avewright/proofwiki-math huggingface dataset:
Theorem: Translation of Convex Set in Vector Space is Convex

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Let $x \in X$.
Then $C + x$, the translation of $C$ by $x$, is convex.

Proof:
Let $t \in \closedint 0 1$ and $u, v \in C + x$.
Then there exists $u', v' \in C$ such that: 
:$u = u' + x$
and:
:$v = v' + x$ 
Then, we have:
{{begin-eqn}}
{{eqn	| l = t u + \paren {1 - t} v
		| r = t u' + t x + \paren {1 - t} v' + \paren {1 - t} x
}}
{{eqn	| r = \paren {t u' + \paren {1 - t} v'} + \paren {t + \paren {1 - t} } x
}}
{{eqn	| r = \paren {t u' + \paren {1 - t} v'} + x
}}
{{end-eqn}}
Since $C$ is convex, we have:
:$t u' + \paren {1 - t} v' \in C$
so:
:$t u + \paren {1 - t} v \in C + x$
So we have that $C + x$ is convex.
{{qed}}
Category:Translation of Subsets of Vector Spaces
Category:Convex Sets (Vector Spaces)
3pf7mh1yoq1idz6b8guml223qog0g13
Theorem: Sum of Convex Sets in Vector Space is Convex

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Then: 
:$A + B$ is convex.

Proof:
Let $x, y \in A + B$ and $t \in \closedint 0 1$. 
Then there exists $a, a' \in A$ and $b, b' \in B$ such that: 
:$x = a + b$
and:
:$y = a' + b'$
Then:
{{begin-eqn}}
{{eqn	| l = t x + \paren {1 - t} y
		| r = t a + t b + \paren {1 - t} a' + \paren {1 - t} b'
}}
{{eqn	| r = \paren {t a + \paren {1 - t} a'} + \paren {t b + \paren {1 - t} b'}
}}
{{end-eqn}}
Since $A$ and $B$ are convex, we have:
:$t a + \paren {1 - t} a' \in A$
and:
:$t b + \paren {1 - t} b' \in B$
so that:
:$t x + \paren {1 - t} y \in A + B$
So $A + B$ is convex.
{{qed}}
Category:Convex Sets (Vector Spaces)
Category:Sum of Convex Sets in Vector Space is Convex
k7kc7vf4isu485halwjrddat7vqh3tr
Theorem: Sum of Convex Sets in Vector Space is Convex/Corollary

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $A$ and $B$ be convex subsets of $X$.
Let $\lambda, \mu \in \Bbb F$.
Then:
:$\lambda A + \mu B$ is convex.

Proof:
From Dilation of Convex Set in Vector Space is Convex, we have: 
:$\lambda A$ and $\mu B$ are convex.
From Sum of Convex Sets in Vector Space is Convex, we have:
:$\lambda A + \mu B$ is convex.
{{qed}}
Category:Sum of Convex Sets in Vector Space is Convex
nzn7p472agtl9fr99dz8pqzkp5rtt93
Theorem: Closed Convex Set in terms of Bounded Linear Functionals

Let $X$ be a normed vector space over $\R$.
Let $X^\ast$ be the normed dual of $X$. 
Let $C$ be a closed convex subset of $X$.
Then: 
:$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$

Proof:
{{finish|A few of these links only apply to finite sups/infs so will need to be replaced. they remain as placeholders}}
Clearly if $x \in C$ then: 
:$\ds \map f x \le \sup_{c \in C} \map f c$
for all $f \in X^\ast$, by the definition of supremum.
That is: 
:$\ds C \subseteq \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
To show that: 
:$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$
we prove that if $x_0 \not \in C$ then: 
:$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
The Rule of Transposition will then give us the result. 
Let $x_0 \not \in C$.
From Finite Topological Space is Compact, we have that $\set {x_0}$ is compact. 
Clearly $\set {x_0}$ is disjoint from $C$. 
We can therefore apply Hahn-Banach Separation Theorem: Compact Convex Set and Closed Convex Set, which gives the existence of $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that: 
:$\map f {x_0} \le c - \epsilon < c + \epsilon \le \map f x$ for each $x \in C$.
From the definition of infimum, we therefore have: 
:$\ds c + \epsilon \le \inf_{x \in C} \map f x$
so that: 
:$\ds \map f {x_0} \le c - \epsilon < c + \epsilon \le \inf_{x \in C} \map f x$
Then, from Negative of Supremum is Infimum of Negatives, we have:
:$\ds -\map f {x_0} \ge \epsilon - c > -\paren {c + \epsilon} \ge \sup_{x \in C} \paren {-\map f x}$
Setting $g = -f \in X^\ast$ we have: 
:$\ds \map g {x_0} > \sup_{x \in C} \map g x$
So:
:$\ds x_0 \not \in \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
as required.
Hence we obtain: 
:$\ds \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c} \subseteq C$
and:
:$\ds C = \bigcap_{f \in X^\ast} \set {x \in X : \map f x \le \sup_{c \in C} \map f c}$
{{qed}}
Category:Normed Dual Spaces
Category:Convex Sets (Vector Spaces)
9fzwy9rx19rpt7tuu915y7x2hxm9caq
Theorem: Transfinite Recursion Theorem/Formulation 2

Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
Then there exists a '''unique''' mapping $F$ on $\On$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
where $F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$.

Proof:
First we establish the following
Theorem: Transfinite Recursion Theorem/Formulation 4

Let $\On$ denote the class of all ordinals.
Let $g$ be a mapping defined for all sets.
Let $c$ be a set.
Then there exists a '''unique''' $\On$-sequence $S_0, S_1, \dots, S_\alpha, \dots$ such that:
{{begin-axiom}}
{{axiom | n = 1
        | q = 
        | ml= S_0
        | mo= =
        | mr= c
}}
{{axiom | n = 2
        | q = \forall \alpha \in \On
        | ml= S_{\alpha + 1}
        | mo= =
        | mr= \map g {S_\alpha}
}}
{{axiom | n = 3
        | q = \forall \lambda \in K_{II}
        | ml= S_\lambda 
        | mo= =
        | mr= \bigcup_{\alpha \mathop < \lambda} S_\alpha
}}
{{end-axiom}}
where $K_{II}$ denotes the class of all limit ordinals.

Proof:
This is a special case of the Transfinite Recursion Theorem: Formulation $3$.
Recall:
{{:Transfinite Recursion Theorem/Formulation 3}}{{qed|lemma}}
Set $\map f x = \bigcup x$, so that:
:$\map f {F \sqbrk \lambda}$
is then:
:$\ds \bigcup_{\alpha \mathop < \lambda} \map F \alpha$
Let $S_\alpha$ be the set $\map F \alpha$
and the given form of the Transfinite Recursion Theorem follows.
It remains to demonstrate uniqueness of $S_0, S_1, \dots, S_\alpha, \dots$.
{{finish|"an obvious transfinite argument"}}
Theorem: Intersection of Zero Loci is Zero Locus

Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $\mathbb S \subseteq \powerset A$ be a subset of the power set of $A$.
Then:
:$\ds \bigcap _{S \mathop \in \mathbb S} \map V S = \map V {\bigcup \mathbb S}$
where $\map V \cdot$ denotes the zero locus.

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \bigcap_{S \mathop \in \mathbb S} \map V S
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall S \in \mathbb S
      | l = x
      | o = \in
      | r = \map V S
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall S \in \mathbb S: \forall f \in S
      | l = \map f x
      | r = 0
      | c = {{Defof|Zero Locus of Set of Polynomials}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall f \in \bigcup \mathbb S
      | l = \map f x
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \in
      | r = \map V {\bigcup \mathbb S}
}}
{{end-eqn}}
{{qed}}
Category:Algebraic Geometry
musfmgbcfcmvo73roqdlm062pisk0vi
Theorem: Union of two Zero Loci is Zero Locus

Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $S,T \subseteq A$.
Then:
:$\ds {\map V S} \cup \map V T = \map V {S \cdot T}$
where:
:$\map V \cdot$ denotes the zero locus
:$S \cdot T := \set { fg : f \in S, g \in T }$

Proof:
Let $x \in {\map V S} \cup \map V T$.
That is, either $x \in \map V S$ or $x \in \map V T$.
Then, if $x \in \map V S$:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = 0 \map g x  = 0$
else:
:$\forall \paren {f,g} \in S \times T : \map f x \map g x = \map f x 0  = 0$
Therefore:
:$x \in \map V {S \cdot T}$
On the other hand:
{{begin-eqn}}
{{eqn | l = x
      | o = \not \in
      | r = {\map V S} \cup \map V T
}}
{{eqn | ll = \leadsto
      | l = x
      | o = \in
      | r = \paren {k^n \setminus \map V S} \cap \paren {k^n \setminus \map V T}
      | c = De Morgan's Laws
}}
{{eqn | ll = \leadsto
      | l = \exists f \in S : \map f x \ne 0
      | o = \land
      | r = \exists g \in T : \map g x \ne 0
}}
{{eqn | ll = \leadsto
      | q = \exists \paren {f,g} \in S \times T
      | l = \map f x \map g x
      | o = \ne
      | r = 0
}}
{{eqn | ll = \leadsto
      | l = x
      | o = \not \in
      | r = \map V {S \cdot T}
}}
{{end-eqn}}
{{qed}}
Category:Algebraic Geometry
tcfx1dd048y3wpepxxxjcykfj8wdvv4
Theorem: Zariski Topology is Topology

Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $\tau$ be the Zariski topology on $k^n$.
Then $\tau$ is indeed a topology on $k^n$.

Proof:
Let $A := k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials.
Recall that by definition of Zariski topology:
:$\forall U \in \tau: \exists T_U \subseteq A: U = X \setminus \map V {T_U}$
where $\map V {T_U}$ denotes the zero locus of $T_U$.
Each of the open set axioms is examined in turn:
Theorem: Pointwise Maximum of Finite Family of Seminorms is Seminorm

Let $\struct {K, \norm {\,\cdot\,}_K}$ be a normed division ring.
Let $X$ be a vector space over $K$. 
Let $\II$ be a set of seminorms on $X$.
Define: 
:$\ds q =: \max_{p \mathop \in \II} p$
where $\max$ denotes the pointwise maximum over $\II$.

Proof:
=== Proof of $(\text N 2)$ ===
Let $\lambda \in K$ and $x \in X$. 
Then: 
{{begin-eqn}}
{{eqn	| l = \map q {\lambda x}
		| r = \max_{p \mathop \in \II} \map p {\lambda x}
        | c = Definition of $q$
}}
{{eqn	| r = \max_{p \mathop \in \II} \norm \lambda_K \map p x
		| c = $(\text N 2)$ in the definition of the seminorm
}}
{{eqn	| r = \norm \lambda_K \max_{p \mathop \in \II} \map p x
}}
{{eqn	| r = \norm \lambda_K \max_{p \mathop \in \II} \map p x
}}
{{end-eqn}}
proving $(\text N 2)$.
{{qed|lemma}}
Theorem: Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin

Let $K$ be a topological field.
Let $X$ and $Y$ be topological vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Then $T$ is everywhere continuous {{iff}} it is continuous at ${\mathbf 0}_X$.

Proof:
=== Necessary Condition ===
Suppose that $T$ is everywhere continuous.
Then $T$ is continuous at every point.
In particular, $T$ is continuous at ${\mathbf 0}_X$.
{{qed|lemma}}
Theorem: Topological Vector Space as Union of Dilations of Open Neighborhood of Origin

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$. 
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
Let $\sequence {r_n}_{n \mathop \in \N}$ be a sequence of positive real numbers such that $r_n \to \infty$. 
Then: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$
where $r_n V$ denotes the dilation of $V$ by $r_n$.

Proof:
Clearly: 
:$\ds \bigcup_{n \mathop = 1}^\infty r_n V \subseteq X$
We show that: 
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
Let $x \in X$. 
From the definition of a topological vector space, the mapping $K \times X \to X$ defined by $\tuple {\lambda, x} \mapsto \lambda x$ is continuous.
From Horizontal Section of Continuous Function is Continuous, it follows that the $x$-horizontal section with $\lambda \mapsto \lambda x$ is continuous.
Hence, the set:
:$U = \set {\lambda \in \Bbb F : \lambda x \in V}$ 
is open.
Since $0 x = {\mathbf 0}_X$ and $V$ is an open neighborhood of ${\mathbf 0}_X$, we have: 
:$U$ is an open neighborhood of $0 \in \Bbb F$.
So, there exists $\epsilon > 0$ such that if $\cmod \lambda < \epsilon$, we have $\lambda x \in U$.
From the definition of a real sequence tending to $+\infty$, there exists $N \in \N$ such that: 
:$\ds r_n > \frac 1 \epsilon$ for $n \ge N$. 
Then: 
:$\ds 0 < \frac 1 {r_N} < \epsilon$
so that: 
:$\dfrac 1 {r_N} x \in V$
giving:
:$x \in r_N V$
So, we have: 
:$\ds x \in \bigcup_{n \mathop = 1}^\infty r_n V$
Since $x \in X$ was arbitrary we have: 
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
giving:
:$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$
{{qed}}
Theorem: Condition for Point being in Closure/Topological Vector Space

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$. 
Let $A^-$ denote the closure of $A$ in $X$.
Let $x \in X$. 
Then $x \in A^-$ {{iff}}:
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$.

Proof:
From the definition of the closure of $A$, $A^-$ is the set of adherent points of $A$.
So $x \in A^-$ {{iff}} for each open neighborhood $U$ of $x$ we have: 
:$U \cap A \ne \O$ 
From Classification of Open Neighborhoods in Topological Vector Space, every such $U$ has the form $U = x + V$ for some open neighborhood $V$ of ${\mathbf 0}_X$, and conversely every set of this form is an open neighborhood of $x$. 
So we have $x \in A^-$ {{iff}}:
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. 
{{qed}}
Category:Condition for Point being in Closure
Category:Topological Vector Spaces
7vll0cqjzkhidq10idgxzqlka7ci9a9
Theorem: Expression for Closure of Set in Topological Vector Space

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$. 
Let $\mathcal V$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Then: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$
where $A^-$ is the closure of $A$.

Proof:
From Condition for Point being in Closure: Topological Vector Space, we have $x \in A^-$ {{iff}}:
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. 
We have that $\paren {x + V} \cap A \ne \O$ {{iff}} there exists some $u \in X$ with $u \in A$ and $u \in x + V$.
This is equivalent to, $u \in A$ and there exists $v \in V$ such that $u = x + v$. 
This is equivalent to, $x = u - v$ for $u \in A$ and $v \in V$.
That is, equivalent to $x \in A - V$.
So, we have that $x \in A^-$ {{iff}} $x \in A - V$ for all open neighborhoods $V$ of ${\mathbf 0}_X$.
So, we have: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A - V}$
From Dilation of Open Set in Topological Vector Space is Open, we have that $-V$ is open {{iff}} $V$ is open.
Writing $A - V = A + \paren {-V}$, we therefore see that: 
:$\ds A^- = \bigcap_{V \in \mathcal V} \paren {A + V}$
{{qed}}
Theorem: Union of Balanced Sets in Vector Space is Balanced

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$. 
Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of balanced subsets of $X$.
Then:
:$\ds E = \bigcup_{\alpha \mathop \in I} E_\alpha$ is balanced.

Proof:
Let $x \in E$.
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$.
We aim to show that $\lambda x \in E$. 
Since $x \in E$, there exists $\alpha \in I$ with $x \in E_\alpha$. 
Since $E_\alpha$ is balanced, we have $\lambda x \in E_\alpha$. 
So:
:$\ds \lambda x \in \bigcup_{\alpha \mathop \in I} E_\alpha = E$
Since $x$ and $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$ were arbitrary, $E$ is balanced.
{{qed}}
Category:Balanced Sets
Category:Set Union
97idbwklfyt0rnc0qio33zd4ikqzlvi
Theorem: Transfinite Recursion Theorem/Formulation 5

Let $\On$ denote the class of all ordinals.
Let $h$ be a mapping.
Then there exists a '''unique''' mapping $F$ such that:
:$\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$

Proof:
This is a special case of the Transfinite Recursion Theorem: Formulation $2$.
Recall:
{{:Transfinite Recursion Theorem/Formulation 2}}{{qed|lemma}}
Consider $\map g x$.
If $x$ is a mapping, take $\map g x = \map h {\Img x}$.
If $x$ is not a mapping, take $\map g x = x$.
The result follows.
{{qed}}
Theorem: Class Mapping has Minimally Superinductive Class

Let $g$ be a (class) mapping.
Then there exists a class $M$ that is minimally superinductive under $g$.

Proof:
{{ProofWanted|According to the rubric of the exercise, this is to be proved using one of the versions of the Transfinite Recursion Theorem that appears in S&F}}
Theorem: Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\Bbb F$. 
Let ${\mathbf 0}_X$ be the origin of $X$.
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
Then there exists a balanced open neighborhood $W$ of ${\mathbf 0}_X$ with $W \subseteq U$.

Proof:
Equip $\Bbb F \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $M : \Bbb F \times X \to X$ by: 
:$\map M {\lambda, x} = \lambda x$
for each $\tuple {\lambda, x} \in \Bbb F \times X$.
From the definition of a topological vector space, $M$ is continuous.
In particular, $M$ is continuous at $\tuple {0, {\mathbf 0}_X}$.
So there exists an open neighborhood $A$ of $\tuple {0, {\mathbf 0}_X}$ such that: 
:$M \sqbrk A \subseteq U$
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $0$ and an open neighborhood $V$ of ${\mathbf 0}_X$ such that: 
:$D \times V \subseteq A$
Using:
:the definition of an open set in $\R$ if $\Bbb F = \R$
:the definition of an open set in $\C$ if $\Bbb F = \C$
there exists $r > 0$ such that if $\alpha \in \Bbb F$ has $\cmod \alpha < r$ we have $\alpha \in D$.
Then for $\alpha \in \Bbb F$ with $\cmod \alpha < r$ we have: 
:$\set \alpha \times V \subseteq D \times V$
so that: 
:$M \sqbrk {\set \alpha \times V} \subseteq U$
from Image of Subset under Mapping is Subset of Image.
We have: 
:$M \sqbrk {\set \alpha \times V} = \alpha V$
so we have: 
:$\alpha V \subseteq U$
for each $\alpha \in \Bbb F$ with $\cmod \alpha < r$. 
Let: 
:$\ds W = \bigcup_{\cmod \alpha < r} \alpha V$
From Dilation of Open Set in Topological Vector Space is Open, $\alpha V$ is open for each $\alpha \in \Bbb F$.
Since $W$ is the union of open sets, $W$ is open.
From Union of Subsets is Subset, we have $W \subseteq U$.
Since $0 \in \alpha V$ for all $\alpha \in \Bbb F$ with $\cmod \alpha < r$, we also have $0 \in W$.
So $W$ is an open neighborhood of $0$.
It remains to show that $W$ is balanced.
Let $x \in W$ and $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$. 
Then there exists $\alpha \in \Bbb F$ with $\cmod \alpha < r$ such that $x \in \alpha V$. 
Then we have $\lambda x \in \paren {\alpha \lambda} V$.
We have $\cmod {\alpha \lambda} < r$, so $\paren {\alpha \lambda} V \subseteq W$.
So we have $\lambda x \in W$.
So we have $\lambda W \subseteq W$ for all $\lambda \in \Bbb F$ with $\cmod \lambda \le 1$, so $W$ is balanced.
{{qed}}
Theorem: Finite Union of Dilations of Balanced Set

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$. 
Let $\lambda_1, \lambda_2, \ldots, \lambda_n$ be real numbers.
Let $W$ be a balanced set. 
Then: 
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
where $\lambda \in \set {\lambda_1, \lambda_2, \ldots, \lambda_n}$ is picked so that $\cmod \lambda$ is maximized.

Proof:
If $\lambda = 0$, then $\lambda_i = 0$ for each $i \in \set {1, 2, \ldots, n}$, so that:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \set {\mathbf 0_X} = \lambda W$
Now suppose that $\lambda \ne 0$.
Then for each $i$ we have:
:$\ds \cmod {\frac {\lambda_i} \lambda} = \frac {\cmod {\lambda_i} } {\cmod \lambda} \le 1$
since we picked $\lambda$ among the $\lambda_i$ to maximise $\cmod \lambda$.
Since $W$ is balanced set, we then have:
:$\ds \frac {\lambda_i} \lambda W \subseteq W$
so that:
:$\lambda_i W \subseteq \lambda W$
for each $i$.
So, from Union of Subsets is Subset, we have:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W \subseteq \lambda W$
From Set is Subset of Union, we also have:
:$\ds \lambda W \subseteq \bigcup_{i \mathop = 1}^n \lambda_i W$
From definition of set equality:
:$\ds \bigcup_{i \mathop = 1}^n \lambda_i W = \lambda W$
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Balanced Sets
t1qjs7m8knsv0z0o7cllpps5xal7jw6
Theorem: Compact Subspace 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 $K$ be a compact subspace (where subspace is meant in the topological sense) of $X$.
Then $K$ is von Neumann-bounded.

Proof:
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
We aim to find $s > 0$ such that $E \subseteq t V$ for $t > s$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists $W \subseteq V$ such that $W$ is a balanced open neighborhood of ${\mathbf 0}_X$.
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, we have: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty n W$
In particular: 
:$\ds K \subseteq \bigcup_{n \mathop = 1}^\infty n W$
Since $K$ is compact, there exists $n_1, n_2, \ldots, n_k$ such
:$\ds K \subseteq \bigcup_{i \mathop = 1}^k n_i W$
Relabel these $n_i$ so that: 
:$n_1 < n_2 < \ldots < n_k$
From Finite Union of Dilations of Balanced Set, we then have: 
:$\ds \bigcup_{i \mathop = 1}^k n_i W = n_k W$
so that:
:$K \subseteq n_k W$
Since $W$ is balanced, for $t > n_k$ we have: 
:$\ds \frac {n_k} t W \subseteq W$
so that: 
:$n_k W \subseteq t W$
Then, we have: 
:$K \subseteq t W$
Since $W \subseteq V$, we obtain: 
:$K \subseteq t V$
for $t > n_k$.
Since $V$ was arbitrary, we have that $K$ is von Neumann-bounded.
{{qed}}
Theorem: Dilations of von Neumann-Bounded Neighborhood of Origin in Topological Vector Space form Local Basis for Origin

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$. 
Let $\sequence {\delta_n}_{n \mathop \in \N}$ be a strictly decreasing real sequence with $\delta_n \to 0$.
Let $V$ be a von Neumann-bounded open neighborhood of ${\mathbf 0}_X$.
Then: 
:$\BB = \set {\delta_n V : n \in \N}$ is a local basis for ${\mathbf 0}_X$.

Proof:
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
We show that there exists $N \in \N$ such that: 
:$\delta_N V \subseteq U$ 
Since $V$ is von Neumann-bounded, there exists $s > 0$ such that: 
:$V \subseteq t U$ for $t > s$.
Since $\delta_n \to 0$ and $\sequence {\delta_n}_{n \mathop \in \N}$ is a strictly decreasing real sequence, there exists $N \in \N$ with: 
:$\ds 0 < \delta_N < \frac 1 s$
so that: 
:$\ds \frac 1 {\delta_N} > s$ 
giving:
:$\ds V \subseteq \frac 1 {\delta_N} U$
so that: 
:$\delta_N V \subseteq U$
Since $U$ was arbitrary, we have that $\BB$ is a local basis for ${\mathbf 0}_X$.
{{qed}}
Theorem: Characterization of Continuous Linear Functionals on Topological Vector Space

Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\Bbb F$.
Let $f$ be a linear functional on $X$ such that $\map f x \ne 0$ for some $x \in X$. 
{{TFAE}}
:$(1) \quad$ $f$ is continuous
:$(2) \quad$ $\ker f$ is closed, where $\ker f$ is the kernel of $f$
:$(3) \quad$ $\ker f$ is not everywhere dense in $X$
:$(4) \quad$ $f$ is bounded in a open neighborhood of ${\mathbf 0}_X$.

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that $f$ is continuous.
From the definition of the kernel, we have: 
:$\ker f = f^{-1} \sqbrk {\set 0}$
Since $\set 0$ is closed in $\GF$, we have that: 
:$f^{-1} \sqbrk {\set 0}$ is closed
from Continuity Defined from Closed Sets.
{{qed|lemma}}