avewright/proofwiki-math huggingface dataset:
Theorem: Sequence of Natural Powers of Right Shift Operator in 2-Sequence Space does not Converge in Strong Operator Topology

Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence normed vector space.
Let $\map {CL} {\ell^2} := \map {CL} {\ell^2, \ell^2}$ be the continuous linear transformation space.
Let $R \in \map {CL} {\ell^2}$ be the right shift operator over $\ell^2$.
Let $\sequence {R^n}_{n \mathop \in \N}$ be a sequence.
Let $\mathbf 0 \in \map {CL} {\ell^2}$ be the zero mapping.
Then $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.

Proof:
Let $\mathbf e_1 = \tuple {1, 0, \ldots} \in \ell^2$
Then $R^n \mathbf e_1 = \tuple {\underbrace {\ldots, 0}_{n \text{ terms} }, 1, 0, \ldots}$
So:
:$\forall n \in \N : \norm {R^n \mathbf e_1}_2 = 1$
Therefore:
:$\ds \lim_{n \mathop \to \infty} \norm {R^n \mathbf e_1}_2 = 1$
Hence, $\sequence {R^n}_{n \mathop \in \N}$ does not converge to $\mathbf 0$ in the strong operator topology.
{{qed}}
Theorem: Pointwise Minimum of Metric and Positive Real Number is Topologically Equivalent Metric

Let $\struct {X, d}$ be a metric space.
Let $c > 0$ be a real number. 
For each $x, y \in X$, define: 
:$\map {d'} {x, y} = \min \set {\map d {x, y}, c}$ 
Then $d'$ is a metric that is topologically equivalent to $d$.

Proof:
=== {{Metric-space-axiom|1|nolink}} ===
Suppose that $x, y \in X$ are such that: 
:$\map {d'} {x, y} = 0$
Since $c > 0$, this implies that: 
:$\map d {x, y} = 0$
Since $d$ is a metric, we have $x = y$ by {{Metric-space-axiom|1}} for $d$. 
Hence {{Metric-space-axiom|1}} is fulfilled.
{{qed|lemma}}
Theorem: Composition of Continuous Linear Transformations is Continuous Linear Transformation

Let $K$ be a field.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct{Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces over $K$.
$\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $S \circ T : X \to Z$ be the composition of mappings such that:
:$\forall T \in \map {CL} {X, Y} : \forall S \in \map {CL} {Y, Z} : \forall x \in X : \map {S \circ T} x := \map S {\map T x}$
Then $S \circ T \in \map {CL} {X, Z}$.

Proof:
=== Linearity ===
Follows from Composition of Linear Transformations is Linear Transformation.
Theorem: Characterization of Fineness of Topology in terms of Topological Bases

Let $X$ be a set.
Let $\tau$ and $\tau'$ be topologies on $X$. 
Let $\BB \subseteq \tau$ be a basis for $\tau$ and $\BB' \subseteq \tau'$ be a basis for $\tau'$.
{{TFAE}}
:$(1): \quad$ $\tau'$ is finer than $\tau$
:$(2): \quad$ for each $x \in X$ and each $B \in \BB$ containing $x$, there exists $B' \in \BB'$ such that $x \in B' \subseteq B$.

Proof:
=== $(1)$ implies $(2)$ ===
Let $x \in X$ and $B \in \BB$ contain $x$.
Since $\tau'$ is finer than $\tau$, we have that $B \in \tau'$. 
Since $\BB'$ is a basis for $\tau'$, there exists a subset $\set {B'_\alpha : \alpha \in I} \subseteq \BB'$ such that: 
:$\ds B = \bigcup_{\alpha \mathop \in I} B'_\alpha$
from Open Set is Union of Elements of Basis.
Since $x \in B$, we have $x \in B'_\alpha$ for some $\alpha \in I$.
We have $B'_\alpha \subseteq B$ from Set is Subset of Union. 
Setting $B = B'_\alpha$, we have therefore found $B' \in \BB'$ such that $x \in B' \subseteq B$, proving $(2)$. 
{{qed|lemma}}
Theorem: Modulus of Linear Functional on Vector Space is Seminorm

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $f : X \to \GF$ be a linear functional.
Define $p_f : X \to \R_{\ge 0}$ by: 
:$\map {p_f} x = \cmod {\map f x}$ 
for each $x \in X$.
Then $p_f$ is a seminorm.

Proof:
=== Proof of {{SeminormAxiom|2}} ===
For each $\lambda \in \GF$ and $x \in X$, we have: 
{{begin-eqn}}
{{eqn	| l = \map {p_f} {\lambda x}
		| r = \cmod {\map f {\lambda x} }
}}
{{eqn	| r = \cmod {\lambda \map f x}
		| c = since $f$ is linear
}}
{{eqn	| r = \cmod \lambda \cmod {\map f x}
}}
{{eqn	| r = \cmod \lambda \map {p_f} x
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex

Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $F$ be a set of linear functionals on $X$ that form a vector space over $\GF$.
That is, for each $\lambda, \mu \in \GF$ and $f, g \in F$, we have: 
:$\lambda f + \mu g \in F$
Let $\tau$ be the initial topology on $X$ generated by $F$.
For each $f \in F$, define $p_f : X \to \R_{\ge 0}$ by: 
:$\map {p_f} x = \cmod {\map f x}$ 
and let: 
:$\PP = \set {p_f : f \in F}$
Then the standard topology on the locally convex space $\struct {X, \PP}$ is precisely $\tau$.

Proof:
Let $\tau'$ be the standard topology on the locally convex space.
From definition, a sub-basis for $\tau'$ is given by: 
:$\SS' = \set {\map {B_{p_f} } {\epsilon, x} : f \in F, \, \epsilon > 0, \, x \in X}$
where: 
:$\map {B_{p_f} } {\epsilon, x} = \set {y \in X : \map {p_f} {y - x} < \epsilon}$
For each $u \in \GF$, let: 
:$\map {B_\epsilon} {u, \GF} = \set {v \in \GF : \cmod {u - v} < \epsilon}$
From Open Balls form Basis for Open Sets of Metric Space:
:$\BB_\GF = \set {\map {B_\epsilon} {u, \GF} : u \in \GF, \, \epsilon > 0}$ 
forms a basis for $\GF$.
From Sub-Basis for Initial Topology in terms of Sub-Bases of Target Spaces:
:$\SS = \set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0}$
is a sub-basis for $\tau$.
We show that $\SS = \SS' \cup \set \O$. 
First, take $f = 0$. 
Then: 
:$\ds f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } = \begin{cases}X & 0 \in \map {B_\epsilon} {u, \GF} \\ \O & \text{otherwise}\end{cases} \in \SS' \cup \set \O$ 
On the other hand, if $f = 0$ and $x \in X$ then: 
{{begin-eqn}}
{{eqn	| l = \map {B_{p_f} } {\epsilon, x}
		| r = \set {y \in X : \map {p_f} {y - x} < \epsilon}
}}
{{eqn	| r = \set {y \in X : \cmod {\map f {y - x} } < \epsilon}
}}
{{eqn	| r = \set {y \in X : 0 < \epsilon} 
}}
{{eqn	| r = X
}}
{{eqn	| o = \in
		| r = \SS
}}
{{end-eqn}}
Now suppose that $f \ne 0$. 
Then from Linear Functional on Vector Space is Zero or Surjective, $f$ is surjective.
So, for each $u \in \GF$ there exists $x \in X$ such that $u = \map f x$.
Then we have: 
{{begin-eqn}}
{{eqn	| l = f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} }
		| r = f^{-1} \sqbrk {\map {B_\epsilon} {\map f x, \GF} } 
}}
{{eqn	| r = \set {y \in X : \map f y \in \map {B_\epsilon} {\map f x, \GF} }
}}
{{eqn	| r = \set {y \in X : \cmod {\map f y - \map f x} < \epsilon}
}}
{{eqn	| r = \set {y \in X : \cmod {\map f {y - x} } < \epsilon}
}}
{{eqn	| r = \set {y \in X : \map {p_f} {y - x} < \epsilon}
}}
{{eqn	| r = \map {B_{p_f} } {\epsilon, x}
}}
{{end-eqn}}
So we have: 
:$\set {f^{-1} \sqbrk {\map {B_\epsilon} {u, \GF} } : u \in \GF, \, \epsilon > 0} = \set {\map {B_{p_f} } {\epsilon, x} : f \in F \setminus \set 0, \, \epsilon > 0}$ 
Combining with our result for $f = 0$, we obtain the desired: 
:$\SS = \SS' \cup \set \O$
Since $\O$ is contained in any topology on $X$ we have: 
:$\map \tau \SS = \map \tau {\SS'}$
and hence:
:$\tau = \tau'$
{{qed}}
Category:Locally Convex Spaces
Category:Linear Functionals
Category:Initial Topology
ts0esur8k7v0rg26vsvah3fpghkgb3k
Theorem: Linearly Independent Set is Contained in some Basis/Infinite Dimensional Case

Let $K$ be a field. 
Let $E$ be a vector space over $K$. 
Let $H$ be a linearly independent subset of $E$.
There exists a basis $B$ for $E$ such that $H \subseteq B$.

Proof:
Let: 
:$\SS = \set {L \supseteq H : L \subseteq E \text { is linearly independent} }$
We have $H \in \SS$, so certainly $\SS \ne \O$.
With view to apply Zorn's Lemma, we show that every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
Let $\CC$ be a non-empty $\subseteq$-chain in $\SS$.
Let: 
:$\ds C = \bigcup \CC$
Let $x_1, x_2, \ldots, x_n \in C$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ such that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k x_k = 0$ 
For each $1 \le i \le n$, let $C_i \in \CC$ be such that $x_i \in C_i$. 
Since $\CC$ is a $\subseteq$-chain, there exists $1 \le j \le n$ such that: 
:$C_i \subseteq C_j$ for all $1 \le i \le n$. 
Then $x_i \in C_j$ for each $1 \le i \le n$. 
Since $C_j$ is linearly independent, we have: 
:$\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$
So $C$ is linearly independent, with $H \subseteq \CC$. 
So $C \in \SS$, while $F \subseteq C$ for each $F \in \CC$.
So $C$ is an upper bound for $\CC$.
So every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
So by Zorn's Lemma, $\SS$ has a $\subseteq$-maximal element $\BB$. 
We show that $\BB$ is a basis for $E$.
From definition 2 of a basis, it suffices to argue that $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$. 
Suppose that $\BB'$ is a linearly independent subset of $G$ such that $\BB \subseteq \BB'$. 
Then since $H \subseteq \BB$, we have $H \subseteq \BB'$.
So $\BB' \in \SS$. 
So by the maximality of $\BB$ in $\SS$, we have $\BB = \BB'$.
So $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$, and hence is a basis.
{{qed}}
{{AoC|Zorn's Lemma}}
Category:Linearly Independent Set is Contained in some Basis
pwuauqicnvt3yk91vfgv3hlnwck5eeg
Theorem: Linearly Independent Set is Contained in some Basis/Finite Dimensional Case

Let $E$ be a vector space of $n$ dimensions.
Let $H$ be a linearly independent subset of $E$.
There exists a basis $B$ for $E$ such that $H \subseteq B$.

Proof:
By hypothesis there is a basis $B$ of $E$ with $n$ elements.
Then $H \cup B$ is a generator for $E$.
So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $E$ such that $H \subseteq C \subseteq H \cup B$.
{{Qed}}
Theorem: Linearly Independent Set is Contained in some Basis/Finite Dimensional Case

Let $E$ be a vector space of $n$ dimensions.
Let $H$ be a linearly independent subset of $E$.
There exists a basis $B$ for $E$ such that $H \subseteq B$.

Proof:
Let $H = \set {\xi_1, \xi_2, \ldots, \xi_r}$.
Consider the basis $B = \set {\alpha_1, \alpha_2, \ldots, \alpha_n}$ of $E$.
Consider the set $G = H \cup B = \set {\xi_1, \xi_2, \ldots, \xi_r, \alpha_1, \alpha_2, \ldots, \alpha_n}$.
We have that $G$ is a generator of $E$.
As $B$ is a basis, it follows that each of $H$ is a linear combination of $B$.
Thus $G = H \cup B$ is linearly dependent.
Thus one of the elements $\alpha_i$ of $B$ is a linear combination of the preceding elements of $G$.
Eliminate this one, and do the same thing with with $G \setminus \set {\alpha_i}$. 
Eventually there will exist a set which is a basis of $E$ containing all the elements of $H$.
{{Qed}}
Theorem: Generator of Vector Space Contains Basis/Infinite Dimensional Case

Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $G$ be a generator of $X$.
Then: 
:$G$ contains a basis for $X$.

Proof:
If $X = \set { {\mathbf 0}_X}$, then we must have $G \subseteq \set { {\mathbf 0}_X}$, while the only basis for $X$ is $\O$.
So in this case, we have the claim immediately. 
Now take $X \ne \set { {\mathbf 0}_X}$. 
Let: 
:$\SS = \set {L \subseteq G : L \text { is linearly independent} }$
Since $G$ generates $X$, it contains some $x \ne {\mathbf 0}_X$.
Then we have $\set x \in \SS$, so $\SS \ne \O$.
With view to apply Zorn's Lemma, we show that every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
Let $\CC$ be a non-empty $\subseteq$-chain in $\SS$.
Let: 
:$\ds C = \bigcup \CC$
Let $x_1, x_2, \ldots, x_n \in C$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ such that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k x_k = 0$ 
For each $1 \le i \le n$, let $C_i \in \CC$ be such that $x_i \in C_i$. 
Since $\CC$ is a $\subseteq$-chain, there exists $1 \le j \le n$ such that: 
:$C_i \subseteq C_j$ for all $1 \le i \le n$. 
Then $x_i \in C_j$ for each $1 \le i \le n$. 
Since $C_j$ is linearly independent, we have: 
:$\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$
So $C$ is linearly independent with $C \subseteq G$. 
So $C \in \SS$, while $F \subseteq C$ for all $F \in \CC$. 
So $C$ is an upper bound for $\CC$.
So every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
So by Zorn's Lemma, $\SS$ has a $\subseteq$-maximal element $\BB$. 
We show that $\BB$ is a basis for $E$.
Let: 
:$V = \map \span \BB$ 
Suppose that $V \ne X$.
We then have that $V$ cannot contain $G$, otherwise we would have $V = X$, since $G$ generates $X$. 
So there exists $y \in G$ such that $y \not \in V$.
From Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set, $\BB \cup \set y$ is linearly independent, contradicting the maximality of $\BB$.
So there exists no such $y$ and we have $V = X$.
Since $\BB$ is a linearly independent generator for $X$, it is a basis.
Since $\BB \in \SS$, we have $\BB \subseteq G$ as desired.
{{qed}}
Category:Generator of Vector Space Contains Basis
b4cz82wv6ex941xfjho9ss4p3wv2k6b
Theorem: Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set

Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $L$ be a linearly independent set of $X$ such that: 
:$U = \map \span L \ne X$
Let: 
:$x \in X \setminus U$
Then $L \cup \set x$ is linearly independent.

Proof:
Take $x_1, \ldots, x_n \in L$ and take $\alpha_1, \ldots, \alpha_n, \alpha_{n + 1} \in K$ such that: 
:$\ds \alpha_{n + 1} x + \sum_{k \mathop = 1}^n \alpha_i x_i = 0$
If $\alpha_{n + 1} = 0$, then we have: 
:$\ds \sum_{k \mathop = 1}^n \alpha_i x_i = 0$
and so $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$ from the linear independence of $L$.
{{AimForCont}} $\alpha_{n + 1} \ne 0$, then we would have: 
:$\ds x = -\alpha^{-1}_{n + 1} \sum_{k \mathop = 1}^n \alpha_i x_i$
This would mean that $x \in U$, (from Linear Span is Linear Subspace) contradicting that $x \in X \setminus U$.
So we have $\alpha_{n + 1} = 0$, and hence $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$.
So $L \cup \set x$ is linearly independent.
{{qed}}
Category:Linear Independence
eysvkuc4vv4g0ryowkik1t1ci8zgk7i
Theorem: Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space

Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $T : X \to Y$ be a linear transformation.
Then: 
:$\dim T \sqbrk X \le \dim X$

Proof:
From Vector Space has Basis, there exists a basis $\BB$ for $X$.
By Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image, $T \sqbrk \BB$ is a generator for $T \sqbrk X$.
From Generator of Vector Space Contains Basis, there exists a basis $\BB'$ for $T \sqbrk X$ such that $\BB' \subseteq T \sqbrk \BB$.
From Cardinality of Image of Mapping not greater than Cardinality of Domain, we have $\card {T \sqbrk \BB} \le \card \BB$. 
Also, since $\BB' \subseteq T \sqbrk \BB$, we have: 
:$\card {\BB'} \le \card {T \sqbrk \BB}$
Since $\BB'$ is a basis for $T \sqbrk X$, we have: 
:$\card {\BB'} = \dim T \sqbrk X$
and since $\BB$ is a basis for $X$ we have: 
:$\card \BB = \dim X$
So we obtain: 
{{begin-eqn}}
{{eqn	| l = \dim T \sqbrk X
		| r = \card {\BB'}
}}
{{eqn	| o = \le
		| r = \card {T \sqbrk \BB}
}}
{{eqn	| o = \le
		| r = \card \BB
}}
{{eqn	| r = \dim X
}}
{{end-eqn}}
{{qed}}
Category:Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space
Category:Linear Transformations
44khskkh5x47h70yc2epdzxqb40wvs1
Theorem: Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image

Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $G$ be a generating set for $X$. 
Then $T \sqbrk G$ be a generating set for $T \sqbrk X$.

Proof:
Let $y \in T \sqbrk X$.
Then there exists $x \in X$ such that $y = T x$.
Since $G$ generates $X$, there exists $n \in \N$, $x_1, \ldots, x_n \in G$ and $\alpha_1, \ldots, \alpha_n \in K$ such that: 
:$\ds x = \sum_{i \mathop = 1}^n \alpha_i x_i$ 
Then from the linearity of $T$ we have: 
:$\ds y = T x = \sum_{i \mathop = 1}^n \alpha_i T x_i$ 
with $T x_i \in T \sqbrk G$ for each $1 \le i \le n$. 
So:
:$T \sqbrk X \subseteq \map \span {T \sqbrk G}$
and so:
:$T \sqbrk X = \map \span {T \sqbrk G}$
So $T \sqbrk G$ is a generating set for $T \sqbrk X$. 
{{qed}}
Category:Linear Transformations
k4iyxqkcazebisyz0drowxup9otw2kr
Theorem: Linear Functional on Vector Space is Zero or Surjective

Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $f : X \to K$ be a linear functional.
Then either: 
:$\map f x = 0$ for each $x \in X$
or:
:$f$ is surjective.

Proof:
Suppose that $\map f {x_0} \ne 0$ for $x_0 \in X$. 
Take $c \in K$. 
Then we have, from linearity:
{{begin-eqn}}
{{eqn	| l = \map f {c \paren {\map f {x_0} }^{-1} x_0}
		| r = c \paren {\map f {x_0} }^{-1} \map f {x_0}
}}
{{eqn	| r = c
}}
{{end-eqn}}
Since $c \in K$ was arbitrary, we have that $f$ is surjective.
{{qed}}
Category:Linear Functionals
ahx4ew8ufhs9jtsz1kvasvc40sp72im
Theorem: Continuity of Linear Functionals in Weak Topology Induced by Pair of Vector Spaces with Bilinear Mapping

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $X'$ be vector spaces over $\GF$.
Let $\innerprod \cdot \cdot : X \times X' \to \GF$ be a bilinear mapping.
For each $x' \in X'$, define $f_{x'} : X \to \GF$ by:
:$\map {f_{x'} } x = \innerprod x {x'}$ for each $x \in X$.
Let:
:$F = \set {f_{x'} : x' \in X'}$
Let $\map \sigma {X, X'}$ be the initial topology on $X$ induced by $F$.
Let $g : X \to \GF$ be a linear functional.
Then $g$ is $\map \sigma {X, X'}$-continuous {{iff}} $g \in F$.

Proof:
=== Sufficient Condition ===
From the definition of the weak topology, $\map \sigma {X, X'}$ is the coarsest topology making all $g \in F$ continuous.
So if $g \in F$ then $g$ is $\map \sigma {X, X'}$-continuous.
{{qed|lemma}}
Theorem: Frink's Metrization Theorem

Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is metrizable
{{iff}}:
:for all $s \in S$ there exists a countable neighborhood basis, denoted $\set {U_{s, n} : n \in \N}$, such that:
::$(1): \quad \forall s \in S, n \in \N : U_{s, n + 1} \subseteq U_{s, n}$
::$(2): \quad \forall s \in S, n \in \N : \exists m > n : \forall t \in S : U_{t, m} \cap U_{s ,m} \ne \O \implies U_{t, m} \subseteq U_{s, n}$

Proof:
{{proof wanted}}
{{Namedfor|Aline Huke Frink|cat = Frink}}
Theorem: Linear Functional on Complex Vector Space is Uniquely Determined by Real Part

Let $X$ be a vector space over $\C$.
Let $f : X \to \C$ be a linear functional.
Define a function $g : X \to \R$: 
:$\map g x = \map \Re {\map f x}$
for each $x \in X$. 
Then: 
:$\map f x = \map g x - i \map g {i x}$
for each $x \in X$.

Proof:
For brevity, define a function $h : X \to \R$ by: 
:$\map h x = \map \Im {\map f x}$
for each $x \in X$. 
Note that: 
:$\map f x = \map \Re {\map f x} + i \map \Im {\map f x} = \map g x + i \map h x$
so that:
:$\map f {i x} = \map g {i x} + i \map h {i x}$
for each $x \in X$.
On the other hand, by the linearity of $f$, we have: 
{{begin-eqn}}
{{eqn	| l = \map f {i x}
		| r = i \map f x
}}
{{eqn	| r = i \paren {\map g x + i \map h x}
}}
{{eqn	| r = i \map g x - \map h x
}}
{{end-eqn}}
for each $x \in X$.
Comparing real parts, we have: 
:$-\map h x = \map g {i x}$ 
so $\map h x = -\map g {i x}$. 
So we have: 
:$\map f x = \map g x - i \map g {i x}$ 
for each $x \in X$.
{{qed}}
Category:Linear Functionals
9rkg49go4vtzu1k11rfgoafsrksgp1r
Theorem: Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube/Lemma 1

:$\AA$ is countable

Proof:
We have:
:$\AA \subseteq \BB \times \BB$
where $\BB \times \BB$ is the Cartesian product of $\BB$ with itself.
From Cartesian Product of Countable Sets is Countable:
:$\BB \times \BB$ is countable
From Subset of Countable Set is Countable:
:$\AA$ is countable
{{qed}}
Category:Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube
6qrbwkux8tvhwxjhx33w6pcv0w7tx1i
Theorem: Regular Lindelöf Space is Normal

Let $T = \struct {S, \tau}$ be a regular Lindelöf topological space.
Then $T$ is a normal space.

Proof:
{{Recall|Normal Space|normal space|index = 1}}
{{Definition:Normal Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a regular Lindelöf space.
{{Recall|Regular Space|regular space|index = 2}}
{{Definition:Regular Space/Definition 2}}
From $T_3$ Lindelöf Space is $T_4$:
:$T$ is a $T_4$ space
By definition, $T$ is a normal space.
{{qed}}
Category:Regular Spaces
Category:Lindelöf Spaces
Category:Normal Spaces
6oblydcpkhhzw5mqjgtsjzoyruko21d
Theorem: Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube

Let $T = \struct {S, \tau}$ be a topological space which is regular and second-countable.
Then $T$ is homeomorphic to a subspace of the Hilbert cube.

Proof:
From Second-Countable Space is Lindelöf: 
:$T$ is a Lindelöf space
From Regular Lindelöf Space is Normal:
:$T$ is a normal space
By definition of second-countable:
:there exists a countable basis $\BB$ for $\tau$
Let:
:$\AA = \set {\tuple {U, V} : U, V \in \BB : U^- \subseteq V}$
where $U^-$ denotes the closure of $U$ in $T$.
Theorem: Equivalence Classes of Diagonal Relation

Let $S$ be a set.
Let $\Delta_S$ denote the diagonal relation on $S$.
The set $\EE_S$ of equivalence classes of $S$ can be expressed as:
:$\EE_S = \set {\set x: x \in S}$
That is, it is the set of all singletons of $S$.

Proof:
Let $x \in S$.
Then by definition of the diagonal relation:
:$y \mathrel {\Delta_S} x \iff y = x$
Hence:
:$y \in \eqclass x {\Delta_S} \iff y = x$
That is:
:$\eqclass x {\Delta_S} = \set x$
Hence the result.
{{qed}}
Category:Diagonal Relation
Category:Examples of Equivalence Classes
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Theorem: Set of Ordered Pairs of Integers is Countable Infinite

The set of all ordered pairs of integers $\Z$ is countably infinite.

Proof:
The set of all ordered pairs of a set $S$ is by definition the Cartesian product $S \times S$.
In this context we are determining the cardinality of $\Z \times \Z$.
From Integers are Countably Infinite, we have that $\Z$ is a countably infinite set.
The result then follows from Cartesian Product of Countable Sets is Countable.
{{qed}}
Theorem: Union of Relation with Inverse is Symmetric Relation

Let $\RR$ be a relation on a set $S$.
Let $\RR^{-1}$ denote the inverse of $\RR$.
Then $\RR \cup \RR^{-1}$, the union of $\RR$ with $\RR^{-1}$, is symmetric.

Proof:
Let $\tuple {a, b} \in \RR \cup \RR^{-1}$.
By definition of union, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
;Case 1
If $\tuple {a, b} \in \RR$, then by definition of inverse relation:
:$\tuple {b, a} \in \RR^{-1}$
But from Set is Subset of Union:
:$\tuple {b, a} \in \RR \cup \RR^{-1}$
{{qed|lemma}}
;Case 2
If $\tuple {a, b} \in \RR^{-1}$, then by definition of inverse relation:
:$\tuple {b, a} \in \paren {\RR^{-1} }^{-1}$
From Inverse of Inverse Relation:
:$\tuple {b, a} \in \RR$
But from Set is Subset of Union:
:$\tuple {b, a} \in \RR \cup \RR^{-1}$
{{qed|lemma}}
Hence by Proof by Cases:
:$\tuple {a, b} \in \RR \cup \RR^{-1} \implies \tuple {b, a} \in \RR \cup \RR^{-1}$
and $\RR \cup \RR^{-1}$ is symmetric by definition.
{{qed}}
Category:Set Union
Category:Inverse Relations
Category:Symmetric Relations
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Theorem: Equivalence of Definitions of Symmetric Closure

{{TFAE|def = Symmetric Closure}}
Let $\RR$ be a relation on a set $S$.

Proof:
First we note that from Union of Relation with Inverse is Symmetric Relation, $\RR \cup \RR^{-1}$ is a symmetric relation.