avewright/proofwiki-math huggingface dataset:
Theorem: Characterization of Continuous Linear Transformations between Locally Convex Spaces

Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ and $\struct {Y, \mathcal Q}$ be locally convex spaces over $\Bbb F$.
Let $T : X \to Y$ be a linear transformation.
{{TFAE}}
:$(1) \quad$ $T$ is everywhere continuous
:$(2) \quad$ $T$ is continuous at ${\mathbf 0}_X$
:$(3) \quad$ for each $q \in \mathcal Q$ there exists a real number $C_q \ge 0$ and a finite set ${\mathcal I}_q \subseteq \mathcal P$ such that: 
::::$\ds \map q {T x} \le C_q \max_{p \in {\mathcal I}_q} \map p x$ 
:::for each $x \in X$
where we consider continuity in the standard topologies of $\struct {X, \mathcal P}$ and $\struct {Y, \mathcal Q}$.

Proof:
=== $(1)$ iff $(2)$ ===
This follows from Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin.
{{qed|lemma}}
Theorem: Translation of Open Set in Topological Vector Space is Open

Let $\struct {X, \tau}$ be a topological vector space.
Let $U \subseteq X$ be an open set.
Let $x \in X$.
Then:
:$U + x$ is open.

Proof:
Let $T_{-x}$ be the translation by $x$ mapping. 
From Translation Mapping on Topological Vector Space is Homeomorphism, we have: 
:$T_{-x}$ is a homeomorphism.
So, since $U$ is open, we have: 
:$T_{-x} \sqbrk U$ is open.
That is: 
:$U - \paren {-x} = U + x$ is open.
{{qed}}
{{explain|Review notation -- there is a caveat that $T_{-x}$ and $T_x$ get confused, and on the translation by $x$ mapping this is specifically defined as $T_x$ not $T_{-x}$.}}
Theorem: Metric Induced by Norm is Invariant Metric

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $d$ be the metric induced by $\norm {\, \cdot \,}$.
Then $d$ is invariant.

Proof:
Let $x, y, z \in X$. 
Then, we have: 
{{begin-eqn}}
{{eqn	| l = \map d {x + z, y + z}
		| r = \norm {\paren {x + z} - \paren {y + z} }
		| c = {{Defof|Metric Induced by Norm}}
}}
{{eqn	| r = \norm {x - y + z - z}
}}
{{eqn	| r = \norm {x - y}
}}
{{eqn	| r = \map d {x, y}
		| c = {{defof|Metric Induced by Norm}}
}}
{{end-eqn}}
{{qed}}
Category:Metric Spaces
Category:Normed Vector Spaces
annay4pshquglg2mc4ke6qwjzuhc1lf
Theorem: Equivalence of Definitions of Primary Ideal

Let $R$ be a commutative ring with unity.
{{TFAE|def = Primary Ideal}}

Proof:
=== Definition 1 implies Definition 2 ===
Let $x + \mathfrak q$ be a zero-divisor of $R / \mathfrak q$.
That is, there is a $y \not \in \mathfrak q$ such that:
:$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$
Thus:
:$xy + \mathfrak q = 0 + \mathfrak q$
which means:
:$xy \in \mathfrak q$
Then, {{hypothesis}}:
:$\exists n \in \N_{>0} : x^n \in \mathfrak q$
so that:
:$\paren {x + \mathfrak q}^n = x^n + \mathfrak q = 0 + \mathfrak q$
Therefore $x + \mathfrak q$ is nilpotent.
{{qed|lemma}}
Theorem: Minimally Superinductive Class is Well-Ordered under Subset Relation

Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be minimally superinductive under $g$.
Then $M$ is well-ordered under the subset relation.

Proof:
We have {{apriori}} that $M$ is a $g$-tower.
The result follows from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}}
Category:Minimally Superinductive Classes
Category:Well-Orderings
Category:Subset Relation
8ynmqqm4swiiis0usxx5rbpz7h0s5tx
Theorem: Balanced Subset of Real Numbers is Bounded or Entire Space

Consider $\R$ as a vector space over $\R$.
Let $E$ be a balanced subset of $\R$.
Then $E$ is bounded, or $E = \R$.

Proof:
Suppose that $E$ is not bounded.
Then, for each $M > 0$ there exists some $x_M \in E$ such that $\size {x_M} > M$.
We show that: 
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
Let:
:$t \in \closedint {-M} M$
Then, we have: 
:$\ds \size {\frac t {x_M} } < 1$
So, since $E$ is balanced, we have: 
:$\ds x_M \cdot \paren {\frac t {x_M} } = t \in E$
So:
:$\closedint {-M} M \subseteq E$ for each $M > 0$.
From Union of Subsets is Subset, we then have: 
:$\ds \bigcup_{M > 0} \closedint {-M} M \subseteq E$
So:
:$\R \subseteq E$
Since we also have $E \subseteq \R$, we obtain $E = \R$.
{{qed}}
CAtegory:Balanced Sets
nnqu15dg41idimeiazngc1tx17izgs3
Theorem: Balanced Subset of Complex Plane is Bounded or Entire Space

Consider $\C$ as a vector space over $\C$.
Let $E$ be a balanced subset of $\C$.
Then $E$ is bounded, or $E = \C$.

Proof:
Suppose that $E$ is not bounded.
Then, for each $R > 0$ there exists some $z_R \in E$ such that $\size {z_R} > R$.
We show that: 
:$\map {B_R} 0 \subseteq E$ for each $R > 0$
where $\map {B_R} 0$ is the open ball centered at $0$ with radius $R$.
Let:
:$w \in \map {B_R} 0$
Then: 
:$\ds \cmod {\frac w {z_R} } < 1$
So, since $E$ is balanced, we have: 
:$\ds z_R \cdot \paren {\frac w {z_R} } = w \in E$
giving:
:$\map {B_R} 0 \subseteq E$ for each $R > 0$.
From Union of Subsets is Subset, we have: 
:$\ds \bigcup_{R > 0} \map {B_R} 0 \subseteq E$
giving:
:$\C \subseteq E$
Since we also have $E \subseteq \C$, we obtain $E = \C$.
{{qed}}
Category:Balanced Sets
bs4haeee57gulamkxg7jbp7smb1kswk
Theorem: Image of Balanced Set under Linear Transformation is Balanced

Let $\Bbb F \in \set {\R, \C}$.
Let $X$ and $Y$ be vector spaces over $\Bbb F$.
Let $E \subseteq X$ be balanced.
Let $T : X \to Y$ be a linear transformation.
Then $\map T E \subseteq Y$ is balanced.

Proof:
We aim to show that for all $s \in \R$ with $\cmod s \le 1$, we have: 
:$s \map T E \subseteq \map T E$
Let $y \in s \map T E$.
Then there exists $x \in E$ such that $y = s T x$.
From the linearity of $T$, we have: 
:$y = \map T {s x}$
Since $x \in E$, we have $s x \in s E$ from the definition of dilation of $E$ by $s$.
Since $E$ is balanced, we have $s E \subseteq E$ and so: 
:$s x \in E$
giving:
:$y = \map T {s x} \in \map T E$
Since $y \in s \map T E$ was arbitrary, we have $s \map T E \subseteq \map T E$, giving that $\map T E$ is balanced.
{{qed}}
Category:Balanced Sets
Category:Linear Transformations
4i9qlgbpvari53vlwm0laslaikxtqr8
Theorem: Extending Operation is a Slowly Progressing Mapping

Let $S$ denote the class of all ordinal sequences.
Let $E: S \to S$ be an extending operation on $S$.
Then $E$ is a slowly progressing mapping.

Proof:
Let $\theta \in S$ be an $\alpha$-sequence.
By definition of extending operation:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
Thus:
:$\theta \subseteq \map E \theta$
and it is seen that $E$ is by definition a progressing mapping.
It is also seen that:
:$\card \theta = \card \alpha$
while:
:$\card {\map E \theta} = \card \alpha + 1$
demonstrating that $E$ is a slowly progressing mapping.
{{qed}}
Theorem: Extending Operation is a Strictly Progressing Mapping

Let $S$ denote the class of all ordinal sequences.
Let $E: S \to S$ be an extending operation on $S$.
Then $E$ is a strictly progressing mapping.

Proof:
Let $\theta \in S$ be an $\alpha$-sequence.
By definition of extending operation:
:$\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
From Extending Operation is a Slowly Progressing Mapping we have {{afortiori}} that $E$ is a progressing mapping.
By definition of $\alpha$-sequence:
:$\alpha \notin \Dom \theta$
and so:
:$\tuple {\alpha, x} \notin \theta$
Hence:
:$\theta \subsetneqq \map E \theta$
demonstrating that $E$ is a strictly progressing mapping.
{{qed}}
Theorem: Dilation of Union of Subsets of Vector Space

Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then: 
:$\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.

Proof:
We have: 
:$\ds v \in \lambda \bigcup_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to: 
:there exists some $\alpha \in I$ such that $v = \lambda x$ for some $x \in E_\alpha$.
This is equivalent to: 
:there exists some $\alpha \in I$ such that $v \in \lambda E_\alpha$.
This is finally equivalent to: 
:$\ds v \in \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
So by the definition of set equality, we have: 
:$\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Set Union
hoszb7mxqqej7l2ejlv1nerauvbuhae
Theorem: Dilation of Intersection of Subsets of Vector Space

Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.

Proof:
First, if $\lambda = 0_K$ then we have:
:$\lambda E_\alpha = \set { {\mathbf 0}_X}$
and:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \set { {\mathbf 0}_X}$
so that:
:$\ds \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha} = \set { {\mathbf 0}_X} = \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
Now take $\lambda \ne 0_K$.
Let $v \in X$.
Then, we have:
:$\ds v \in \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$\ds \lambda^{-1} v \in \bigcap_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$\lambda^{-1} v \in E_\alpha$ for each $\alpha \in I$
{{iff}}:
:$v \in \lambda E_\alpha$ for each $\alpha \in I$
{{iff}}:
:$\ds v \in \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
So, we have:
:$\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
in the case $\lambda \ne 0_K$ as well.
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Set Intersection
lr72gta0l6e2jhnaa5yk1o9peotj0e2
Theorem: Vertical Section of Continuous Function is Continuous

Let $X$, $Y$ and $T$ be topological spaces.
Equip the Cartesian product $X \times Y$ with the product topology.
Let $f : X \times Y \to T$ be a continuous mapping. 
Let $x \in X$.
Then the $x$-vertical section $f_x : Y \to T$ is continuous.

Proof:
From the definition of the $x$-vertical section, we have: 
:$\map {f_x} y = \map f {x, y}$
for each $y \in Y$. 
Define the map $p_x : Y \to X \times Y$ by: 
:$\map {p_x} y = \tuple {x, y}$ 
for each $y \in Y$. 
We have that: 
:$f_x = f \circ p_x$
From Composite of Continuous Mappings is Continuous, since $f$ is continuous, it suffices to show that $p_x$ is continuous.
From Box Topology on Finite Product Space is Product Topology, the product topology on $X \times Y$ is also the box topology. 
From Continuity Test using Basis, it then suffices to check that: 
:$p_x^{-1} \sqbrk {U \times V}$ is open whenever $U$ is open in $X$ and $V$ is open in $Y$.
{{explain|Continuity Test using Basis requires checking the use of analytic basis for the corresponding topology. The justification for how the analytic basis is used seems to be missing in this proof}}
Let $U$ be open in $X$ and $V$ be open in $Y$.
Note that: 
:$y \in p_x^{-1} \sqbrk {U \times V}$
{{iff}}:
:$\tuple {x, y} \in U \times V$
which is equivalent to: 
:$x \in U$ and $y \in V$. 
Clearly then if $x \not \in U$, we have $p_x^{-1} \sqbrk {U \times V} = \O$.
We can also read off from this equivalence that if $x \in U$, we have $p_x^{-1} \sqbrk {U \times V} = V$.
From Empty Set is Element of Topology, $\O$ is open in $Y$.
Since $V$ is open in $Y$ by hypothesis, we therefore have that $p_x$ is continuous from Continuity Test using Basis.
Hence the result.
{{qed}}
Category:Vertical Section of Functions
Category:Continuous Mappings
Category:Product Topology
2nafg484x168ej3s5vsg7yd3b46ks29
Theorem: Dilation Mapping on Topological Vector Space is Continuous

Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field.
Let $\struct {X, +_X, \circ_X, \tau_X}$ be a vector space over $K$. 
Let $\lambda \in K$. 
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is continuous.

Proof:
Let $f: K \times X \to X$ be the map defined by:
:$\tuple {\lambda, x} \mapsto \lambda \circ_X x$
for any $x \in X$
From the definition of a topological vector space, the mapping $f$ is continuous.
We are  given that $c_\lambda$ is the dilation by $\lambda$ mapping.
That is,  $c_\lambda : X \to X$ is the map:
:$\forall x \in X: \map {c_\lambda} x = \lambda \circ_X x$
So $c_\lambda$ is by definition a $\lambda$-vertical section of $f$ such that:
:$\forall x \in X: \map {c_\lambda} x = \map f {\lambda, x}$
The result follows from Vertical Section of Continuous Function is Continuous.
{{qed}}
Category:Dilation Mappings
Category:Topological Vector Spaces
as3lkzusxeby1qemx5qzh17b92u73ui
Theorem: Dilation Mapping on Topological Vector Space is Homeomorphism

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $\lambda \in K \setminus \set {0_K}$. 
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
Then $c_\lambda$ is a homeomorphism.

Proof:
From Dilation Mapping on Topological Vector Space is Continuous, both $c_{\lambda}$ and $c_{1/\lambda}$ are continuous.
It is therefore sufficient to establish that $c_{1/\lambda}$ is the inverse mapping of $c_\lambda$.
For all $x \in X$, we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {c_\lambda \circ c_{1/\lambda} } } x
		| r = \lambda \paren {\frac 1 \lambda x}
}}
{{eqn	| r = x
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \map {\paren {c_{1/\lambda} \circ c_\lambda} } x
		| r = \frac 1 \lambda \paren {\lambda x}
}}
{{eqn	| r = x
}}
{{end-eqn}}
So both $c_\lambda \circ c_{1/\lambda}$ and $c_{1/\lambda} \circ c_\lambda$ are the identity mapping for $X$.
So $c_{1/\lambda}$ is the inverse mapping of $c_\lambda$, as required.
{{qed}}
Theorem: Dilation of Open Set in Topological Vector Space is Open

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $\lambda \in K \setminus \set {0_K}$. 
Let $U$ be a open set in $X$.
Then $\lambda U$, the dilation of $U$ by $\lambda$ is open.

Proof:
Let $c_\lambda$ be the dilation by $\lambda$ mapping.
From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.
Hence the image $c_\lambda \sqbrk U$ is open.
That is, $\lambda U$ is open.
{{qed}}
Category:Topological Vector Spaces
bszuwbeeyz3zg2eyhh6t4yivewu94pb
Theorem: Translation Mapping on Topological Vector Space is Continuous

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $x \in X$.
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is continuous.

Proof:
From the definition of a topological vector space, the mapping $X \times X \to X$ defined by $\tuple {y, x} \mapsto y + x$ is continuous.
From Horizontal Section of Continuous Function is Continuous, it follows that the $\paren {-x}$-horizontal section $T_x : X \to X$ with $y \mapsto y - x$ is continuous.
{{qed}}
Category:Translation Mappings (Vector Spaces)
Category:Topological Vector Spaces
q6rf2a76cxj89vckzz9bfltmfevnjlu
Theorem: Translation Mapping on Topological Vector Space is Homeomorphism

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $x \in X$. 
Let $T_x$ be the translation by $x$ mapping.
Then $T_x$ is a homeomorphism.

Proof:
From Translation Mapping on Topological Vector Space is Continuous, both $T_x$ and $T_{-x}$ are continuous.
It is therefore sufficient to establish that $T_{-x}$ is the inverse mapping of $T_x$.
For all $y \in X$, we have:
{{begin-eqn}}
{{eqn | l = \map {\paren {T_x \circ T_{-x} } } y
      | r = \map {T_x} {y - \paren {-x} }
}}
{{eqn | r = \map {T_x} {y + x}
}}
{{eqn | r = y + x - x
}}
{{eqn | r = y
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map {\paren {T_{-x} \circ T_x} } y
      | r = \map {T_{-x} } {y - x}
}}
{{eqn | r = y - x - \paren {-x}
}}
{{eqn | r = y - x + x
}}
{{eqn | r = y
}}
{{end-eqn}}
So both $T_x \circ T_{-x}$ and $T_{-x} \circ T_x$ are the identity mapping for $X$.
So $T_{-x}$ is the inverse mapping of $T_x$, as required.
{{qed}}
Theorem: Sum of Set and Open Set in Topological Vector Space is Open

Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A$ and $B$ be subsets of $X$, with $A$ an open set.
Then the linear combination $A + B$ is open.

Proof:
It can be shown that: 
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
If: 
:$\ds v \in \bigcup_{b \mathop \in B} \paren {A + b}$
then $v = a + b$ for some $a \in A$ and $b \in B$.
So $v \in A + B$, giving: 
:$\ds \bigcup_{b \mathop \in B} \paren {A + b} \subseteq A + B$
Conversely, suppose that $v \in A + B$.
Then there exists $a \in A$ and $b \in B$ such that $v = a + b$. 
Then, we have $v \in A + b$, and so $v \in A + B$.
We therefore obtain: 
:$\ds A + B \subseteq \bigcup_{b \mathop \in B} \paren {A + b}$
So, by the definition of set equality, we have: 
:$\ds A + B = \bigcup_{b \mathop \in B} \paren {A + b}$
From Translation of Open Set in Topological Vector Space is Open:
:$A + b$ is open for each $b \in B$.
So $A + B$ is the union of open sets. 
Since topologies are closed under union, we therefore have that $A + B$ is open.
{{qed}}
Category:Topological Vector Spaces
kyr2on04alf75zhea1gwv031t4stlae
Theorem: Image of Translation of Set under Linear Transformation is Translation of Image

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $x \in X$.
Then: 
:$\map T {E + x} = \map T E + T x$
where $E + x$ denotes the translation of $E$ by $x$.

Proof:
We have: 
:$y \in \map T {E + x}$
{{iff}}:
:$y = \map T {u + x}$ for some $u \in E$.
From the linearity of $T$, this is equivalent to: 
:$y = T u + T x$ for some $u \in E$.
This is equivalent to:
:$y \in \map T E + T x$
So by the definition of set equality we have:
:$\map T {E + x} = \map T E + T x$
{{qed}}
Category:Linear Transformations
Category:Translation of Subsets of Vector Spaces
jvjpgtypl1efgn4vobv5007cgurzi3g
Theorem: Image of Dilation of Set under Linear Transformation is Dilation of Image

Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$. 
Let $T : X \to Y$ be a linear transformation.
Let $E \subseteq X$ be a non-empty set.
Let $\lambda \in K$. 
Then: 
:$\map T {\lambda E} = \lambda \map T E$
where $\lambda E$ denotes the dilation of $E$ by $\lambda$.

Proof:
We have: 
:$y \in \map T {\lambda E}$
{{iff}}:
:$y = \map T {\lambda x}$ for some $x \in E$.
From the linearity of $T$, this is equivalent to:
:$y = \lambda T x$
This is equivalent to:
:$y \in \lambda \map T E$
So by the definition of set equality we have:
:$\map T {\lambda E} = \lambda \map T E$
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Linear Transformations
qqfp88f62n066jheihaybessyyto07a
Theorem: Characteristic of Extending Operation

Let $E$ be an extending operation.
Then there exists a mapping $F$ on the class of all ordinals $\On$ such that:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
:$F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$
:$\alpha^+$ denotes the successor ordinal of $\alpha$.

Proof:
Let $M$ be the class which is minimally superinductive under $E$.
Recall the Transfinite Recursion Theorem:
{{begin-axiom}}
{{axiom | n = 1
        | lc= Zeroth Ordinal:
        | q = 
        | m = M_0 = \O
}}
{{axiom | n = 2
        | lc= Successor Ordinal:
        | q = \forall \alpha \in \On
        | m = M_{\alpha^+} = \map E {M_\alpha}
}}
{{axiom | n = 3
        | lc= Limit Ordinal:
        | q = \forall \lambda \in K_{II}
        | m = M_\lambda = \bigcup_{\alpha \mathop < \lambda} M_\alpha
}}
{{end-axiom}}
where:
:for an arbitrary ordinal $\alpha$, $M_\alpha$ denotes the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$
:$K_{II}$ denotes the class of all limit ordinals.
We know that $M_0 = 0$, so $M_0$ is a $0$-sequence.
Let $M_\alpha$ be an arbitrary $\alpha$-sequence.
Then from $(2)$ above, $M_{\alpha^+}$ is an $\alpha^+$-sequence.
Let $\lambda$ be a limit ordinal.
Suppose that, for each $\alpha < \lambda$, $M_\alpha$ is an $\alpha$-sequence.
Then by Length of Union of Chain of Ordinal Sequences, $M_\lambda$ is a $\lambda$-sequence.
Hence by the Second Principle of Transfinite Induction:
:for every $\alpha \in \On$, $M_\alpha$ is an ordinal sequence of length $\alpha$.
Let $F = \bigcup M$ be the union of $M$.
Then by Length of Union of Chain of Ordinal Sequences again, $F$ is a mapping on $\On$.
Because each $M_\alpha \subseteq F$ and $\Dom {M_\alpha} = \alpha$, it follows that $M_\alpha = F \restriction \alpha$.
Hence from Union of Nest of Ordinal Sequences which is Proper Class:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
{{qed}}
Theorem: Translation of Union of Subsets of Vector Space

Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x = \bigcup_{\alpha \mathop \in I} \paren {E_\alpha + x}$
where $E_\alpha + x$ denotes the translation of $E_\alpha$ by $x$.

Proof:
We have:
:$\ds v \in \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x$
{{iff}}:
:$v = u + x$ for some $\ds u \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
This is equivalent to: 
:there exists $\alpha \in I$ and $u \in E_\alpha$ such that $v = u + x$.
That is: 
:there exists $\alpha \in I$ such that $v \in E_\alpha + x$.
So by the definition of set equality, we have: 
:$\ds \paren {\bigcup_{\alpha \mathop \in I} E_\alpha} + x = \bigcup_{\alpha \mathop \in I} \paren {E_\alpha + x}$
{{qed}}
Category:Translation of Subsets of Vector Spaces
Category:Set Union
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Theorem: Distance between Disjoint Compact Set and Closed Set in Metric Space is Positive

Let $\struct {X, d}$ be a metric space.
Let $A$ and $B$ be disjoint subsets of $X$ such that $A$ is compact and $B$ is closed.
Then $\map d {A, B} > 0$, where $\map d {A, B}$ is the distance between $A$ and $B$.

Proof:
Define $f : X \to \R$ by: 
:$\map f x = \map d {x, B}$
for each $x \in X$.
From the definition of the distance between $A$ and $B$, we have: 
:$\ds \map d {A, B} = \inf_{a \mathop \in A} \map f a$ 
From Compact Subspace of Hausdorff Space is Closed and Metric Space is Hausdorff, $A$ is closed and hence contains all its limit points.
From Point at Zero Distance from Subset of Metric Space is Limit Point or Element we then have:
:$\map f a > 0$
for each $a \in A$.
From Distance from Point to Subset is Continuous Function, $f$ is continuous.
From Restriction of Continuous Mapping is Continuous: Metric Spaces, the restriction $f \restriction_A$ is continuous.
From {{Corollary|Continuous Image of Compact Space is Compact|3}}, we have that $f$ attains its bounds. 
That is, there exists $\alpha \in A$ such that: 
:$\ds \map d {\alpha, B} = \inf_{a \mathop \in A} \map f a$
giving:
:$\ds \inf_{a \in A} \map f a > 0$
and so:
:$\map d {A, B} > 0$ 
{{qed}}
Category:Compact Metric Spaces
Category:Distance Function
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Theorem: Dilation of Open Set in Normed Vector Space is Open

Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $U \subseteq X$ be an open set.
Let $\lambda \in \Bbb F \setminus \set 0$. 
Then:
:$\lambda U$ is open.

Proof:
Let:
:$v \in \lambda U$
Then:
:$\dfrac v \lambda \in U$
Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:
:$\ds \norm {\frac v \lambda - \frac {v'} \lambda} < \epsilon$
we have $v'/\lambda \in U$. 
That is, $v' \in \lambda U$. 
So, whenever: 
:$\ds \norm {v - v'} < \epsilon \cmod \lambda$
we have $v' \in \lambda U$.
Since $v$ was arbitrary: 
:$\lambda U$ is open.
{{qed}}
Category:Dilations of Subsets of Vector Spaces
Category:Open Sets (Normed Vector Spaces)
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