avewright/proofwiki-math huggingface dataset:
Theorem: Finite Intersection of Absorbing Sets is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A_1, \ldots, A_n \subseteq X$ be absorbing sets.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is absorbing.

Proof:
Let $x \in X$ and $j \in \set {1, \ldots, n}$.
Since $A_j$ is absorbing, there exists $t_j \in \R_{> 0}$ such that:
:$x \in \alpha A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t_j$. 
Let $t = \max \set {t_1, \ldots, t_n}$.
Then, we have:
:$\ds x \in \bigcap_{j \mathop = 1}^n \paren {\alpha A_j}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
From Dilation of Intersection of Subsets of Vector Space, we have:
:$\ds x \in \alpha \bigcap_{j \mathop = 1}^n A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
So:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is absorbing.
{{qed}}
Category:Absorbing Sets
fqemg8yp4f5age0tszj8lgnfgwuwbhq
Theorem: Non-Zero Dilation of Absorbing Set is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Let $\lambda \in \GF \setminus \set 0$. 
Then $\lambda A$ is absorbing.

Proof:
Let $x \in X$.
Then there exists $t \in \R_{> 0}$ such that: 
:$x \in \alpha A$ for $\cmod \alpha \ge t$.
That is: 
:$\ds x \in \frac \alpha \lambda \paren {\lambda A}$ for $\cmod \alpha \ge t$.
Since the map $\alpha \mapsto \alpha/\lambda$ is a bijection from $\cmod \alpha \ge t$ to $\cmod \alpha \ge t/\cmod \lambda$, we have:
:$\ds x \in \beta \paren {\lambda A}$ for $\cmod \beta \ge \dfrac t {\cmod \lambda}$.
So $\lambda A$ is absorbing.
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Absorbing Sets
otnv0oa4g8gnz8pc6fcxc0z95fq3f4j
Theorem: Sum of Absorbing Sets is Absorbing

Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A, B \subseteq X$ be absorbing sets.
Then $A + B$ is absorbing.

Proof:
Let $u \in A + B$. 
Then there exists $a \in A$, $b \in B$ such that $u = a + b$.
Now, there exists $t_1 \in \R_{> 0}$ such that: 
:$a \in \alpha A$ for $\alpha \in \C$ with $\cmod \alpha \ge t_1$
There also exists $t_2 \in \R_{> 0}$ such that:
:$b \in \alpha B$ for $\alpha \in \C$ with $\cmod \alpha \ge t_2$
Let $t = \max \set {t_1, t_2}$. 
Then for $\alpha \in \C$ with $\cmod \alpha \ge t$, we have:
:$u = a + b \in \alpha A + \alpha B = \alpha \paren {A + B}$
from Dilation of Subset of Vector Space Distributes over Sum.
So $A + B$ is absorbing.
{{qed}}
Category:Absorbing Sets
h3a5kleznzv8ge1hbb8te7yqq2o4dk6
Theorem: Sum of Compact Subsets of Topological Vector Space is Compact

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $A$ and $B$ be compact (topological) subspaces of $X$. 
Then $A + B$ is compact.

Proof:
Since $X$ is a topological vector space, the map $+ : X \times X \to X$ defined by:
:$\map + {x, y} = x + y$
for each $x, y \in X$ is continuous.
We have:
:$A + B = + \sqbrk {A \times B}$
From Tychonoff's Theorem, $A \times B$ is compact.
From Continuous Image of Compact Space is Compact, $+ \sqbrk {A \times B} = A + B$ is therefore compact.
{{qed}}
Category:Compact Topological Spaces
Category:Topological Vector Spaces
oc3epeqva52y64twhs11anfthax8ule
Theorem: Quotient Mapping on Quotient Topological Vector Space is Open Mapping

Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $N$ be a closed linear subspace of $X$.
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
Let $\pi : \struct {X, \tau} \to \struct {X/N, \tau_N}$ be the quotient mapping.
Then $\pi$ is an open mapping.

Proof:
Let $V \in \tau$.
From Quotient Mapping is Linear Transformation, $\pi$ is a linear transformation.
From Preimage of Image of Linear Transformation, we have:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$
From Kernel of Quotient Mapping, we have $\ker \pi = N$ and so:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$
From Sum of Set and Open Set in Topological Vector Space is Open, $N + V \in \tau$.
So $\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$ is open in $\struct {X, \tau}$.
From the definition of the quotient topology, it follows that $\pi \sqbrk V$ is open in $\struct {X/N, \tau_N}$.
{{qed}}
Theorem: Open Neighborhood of Dilation of Point in Topological Vector Space contains Pointwise Scalar Multiplication of Open Neighborhood of Scalar with Open Neighborhood of Vector

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $\lambda \in K$ and $x \in X$. 
Let $U$ be an open neighborhood of $\lambda x$.
Then there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ in $X$ such that:
:$D V \subseteq U$
where:
:$D V = \set {\mu y : \mu \in D, \, y \in V}$

Proof:
Equip $K \times X$ with the product topology.
From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.
Define $m : K \times X \to X$ by: 
:$\map m {\mu, y} = \mu y$
for each $\tuple {\mu, y} \in K \times X$.
From the definition of a topological vector space, $m$ is continuous.
In particular, $m$ is continuous at $\tuple {\lambda, x}$.
Hence, there exists an open neighborhood $W$ of $\tuple {\lambda, x}$ such that: 
:$m \sqbrk W \subseteq U$
From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ of $x$ in $X$ such that:
:$D \times V \subseteq W$
Then we have, from Image of Subset under Mapping is Subset of Image:
:$m \sqbrk {D \times V} \subseteq m \sqbrk W \subseteq U$
Since:
:$m \sqbrk {D \times V} = D V$
we have:
:$D V \subseteq U$
{{qed}}
Category:Topological Vector Spaces
105e5kgeiiyiy8yzy5h7g3kj95dzezs
Theorem: Image of Pointwise Scalar Multiplication of Subset of Scalars with Subset of Vectors under Linear Transformation

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation. 
Let $S \subseteq K$ and $D \subseteq X$ be non-empty sets.
Then:
:$T \sqbrk {S D} = S T \sqbrk D$
where:
:$S D = \set {\lambda x : \lambda \in S, \, x \in D}$

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = T \sqbrk {S D}
		| r = T \sqbrk {\bigcup_{s \mathop \in S} s D}
}}
{{eqn	| r = \bigcup_{s \mathop \in S} T \sqbrk {s D}
		| c = Image of Union under Mapping
}}
{{eqn	| r = \bigcup_{s \mathop \in S} s T \sqbrk D
		| c = Image of Dilation of Set under Linear Transformation is Dilation of Image
}}
{{eqn	| r = S T \sqbrk D
}}
{{end-eqn}}
{{qed}}
Category:Vector Spaces
sk3mv4x44px9briilxgguho5q6vdgie
Theorem: Existence of Banach Limits

Let $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$ be the normed vector space of bounded sequences on $\R$. 
Let $\struct {\paren {\map {\ell^\infty} \R}^\ast, \norm \cdot_{\paren {\ell^\infty}^\ast} }$ be the normed dual space of $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$.
Then there exists a Banach limit $L \in \paren {\map {\ell^\infty} \R}^\ast$.

Proof:
Let $\map c \R$ be the set of the convergent real sequences.
Then:
:$\map c \R \subseteq \map {\ell^\infty} \R$
is a linear subspace in view of:
:Convergent Real Sequence is Bounded
:Linear Combination of Convergent Sequences in Topological Vector Space is Convergent 
Define a mapping $f_0 : \map c \R \to \R$ by:
:$\ds \map {f_0} {\sequence {x_n} } := \lim _{n \mathop \to \infty} x_n$
{{refactor|level = basic|be good to get the lemmata into their own pages. It's on my list.}}
Theorem: Properties of Product of Identity plus Operator Raised to Powers of 2

Let $X$ be a Banach space.
Let $\map \LL X$ be the set of all linear transformations.
Let $\map {CL} X$ be a continuous linear transformation sapce.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $A \in \map {CL} X$ be such that $\norm A < 1$.
Let $I$ be the identity mapping.
Let $\circ$ be the composition of mappings.
For all $n \in \N$ let $A^n := \underbrace {A \circ A \circ \ldots \circ A}_{n \text{ times}}$, with $A^0 := I$.
For all $n \in \N$ let $P_n = \underbrace{\paren {I + A} \circ \paren {I + A^2} \circ \ldots \circ \paren {I + A^{2^n} } }_{n + 1 \text{ terms} }$
Then:
:$\forall n \in \N : \paren {I - A} \circ P_n = I - A^{2^{n + 1} }$
:the sequence $\sequence {P_n}_{n \mathop \in \N}$ converges in $\map {CL} X$ to $\paren {I - A}^{-1}$

Proof:
=== $\paren {I - A} \circ P_n = I - A^{2^{n + 1} }$ ===
This will be a proof by induction.
Theorem: Locally Convex Space is Topological Vector Space/Corollary

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\GF$ equipped with the standard topology $\tau$. 
Then $\struct {X, \tau}$ is a Hausdorff topological vector space.

Proof:
From Locally Convex Space is Topological Vector Space, $\struct {X, \tau}$ is a topological vector space.
From Locally Convex Space is Hausdorff iff induces Hausdorff Topology, $\struct {X, \tau}$ is a Hausdorff space.
So $\struct {X, \tau}$ is a Hausdorff topological vector space.
{{qed}}
Category:Hausdorff Topological Vector Spaces
Category:Locally Convex Space is Topological Vector Space
dywgrg4gaf3vcxtoj1vyvupi3a0lt21
Theorem: Topological Vector Space is Hausdorff iff T1

Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
{{TFAE}}
:$(1): \quad \struct {X, \tau}$ is Hausdorff
:$(2): \quad \struct {X, \tau}$ is a $T_1$ space
:$(3): \quad$ for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $x \not \in U_x$.

Proof:
=== $(1)$ implies $(2)$ ===
This is precisely $T_2$ Space is $T_1$ Space.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Hausdorff Topological Vector Space

Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$. 
{{TFAE|def = Hausdorff Topological Vector Space}}
=== Definition 1 ===
{{:Definition:Hausdorff Topological Vector Space/Definition 1}}
=== Definition 2 ===
{{:Definition:Hausdorff Topological Vector Space/Definition 2}}

Proof:
From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is Hausdorff {{iff}} it is $T_1$ space.
From definition 3 of a $T_1$ space, it follows that $\struct {X, \tau}$ is Hausdorff {{iff}}:
:for each $x \in X$, $\set x$ is closed in $\struct {X, \tau}$.
{{qed}}
Category:Hausdorff Topological Vector Spaces
15md2hywfiyh0hw245wdpifgdhy4m47
Theorem: Characterization of Hausdorff Topological Vector Space

Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\struct {X, \tau}$ is a Hausdorff topological vector space}}
{{item|(2):|the intersection of all open neighborhoods of ${\mathbf 0}_X$ in $\struct {X, \tau}$ is $\set { {\mathbf 0}_X}$}}
{{item|(3):|$\set { {\mathbf 0}_X}$ is closed in $\struct {X, \tau}$. }}
{{end-itemize}}

Proof:
=== $(1)$ implies $(2)$ ===
Suppose that $\struct {X, \tau}$ is a Hausdorff topological vector space.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
From Topological Vector Space is Hausdorff iff $T_1$, $\struct {X, \tau}$ is $T_1$.
Clearly:
:$\set { {\mathbf 0}_X} \subseteq \bigcap \FF$
Now let $x \in X \setminus \set { {\mathbf 0}_X}$. 
Then there exists an open neighborhood $U \in \FF$ of ${\mathbf 0}_X$ such that $x \not \in U$. 
In particular, $x \not \in \bigcap \FF$. 
So we obtain $\bigcap \FF = \set { {\mathbf 0}_X}$.
{{qed|lemma}}
Theorem: Dilation of Complement of Set in Vector Space

Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $\lambda \in K$.
Then:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$

Proof:
It is immediate from the definition of a dilation that if $x \in F$ we have $\lambda x \in \lambda F$. 
Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.
That is, $\lambda \paren {x - y} = 0$. 
Since $\lambda \ne 0$, it follows that $x = y$, and so $x \in F$.
So we have $x \in F$ {{iff}} $\lambda x \in \lambda F$.
So for $x \in X$ we have $x \not \in F$ {{iff}} $\lambda x \not \in \lambda F$. 
That is, $x \in X \setminus F$ {{iff}} $\lambda x \in X \setminus \paren {\lambda F}$.
It follows that $\lambda x \in \lambda \paren {X \setminus F}$ {{iff}} $\lambda x \in X \setminus \paren {\lambda F}$.
We can therefore conclude:
:$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$
{{qed}}
Category:Dilations of Subsets of Vector Spaces
sm2zewho61r1r0dzvy5xoqgdsz7fplb
Theorem: Translation of Complement of Set in Vector Space

Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $F \subseteq X$ be a non-empty set.
Let $x \in X$.
Then:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$

Proof:
It is immediate that if:
:$y \in F$
we have:
:$y + x \in F + x$.
Conversely, if:
:$y + x \in F + x$
then:
:$y + x = u + x$
for some $u \in F$.
That is:
:$y = u$
and so:
:$y \in F$.
Hence we have:
:$y \in F$ {{iff}} $y + x \in F + x$
Hence for $y \in X$ we have:
:$y + x \not \in F + x$ {{iff}} $y \not \in F$
That is:
:$y + x \in X \setminus \paren {F + x}$ {{iff}} $y \in X \setminus F$ {{iff}} $y + x \in \paren {X \setminus F} + x$.
So:
:$y + x \in X \setminus \paren {F + x}$ {{iff}} $y + x \in \paren {X \setminus F} + x$.
We can therefore conclude:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$
{{qed}}
Category:Translation of Subsets of Vector Spaces
thnbc87d7zjoygkui5igutzgkkkubkg
Theorem: Translation 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 $x \in X$.
Then $F + x$ is a closed set in $X$.

Proof:
We aim to show that $X \setminus \paren {F + x}$ is open.
We are given $F$ is closed in $X$.
So, $X \setminus F$ is open.
It follows from Translation of Open Set in Topological Vector Space is Open that $\paren {X \setminus F} + x$ is open.
From Translation of Complement of Set in Vector Space, we have:
:$X \setminus \paren {F + x} = \paren {X \setminus F} + x$
Since we have established that $\paren {X \setminus F} + x$ is open, we have that $X \setminus \paren {F + x}$ is open.
Hence by definition, $F + x$ is closed in $X$.
{{qed}}
Theorem: Translation 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 $x \in X$.
Then $F + x$ is a closed set in $X$.

Proof:
Define a mapping $T_{-x} : X \to X$ by:
:$\map {c_\lambda} y = y + x$
for each $y \in X$.
From Translation Mapping on Topological Vector Space is Homeomorphism, $T_{-x}$ is a homeomorphism.
From Definition 4 of a homeomorphism, $T_{-x}$ is therefore a closed mapping.
Hence $T_{-x} \sqbrk F = F + x$ is closed.
{{qed}}
Theorem: Image of Linear Combination of Subsets of Vector Space under Linear Transformation

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $\lambda, \mu \in K$.
Let $A, B \subseteq X$.
Then:
:$T \sqbrk {\lambda A + \mu B} = \lambda T \sqbrk A + \mu T \sqbrk B$

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = T \sqbrk {\bigcup_{x \in A} \paren {\lambda x + \mu B} }
		| r = \bigcup_{x \in A} T \sqbrk {\lambda x + \mu B}
		| c = Image of Union under Mapping
}}
{{eqn	| r = \bigcup_{x \in A} \paren {\lambda T x + T \sqbrk {\mu B} }
		| c = Image of Translation of Set under Linear Transformation is Translation of Image
}}
{{eqn	| r = \bigcup_{x \in A} \paren {\lambda T x} + T \sqbrk {\mu B}
		| c = Translation of Union of Subsets of Vector Space
}}
{{eqn	| r = \lambda \bigcup_{x \in A} \paren {T x} + T \sqbrk {\mu B}
		| c = Image of Dilation of Set under Linear Transformation is Dilation of Image
}}
{{eqn	| r = \lambda T \sqbrk {\bigcup_{x \in A} x} + T \sqbrk {\mu B}
		| c = Image of Union under Mapping
}}
{{eqn	| r = \lambda T \sqbrk A + T \sqbrk {\mu B}
}}
{{eqn	| r = \lambda T \sqbrk A + \mu T \sqbrk B
		| c = Image of Dilation of Set under Linear Transformation is Dilation of Image
}}
{{end-eqn}}
{{qed}}
Category:Linear Combination of Subsets of Vector Spaces
06xtokwphh6yuyrl2ok0cq230go6b10
Theorem: Quotient Topological Vector Space is Hausdorff iff Linear Subspace is Closed

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $N$ be a linear subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\tau_N$ be the quotient topology on $X/N$.
Then $\struct {X/N, \tau_N}$ is Hausdorff {{iff}}:
:$N$ is a closed linear subspace.

Proof:
Let $\pi : X \to X/N$ be the quotient mapping.
=== Necessary Condition ===
Suppose that $\struct {X/N, \tau_N}$ is Hausdorff.
From Characterization of Hausdorff Topological Vector Space, $\set { {\mathbf 0}_{X/N} }$ is closed in $\struct {X/N, \tau_N}$.
From the definition of the quotient topology, $\pi$ is continuous.
Hence $\pi^{-1} \sqbrk {\set { {\mathbf 0}_{X/N} } }$ is closed in $\struct {X, \tau}$.
From Kernel of Quotient Mapping, we have $\pi^{-1} \sqbrk {\set { {\mathbf 0}_{X/N} } } = N$.
Hence $N$ is closed in $\struct {X, \tau}$.
{{qed|lemma}}
=== Sufficient Condition ===
Suppose that $N$ is closed in $\struct {X, \tau}$.
Then $X \setminus N$ is open in $\struct {X, \tau}$.
From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, $\pi$ is an open mapping.
Hence $\pi \sqbrk {X \setminus N}$ is open in $\struct {X/N, \tau_N}$.
Since $\pi$ is surjective by construction and $\pi^{-1} \sqbrk {\set { {\mathbf 0}_{X/N} } } = N$, we have:
:$\pi \sqbrk {X \setminus N} = \paren {X/N} \setminus \set { {\mathbf 0}_{X/N} }$
So $\paren {X/N} \setminus \set { {\mathbf 0}_{X/N} }$ is open in $\struct {X/N, \tau_N}$.
So $\set { {\mathbf 0}_{X/N} }$ is closed in $\struct {X/N, \tau_N}$.
From Characterization of Hausdorff Topological Vector Space, we have that $\struct {X/N, \tau_N}$ is Hausdorff.
{{qed}}
Theorem: Preimage of Image of Linear Transformation

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $T : X \to Y$ be a linear transformation.
Let $A \subseteq X$ be a non-empty set.
Then $T^{-1} \sqbrk {T \sqbrk A} = \ker T + A$

Proof:
Let $x \in T^{-1} \sqbrk {T \sqbrk A}$.
Then $T x \in T \sqbrk A$.
Then there exists $y \in A$ such that $T x = T y$. 
Hence we have $\map T {x - y} = 0$.
Hence $x - y \in \ker T$.
So $x \in y + \ker T$.
We have $y + \ker T \subseteq A + \ker T$, so we obtain:
:if $x \in \in T^{-1} \sqbrk {T \sqbrk A}$ then $x \in \ker T + A$
so that:
:$T^{-1} \sqbrk {T \sqbrk A} \subseteq \ker T + A$
Now let $x \in \ker T + A$.
Then there exists $v \in \ker T$ and $y \in A$ such that $x = v + y$.
From linearity, we have $T x = \map T {v + y} = T v + T y$.
Since $v \in \ker T$, we have $T x = T y$.
Since $T y \in T \sqbrk A$, we have $T x \in T \sqbrk A$.
So $x \in T^{-1} \sqbrk {T \sqbrk A}$.
{{qed}}
Category:Linear Transformations
kbe2rrkj7qu3b6l28rzkjtkrrqg3mv0
Theorem: Sum of Closed Linear Subspace and Finite-Dimensional Subspace of Hausdorff Topological Vector Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a Hausdorff topological vector space over $\GF$.
Let $N$ be a closed linear subspace of $X$.
Let $F$ be a finite dimensional linear subspace of $X$.
Then $N + F$ is closed in $\struct {X, \tau}$.

Proof:
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
From Characterization of Hausdorff Topological Vector Space, $X/N$ is Hausdorff.
Let $\pi : X \to X/N$ be the quotient mapping.
From Image of Linear Transformation is Submodule and Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space, we have that $\pi \sqbrk F$ is a finite dimensional linear subspace of $\struct {X/N, \tau_N}$.
From Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Closed, $\pi \sqbrk F$ is closed in $\struct {X/N, \tau_N}$.
From the definition of the quotient topology, $\pi$ is continuous.
So $\pi^{-1} \sqbrk F$ is closed in $\struct {X/N, \tau_N}$.
From Preimage of Image of Linear Transformation, we have:
:$\pi^{-1} \sqbrk {\pi \sqbrk F} = \ker \pi + F$
From Kernel of Quotient Mapping, we have $\ker \pi = N$ and so:
:$\pi^{-1} \sqbrk {\pi \sqbrk F} = N + F$
Hence $N + F$ is closed in $\struct {X/N, \tau_N}$.
{{qed}}
Theorem: Consistency Principle for Binary Mess

Let $S$ be a set.
Let $M$ be a binary mess on $S$.
Then there exists a mapping $f : S \to \Bbb B$ such that:
:$f$ is consistent with $M$

Proof:
{{explain|There's a lot going on here. I'm sure that at least some of it is more general than what is needed to prove the result, so should be pulled out and turned into separate results.}}
{{Proofread}}
Let $I$ be the set of all finite subsets of $S$.
Recall that by definition of binary mess:
:$\ds M \subseteq \bigcup_{P \mathop \in I} \Bbb B^P$
is a set of mappings from finite subsets of $S$ to a Boolean domain.
For every $P \in I$, define:
:$M_P = \set {t \in M : \Dom t = P}$
By Binary Mess Axiom $(\text M 2)$ we have that $M_P$ cannot be empty.
For every $J \subseteq I$, define:
:$Z_J = \set {z \in M^J : \paren {\forall P \in \Dom z : \map z P \in M_P} \land \paren {\forall P, Q \in \Dom z : t_1 {\restriction_R} = t_2 {\restriction_R} } }$
where:
:$t_1 = \map z P$
:$t_2 = \map z Q$
:$R = \Dom {t_1} \cap \Dom {t_2}$
:$M^J$ denotes the set of all mappings from $J$ to $M$.
Define:
:$\ds Z = \bigcup_{J \mathop \subseteq I} Z_J$
:$X_P = \set {z \in Z : P \in \Dom z}$
:$F = \set {X_P : P \in I}$
Theorem: Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces

Let $K$ be a topological field.
Let $I$ be a set. 
Let $\family {X_i}_{i \in I}$ be an $I$-indexed family of topological vector spaces over $K$.
Let:
:$\ds X = \prod_{i \mathop \in I} X_i$
be the direct product of $\family {X_i}_{i \in I}$.
Equip $X$ with the product topology.
Then $X$ is Hausdorff {{iff}} $X_i$ is Hausdorff for each $i \in I$.

Proof:
For each $i \in I$, let $\FF_i$ be the set of open neighborhoods of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the zero vector of $X_i$.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Theorem: Radius of Curvature in Whewell Form

The '''radius of curvature''' $\kappa$ of $C$ at a point $P$ can be expressed in the form of a Whewell equation as:
:$\rho = \size {\dfrac {\d s} {\d \psi} }$
where:
:$s$ is the arc length of $C$
:$\psi$ is the turning angle of $C$
:$\size {\, \cdot \,}$ denotes the absolute value function.

Proof:
By definition, the radius of curvature $\rho$ is given by:
:$\rho = \dfrac 1 {\size \kappa}$
where $\kappa$ is the curvature, given in Whewell form as:
:$\kappa = \dfrac {\d \psi} {\d s}$
Hence the result.
{{qed}}
Theorem: Taylor's Theorem/Two Variables

Let $f: \R^2 \to \R$ be a real-valued function which is appropriately differentiable in a neighborhood of the point $\tuple {a, b}$.
Then for $\sqrt {h^2 + k^2} < R$ for some $R \in \R$:
{{begin-eqn}}
{{eqn | l = \map f {a + h, b + k}
      | r = \frac 1 {0!} \map f {a, b}
      | c =
}}
{{eqn | o =
      | ro= +
      | r = \frac 1 {1!} \valueat {\paren {h \dfrac {\partial f} {\partial x} + k \dfrac {\partial f} {\partial y} } } {a, b}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \frac 1 {2!} \valueat {\paren {h^2 \dfrac {\partial^2 f} {\partial x^2} + 2 h k \dfrac {\partial^2 f} {\partial x \partial y} + k^2 \dfrac {\partial^2 f} {\partial y^2} } } {a, b}
      | c = 
}}
{{eqn | o =
      | ro= +
      | r = \cdots
      | c = 
}}
{{end-eqn}}

Proof:
{{ProofWanted}}
{{Namedfor|Brook Taylor}}