu_2} \in \Lambda_1 \times \Lambda_2$ {{iff}}:
:$\lambda_1 \preceq_1 \mu_1$ and $\lambda_2 \preceq_2 \mu_2$
From Product of Directed Sets is Directed Set, $\struct {\Lambda_1 \times \Lambda_2, \sqsubseteq}$ is a directed set.
Define:
:$x'_{\tuple {\lambda, \mu} } = x_\lambda$
and:
:$y'_{\tuple {\lambda, \mu} } = y_\mu$
for each $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$.
We show that $\langle {x'_{\tuple {\lambda, \mu} } \rangle}_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x$.
Fix $\mu_0 \in \Lambda_2$.
Let $U$ be an open neighborhood of $x$. 
Since $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda_1}$, there exists $\lambda_0 \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda \succeq \lambda_0$, we have $x_\lambda \in U$.
Then, if $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$ has $\tuple {\lambda, \mu} \sqsupseteq \tuple {\lambda_0, \mu_0}$, we in particular have $\lambda \succeq \lambda_0$ and so $x'_{\tuple {\lambda, \mu} } \in U$. 
Since $U$ was arbitrary, $\sequence {x'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x$.
Similarly, $\sequence {y'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $y$.
From Sum of Convergent Nets in Topological Vector Space is Convergent, $\sequence {x'_{\tuple {\lambda, \mu} } + y'_{\tuple {\lambda, \mu} } }_{\tuple {\lambda, \mu} \mathop \in \Lambda_1 \times \Lambda_2}$ converges to $x + y$.
Since $x'_{\tuple {\lambda, \mu} } \in A$ and $y'_{\tuple {\lambda, \mu} } \in B$ for all $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$, we have that $x'_{\tuple {\lambda, \mu} } + y'_{\tuple {\lambda, \mu} } \in A + B$ for each $\tuple {\lambda, \mu} \in \Lambda_1 \times \Lambda_2$.
So $x + y$ is the limit of a net valued in $A + B$.
So by Point in Set Closure iff Limit of Net, we have $x + y \in \paren {A + B}^-$.
So, we have $A^- + B^- \subseteq \paren {A + B}^-$.
{{qed}}
Theorem: Vanishing Ideal of Larger Subset of Affine Space is Smaller

Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $S \subseteq T \subseteq \mathbb A^n_k$.
Then:
:$\map I S \supseteq \map I T$
where $\map I S$ and $\map I T$ denote the vanishing ideals of $S$ and $T$, respectively.

Proof:
{{begin-eqn}}
{{eqn | l = f
      | o = \in
      | r = \map I T
}}
{{eqn | ll= \leadsto
      | q = \forall x \in T
      | l = \map f x
      | r = 0
}}
{{eqn | ll= \leadsto
      | q = \forall x \in S
      | l = \map f x
      | r = 0
      | c = since $S \subseteq T$ by assumption
}}
{{eqn | ll= \leadsto
      | l = f
      | o = \in
      | r = \map I S
}}
{{end-eqn}}
{{qed}}
Category:Algebraic Geometry
51vghfhpaugl8uglfeyq0p6knobup6q
Theorem: Equivalence of Definitions of Convex Set in Vector Space

Let $\Bbb F \in \set {\R, \C}$.
Let $V$ be a vector space over $\Bbb F$.
Let $C \subseteq V$.
{{TFAE|def = Convex Set (Vector Space)|view = convexity}}

Proof:
From the definition of the linear combination $t C + \paren {1 - t} C$, we have: 
:$t C + \paren {1 - t} C = \set {t x + \paren {1 - t} y : x, y \in C}$
for $t \in \R$. 
So we have: 
:$t C + \paren {1 - t} C \subseteq C$
{{iff}}:
:$t x + \paren {1 - t} y \in C$ for all $x, y \in C$ and $t \in \closedint 0 1$.
Hence the result.
{{qed}}
Category:Convex Sets (Vector Spaces)
m6dkrdeb1ib2h3luwecbdzy6pjirfog
Theorem: Interior of Convex Set in Topological Vector Space is Convex

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be a convex set.
Then the interior of $C$, $C^\circ$, is convex.

Proof:
Let $t \in \closedint 0 1$. 
Since $C$ is convex, we have: 
:$t C + \paren {1 - t} C \subseteq C$
Since $C^\circ \subseteq C$, we have: 
{{begin-eqn}}
{{eqn	| l = t C + \paren {1 - t} C
		| r = \set {t x + \paren {1 - t} y : x, y \in C}
		| c = {{Defof|Linear Combination of Subsets of Vector Space}}
}}
{{eqn	| o = \supseteq
		| r = \set {t x + \paren {1 - t} y : x, y \in C^\circ}
}}
{{eqn	| r = t C^\circ + \paren {1 - t} C^\circ
		| c = {{Defof|Linear Combination of Subsets of Vector Space}}
}}
{{end-eqn}}
From Set Interior is Largest Open Set we have that $C^\circ$ is open.
From Dilation of Open Set in Topological Vector Space is Open, we have that $t C^\circ$ and $\paren {1 - t} C^\circ$ are open.
From Sum of Set and Open Set in Topological Vector Space is Open, we have that $t C^\circ + \paren {1 - t} C^\circ$ is open.
So $t C^\circ + \paren {1 - t} C^\circ$ is an open set contained in $C$, so we have: 
:$t C^\circ + \paren {1 - t} C^\circ \subseteq C^\circ$
from the definition of interior.
Since this holds for all $t \in \closedint 0 1$, it follows that $C^\circ$ is convex.
{{qed}}
Theorem: Dilation of Closed Set in Topological Vector Space is Closed Set

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $F$ be a closed set in $X$. 
Let $\lambda \in K \setminus \set {0_K}$.
Then $\lambda F$ is a closed set in $X$.

Proof:
We aim to show that $X \setminus \paren {\lambda F}$ is open.
Since $F$ is closed, $X \setminus F$ is open.
It follows from Dilation of Open Set in Topological Vector Space is Open that $\lambda \paren {X \setminus F}$ is open.
From Dilation of Complement of Set in Vector Space, we have: 
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$.
Since we have established that $\lambda \paren {X \setminus F}$ is open, it follows that $X \setminus \paren {\lambda F}$ is open.
So $\lambda F$ is closed.
{{qed}}
Theorem: Dilation of Closure of Set in Topological Vector Space is Closure of Dilation

Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$. 
Let $A \subseteq X$.
Let $\lambda \in F \setminus \set {0_F}$.
Then we have: 
:$\lambda A^- = \paren {\lambda A}^-$
where $A^-$ denotes the closure of $A$.

Proof:
Let $x \in \lambda A^-$.
Then $\lambda^{-1} x \in A^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $A$ converging to $\lambda^{-1} x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda y_\mu}_{\mu \mathop \in \Lambda}$ converges to $x$.
Since $\lambda y_\mu \in \lambda A$ for each $\mu \in \Lambda$, we have that $x \in \paren {\lambda A}^-$ from Point in Set Closure iff Limit of Net.
So $\lambda A^- \subseteq \paren {\lambda A}^-$. 
Now let $x \in \paren {\lambda A}^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $\lambda A$ converging to $x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda^{-1} y_\mu}_{\mu \mathop \in \Lambda}$ converges to $\lambda^{-1} x$.
Since $\lambda^{-1} y_\mu \in A$ for each $\mu \in \Lambda$, we have that $\lambda^{-1} x \in A^-$ from Point in Set Closure iff Limit of Net..
So $x \in \lambda A^-$.
Hence we have $\paren {\lambda A}^- \subseteq \lambda A^-$.
Hence we conclude that $\lambda A^- = \paren {\lambda A}^-$.
{{qed}}
Theorem: Closure of Convex Set in Topological Vector Space is Convex

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Then the closure $C^-$ of $C$ is convex.

Proof:
Let $t \in \closedint 0 1$.
Since $C$ is convex, we have: 
:$t C + \paren {1 - t} C \subseteq C$ 
We show that: 
:$t C^- + \paren {1 - t} C^- \subseteq C^-$
We have: 
{{begin-eqn}}
{{eqn	| l = t C^- + \paren {1 - t} C^-
		| r = \paren {t C}^- + \paren {\paren {1 - t} C}^-
		| c = Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
}}
{{eqn	| o = \subseteq
		| r = \paren {t C + \paren {1 - t} C}^-
		| c = Sum of Closures is Subset of Closure of Sum in Topological Vector Space
}}
{{eqn	| o = \subseteq
		| r = C^-
		| c = applying Topological Closure of Subset is Subset of Topological Closure since $t C + \paren {1 - t} C \subseteq C$
}}
{{end-eqn}}
{{qed}}
Theorem: Closure of Linear Subspace of Topological Vector Space is Linear Subspace

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $Y$ be a linear subspace of $X$.
Then the closure $Y^-$ of $Y$ is a linear subspace of $X$.

Proof:
Since $Y \subseteq Y^-$, we have that $Y^-$ is non-empty.
We now use the One-Step Vector Subspace Test.
We show that for each $\lambda \in K$ and $u, v \in Y^-$ we have: 
:$u + \lambda v \in Y^-$
That is, we want to show that: 
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda \in K$. 
Let $\lambda \in K$.
Then: 
{{begin-eqn}}
{{eqn	| l = Y^- + \lambda Y^-
		| r = Y^- + \paren {\lambda Y}^-
		| c = Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
}}
{{eqn	| o = \subseteq
		| r = \paren {Y + \lambda Y}^-
		| c = Sum of Closures is Subset of Closure of Sum in Topological Vector Space
}}
{{end-eqn}}
Since $Y$ is a linear subspace, for each $u, v \in Y$ we have: 
:$u + \lambda v \in Y$
So:
:$Y + \lambda Y \subseteq Y$
From Topological Closure of Subset is Subset of Topological Closure, we therefore have: 
:$\paren {Y + \lambda Y}^- \subseteq Y^-$
so that: 
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda \in K$ as required.
{{qed}}
Theorem: Closure of Balanced Set in Topological Vector Space is Balanced

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set.
Then the closure $B^-$ of $B$ is balanced.

Proof:
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$. 
Then, we have: 
:$\lambda B \subseteq B$ 
So, from Topological Closure of Subset is Subset of Topological Closure we have: 
:$\paren {\lambda B}^- \subseteq B^-$
From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have $\paren {\lambda B}^- = \lambda B^-$ and so: 
:$\lambda B^- \subseteq B^-$
So $B^-$ is balanced.
{{qed}}
Theorem: Vanishing Ideal of Zero Locus of Ideal is Radical

Let $k$ be an algebraically closed field.
Let $n \ge 0$ be a natural number.
Let $k \sqbrk {X_1, \ldots, X_n}$ be the polynomial ring in $n$ variables over $k$.
Let $\mathfrak a \subseteq k \sqbrk {X_1, \ldots, X_n}$ be an ideal.
Then:
:$\map I {\map V {\mathfrak a} } = \map \Rad {\mathfrak a}$
where:
:$\map V \cdot$ denotes the zero locus
:$\map I \cdot$ denotes the vanishing ideal
:$\map \Rad \cdot$ denotes the radical

Proof:
This is exactly Hilbert's Nullstellensatz.
{{qed}}
Category:Algebraic Geometry
9z1byf2yecuy18dmzg9bfp6okifwdxi
Theorem: Zero Locus of Vanishing Ideal is Closure

Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $k \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $\mathbb A^n_k$ be equipped with the Zariski topology.
Let $S \subseteq \mathbb A^n_k$.
Then:
:$\map V {\map I S} = S^-$
where:
:$\map I \cdot$ denotes the vanishing ideal
:$\map V \cdot$ denotes the zero locus
:$S^-$ is the closure of $S$.

Proof:
{{MissingLinks|still plenty}}
{{explain|Explanatory words needed to bridge some of the gaps still}}
Let $P := k \sqbrk {x_1, \ldots, x_n}$.
By definition of closure, the set $S^-$ is closed in the Zariski topology, therefore $S^-$ is an affine algebraic set.
By definition of an affine algebraic set, there exists some set $T \subseteq P$ such that:
:$S^- = \map V T$
Let $J$ be the ideal generated by $T$.
By Zero Locus of Set is Zero Locus of Generated Ideal:
:$S^- = \map V J$
Let $p \in J$.
We have that $S^- = \map V J$ is the zero locus of $J$.
Therefore $p$ vanishes on $S^-$.
Therefore:
:$p \in \map I {S^-}$
Thus:
:$J \subseteq \map I {S^-}$
By Ideal and Zero Locus are Order Reversing:
:$\map V {\map I {S^-} } \subseteq \map V J = S^-$
{{qed|lemma}}
We have that $S^-$ is the closure of $S$.
Hence by Set is Subset of its Topological Closure:
:$S \subseteq S^-$
By Ideal and Zero Locus are Order Reversing:
:$\map I {S^-} \subseteq \map I S$
By Ideal and Zero Locus are Order Reversing
:$\map V {\map I S} \subseteq \map V {\map I {S^-}} \subseteq S^-$
$\map I S$ is the vanishing ideal of $S$.
Therefore every $p \in \map I S$ vanishes on $S$.
Thus:
:$S \subseteq \map V {\map I S}$
By the definition of the zero locus, $\map V {\map I S}$ is an affine algebraic set. 
Therefore $\map V {\map I S}$ is closed in the Zariski topology.
By the definition of topological closure:
:$S \subseteq \map V {\map I S}$ implies $S^- \subseteq \map V {\map I S}$
as $S^-$ is the smallest closed set containing $S$.
{{qed|lemma}}
So, we have:
:$\map V {\map I S} \subseteq S^-$
and:
:$S^- \subseteq \map V {\map I S}$
Hence by definition of set equality:
:$S^- = \map V {\map I S}$
{{qed}}
Category:Algebraic Geometry
6q2mf29mr99pycevist7f3xxsz6481c
Theorem: Existence of Disjoint Well-Ordered Sets Isomorphic to Ordinals

Let $\alpha$ and $\beta$ be ordinals.
Then there exist two well-ordered sets $a$ and $b$ such that:
:$a$ and $b$ are order isomorphic to $\alpha$ and $\beta$ respectively
:$a$ and $b$ are disjoint.

Proof:
Let $a$ and $b$ be defined as:
{{begin-eqn}}
{{eqn | l = a
      | o = :=
      | r = \alpha \times \set 0
}}
{{eqn | l = b
      | o = :=
      | r = \beta \times \set 1
}}
{{end-eqn}}
Then let $a$ and $b$ be ordered by their first coordinate.
The result follows.
{{qed}}
Theorem: Dilation of Interior of Set in Topological Vector Space is Interior of Dilation

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$.
Let $\lambda \in K \setminus \set 0$.
Then we have: 
:$\lambda A^\circ = \paren {\lambda A}^\circ$
where $A^\circ$ denotes the interior of $A$.

Proof:
From the definition of interior we have:
:$\ds A^\circ = \bigcup \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$
For brevity write: 
:$\ds \SS_1 = \leftset {U \subseteq A: U}$ is open in $\rightset X$
and:
:$\ds \SS_2 = \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$
so that: 
:$\ds A^\circ = \bigcup_{U \mathop \in \SS_1} U$
and:
:$\ds \paren {\lambda A}^\circ = \bigcup_{U \mathop \in \SS_2} U$
From Dilation of Union of Subsets of Vector Space, it now suffices to show that: 
:$\SS_2 = \set {\lambda U : U \in \SS_1}$
Let $U \in \SS_1$.
Then $U$ is open and $U \subseteq A$.
From Dilation of Open Set in Topological Vector Space is Open, $\lambda U$ is open.
Since $\lambda U \subseteq \lambda A$, we have: 
:$\set {\lambda U: U \in \SS_1} \subseteq \SS_2$
Now let $U' \in \SS_2$.
Then $U'$ is open and $U' \subseteq \lambda A$.
Let $U = \lambda^{-1} U'$.
From Dilation of Open Set in Topological Vector Space is Open, $U$ is open.
We have $U \subseteq A$, so $U \in \SS_1$. 
Since we have $U' = \lambda U$ for $U \in \SS_1$, we have: 
:$U' \in \set {\lambda U: U \in \SS_1}$
giving:
:$\SS_2 \subseteq \set {\lambda U : U \in \SS_1}$
So:
:$\SS_2 = \set {\lambda U : U \in \SS_1}$
by the definition of set equality.
So: 
{{begin-eqn}}
{{eqn	| l = \paren {\lambda A}^\circ
		| r = \bigcup_{U' \mathop \in \SS_2} U'
}}
{{eqn	| r = \bigcup_{U \mathop \in \SS_1} \paren {\lambda U}
}}
{{eqn	| r = \lambda \bigcup_{U \mathop \in \SS_1} U
		| c = Dilation of Union of Subsets of Vector Space
}}
{{eqn	| r = \lambda A^\circ
}}
{{end-eqn}}
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Set Interiors
Category:Topological Vector Spaces
r97biuklbn9760dqn4fvgcf8rbu6kf1
Theorem: Interior of Balanced Set containing Origin in Topological Vector Space is Balanced

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set such that: 
:${\mathbf 0}_X \in B^\circ$
where $B^\circ$ is the interior of $B$.
Then $B^\circ$ is balanced.

Proof:
Let $\lambda \in \Bbb F$ have $0 < \cmod \lambda \le 1$. 
Then, we have: 
:$\lambda B \subseteq B$ 
From Interior of Subset, we have: 
:$\paren {\lambda B}^\circ \subseteq B^\circ$
Then from Dilation of Interior of Set in Topological Vector Space is Interior of Dilation we have: 
:$\lambda B^\circ \subseteq B^\circ$
for $\lambda \in \Bbb F$ with $0 < \cmod \lambda \le 1$.
By hypothesis, we have: 
:${\mathbf 0}_X \in B^\circ$
and so:
:$\set { {\mathbf 0}_X} \subseteq B^\circ$
So we have: 
:$\lambda B^\circ \subseteq B^\circ$
if $\lambda = 0$. 
So:
:$\lambda B^\circ \subseteq B^\circ$
for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $B^\circ$ is balanced.
{{qed}}
Theorem: Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $a, b \in X$.
Let $W$ be an open neighborhood of $a + b$.
Then there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that: 
:$W_1 + W_2 \subseteq W$
where $W_1 + W_2$ denotes the sum of $W_1$ and $W_2$.

Proof:
Equip $X \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $S : X \times X \to X$ by: 
:$\map S {x, y} = x + y$
for each $\tuple {x, y} \in X \times X$. 
From the definition of a topological vector space, $S$ is continuous.
In particular, $S$ is continuous at $\tuple {a, b}$.
Hence, there exists an open neighborhood $V$ of $\tuple {a, b}$ such that: 
:$S \sqbrk V \subseteq W$ 
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $W_1$ of $a$ and an open neighborhood $W_2$ of $b$ such that:
:$W_1 \times W_2 \subseteq V$
Then we have, from Image of Subset under Mapping is Subset of Image: 
:$S \sqbrk {W_1 \times W_2} \subseteq S \sqbrk V$
Since: 
:$S \sqbrk {W_1 \times W_2} = W_1 + W_2$
this gives: 
:$W_1 + W_2 \subseteq W$
as required.
{{qed}}
Category:Topological Vector Spaces
Category:Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods
jf0b8qzkugegvqbqja8cv90tu21fzeg
Theorem: Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods/Corollary 1

Let $W$ be an open neighborhood of ${\mathbf 0}_X$. 
Then there exists a symmetric open neighborhood $U$ of ${\mathbf 0}_X$ such that: 
:$U + U \subseteq W$

Proof:
Since ${\mathbf 0}_X + {\mathbf 0}_X = {\mathbf 0}_X$, we can apply Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods to obtain an open neighborhood $V_1$ of ${\mathbf 0}_X$ and an open neighborhood $V_2$ of ${\mathbf 0}_X$ such that: 
:$V_1 + V_2 \subseteq W$
Note at this point we do not necessarily have $V_1 = V_2$.
From Dilation of Open Set in Topological Vector Space is Open, $-V_1$ and $-V_2$ are open neighborhoods of ${\mathbf 0}_X$.
Let:
:$U = V_1 \cap V_2 \cap \paren {-V_1} \cap \paren {-V_2}$
Since $U$ is the finite intersection of open sets, it is itself open.
Since ${\mathbf 0}_X$ is contained in each of the four sets of the intersection, we have ${\mathbf 0}_X \in U$.
So $U$ is an open neighborhood of ${\mathbf 0}_X$.
We show that $U$ is symmetric and that: 
:$U + U \subseteq W$
Let $x \in U$. 
Then $x \in V_1$, $x \in V_2$, $x \in -V_1$ and $x \in -V_2$.
So we have that $-x \in -V_1$, $-x \in -V_2$, $-x \in V_1$ and $-x \in -V_2$.
So:
:$-x \in \paren {-V_1} \cap \paren {-V_2} \cap V_1 \cap V_2 = U$
So we have that $U$ is symmetric.
Now let $x, y \in U$. 
We want to show that $x + y \in W$.
Since $x, y \in U$, we have $x \in V_1$ and $y \in V_2$.
So, we have $x + y \in V_1 + V_2$.
We have $V_1 + V_2 \subseteq W$, so $x + y \in W$.
So, we have $U + U \subseteq W$. 
Hence $U$ is our desired open neighborhood.
{{qed}}
Category:Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods
rhhdqfht8iphugsy8yu6dhhujl2rxep
Theorem: Sum of Union of Subsets of Vector Space and Subset

Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be a $I$-indexed family of subsets of $X$. 
Let $C \subseteq X$. 
Then: 
:$\ds C + \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$

Proof:
We have: 
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
{{iff}} there exists $c \in C$, $\alpha \in I$, $x \in E_\alpha$ such that: 
:$v = c + x$
That is, {{iff}} $v = c + x$ for $c \in C$ and:
:$\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$ 
This is equivalent to: 
:$\ds v \in C + \bigcup_{\alpha \mathop \in I} E_\alpha$
so, we get:
:$\ds C + \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {C + E_\alpha}$
by the definition of set equality.
{{qed}}
Category:Vector Spaces
Category:Set Union
m5a4g7yhjxz75b4jk1djlfq2zgocx8n
Theorem: Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $E$ be a von Neumann-bounded subset of $X$.
Then the closure $E^-$ of $E$ is von Neumann-bounded.

Proof:
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
From Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that: 
:$W^- \subseteq V$
Since $E$ is von Neumann-bounded, there exists $s > 0$ such that: 
:$E \subseteq t W$ for $t \in \Bbb F$ with $t > s$.
From Topological Closure of Subset is Subset of Topological Closure, we have: 
:$E^- \subseteq \paren {t W}^-$ for $t \in \Bbb F$ with $t > s$.
From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have: 
:$\paren {t W}^- = t W^-$
Then we have: 
:$E^- \subseteq t W^- \subseteq t V$ for $t \in \Bbb F$ with $t > s$.
Since $V$ was arbitrary, we have that $E^-$ is von Neumann-bounded.
{{qed}}
Theorem: Function is Convex iff Epigraph is Convex

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $C \subseteq X$ be a convex set.
Consider $X \times \GF$ as the direct product of $X$ and $\GF$.
Let $f : C \to \overline \R$ be a function.
Let $\map {\operatorname {epi} } f$ be the epigraph of $f$.
Then $f$ is convex {{iff}} $\map {\operatorname {epi} } f$ is convex in $X \times \GF$.

Proof:
=== Necessary Condition ===
Suppose that $f$ is convex.
Let $t \in \openint 0 1$. 
Let $\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$.
Then:
:$\map f {x_1} \le \alpha_1 < \infty$
and:
:$\map f {x_2} \le \alpha_2 < \infty$ 
We therefore have, since $t > 0$:
:$t \map f {x_1} + \paren {1 - t} \map f {x_2} \le t \alpha_1 + \paren {1 - t} \alpha_2$
Since $f$ is convex, we have:
:$\map f {t x_1 + \paren {1 - t} x_2} \le t \map f {x_1} + \paren {1 - t} \map f {x_2}$
Hence:
:$\map f {t x_1 + \paren {1 - t} x_2} \le t \alpha_1 + \paren {1 - t} \alpha_2$ 
Hence:
:$\tuple {t x_1 + \paren {1 - t} x_2, t \alpha_1 + \paren {1 - t} \alpha_2} \in \map {\operatorname {epi} } f$
so that:
:$t \tuple {x_1, \alpha_1} + \paren {1 - t} \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$
whenever:
:$\tuple {x_1, \alpha_1}, \tuple {x_2, \alpha_2} \in \map {\operatorname {epi} } f$
Hence $\map {\operatorname {epi} } f$ is convex in $X \times \GF$.
{{qed|lemma}}