avewright/proofwiki-math huggingface dataset:
Theorem: Point in Hilbert Sequence Space has no Compact Neighborhood

Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then no point of $\ell^2$ has a compact neighborhood.

Proof:
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space.
Let $x = \sequence {x_i} \in A$ be a point of $\ell^2$.
Consider the closed $\epsilon$-ball of $x$ in $\ell^2$:
:$\map { {B_\epsilon}^-} x := \set {y \in A: \map {d_2} {x, y} \le \epsilon}$
for some $\epsilon \in \R_{>0}$.
Consider the point:
:$\sequence {y_n} = \tuple {x_1, x_2, \ldots, x_{n - 1}, x_n + \epsilon, x_{n + 1}, \ldots}$
We have that $\sequence {y_n} \in \map { {B_\epsilon}^-} x$.
But for $m \ne n$ we have that:
:$\map {d_2} {y_m, y_n} = \epsilon \sqrt 2$
and so $\sequence {y_n}$ has no convergent subsequence.
Thus $\map { {B_\epsilon}^-} x$ is not compact.
So $x$ has no compact neighborhood.
{{qed}}
Theorem: Compact Subset of Hilbert Sequence Space is Nowhere Dense

Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Let $H$ be a compact subset of $\ell^2$.
Then $H$ is nowhere dense in $\ell^2$.

Proof:
By Compact Subset of Hilbert Sequence Space is Closed, $H$ is a closed set of $\ell^2$.
From Set is Closed iff Equals Topological Closure:
:$H^- = H$
where $H^-$ denotes the closure of $H$.
From Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact:
:$H^\circ = \O$
where $H^\circ$ denotes the interior of $H$.
Thus:
:$\paren {H^-}^\circ = \O$
and the result follows by definition of nowhere dense.
{{qed}}
Theorem: Hilbert Sequence Space is not Sigma-Compact

Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is not $\sigma$-compact.

Proof:
By Compact Subset of Hilbert Sequence Space is Nowhere Dense, a compact subset of $\ell^2$ is nowhere dense in $\ell^2$.
We have that Hilbert Sequence Space is Complete Metric Space.
From Complete Metric Space is Non-Meager, $\ell^2$ is non-meager.
It follows that $\ell^2$ is not $\sigma$-compact.
{{explain|How?}}
{{qed}}
Theorem: Hilbert Sequence Space is Injectively Path-Connected

Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} {x_i}^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is injectively path-connected.

Proof:
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$.
Consider the mapping $f: \closedint 0 1 \to \ell^2$ defined as:
:$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop \ge 0} \paren {t x_i + \paren {1 - t} y_i}^2
      | r = \sum_{i \mathop \ge 0} \paren {t^2 {x_i}^2 + \paren {1 - t}^2 {y_i}^2 + 2 t \paren {1 - t} x_i y_i}
      | c = 
}}
{{eqn | r = t^2 \sum_{i \mathop \ge 0} {x_i}^2 + \paren {1 - t}^2 \sum_{i \mathop \ge 0} {y_i}^2 + 2 t \paren {1 - t} \sum_{i \mathop \ge 0} x_i y_i
      | c = 
}}
{{end-eqn}}
which is convergent.
{{explain|Find the appropriate rules to prove the above statement.}}
Then $f$ is an injective path joining $x$ to $y$.
Hence the result.
{{qed}}
Theorem: Fréchet Product Space is Metric Space

Let $\struct {\R^\omega, d}$ be the '''Fréchet product space on $\R^\omega$'''.
Then $\struct {\R^\omega, d}$ is a metric space.

Proof:
It is to be demonstrated that $d$ satisfies all the metric space axioms.
Recall from the definition of the '''Fréchet product space''' that the distance function $d: \R^\omega \times \R^\omega \to \R$ is defined on $\R^\omega$ as:
:$\forall x, y \in \R^\omega: \map d {x, y} = \ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$
where:
:$x := \sequence {x_i}_{i \mathop \in \N} = \tuple {x_0, x_1, x_2, \ldots}$
:$y := \sequence {y_i}_{i \mathop \in \N} = \tuple {y_0, y_1, y_2, \ldots}$
denote arbitrary elements of $\R^\omega$.
First it is confirmed that Fréchet Product Metric is Absolutely Convergent on arbitrary $x$ and $y$, as follows.
{{Explain|Either prove it in a linked page or prove it here, but doing it in both places is contrary to the minimalist {{ProofWiki}} approach}}
First note that:
:$\dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } = 1 - \dfrac 1 {1 + \size {x_i - y_i} }$
implies that we have:
:$0 \le \dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } < 1$
By the Ratio Test we have that the series is absolutely convergent.
An important property of absolutely convergent series is that any rearrangement of the terms of the series is also convergent and equal to the original series.
This will be used in the proof of {{Metric-space-axiom|2}}.
Theorem: Fréchet Product Space is Complete Metric Space

Let $\struct {\R^\omega, d}$ be the '''Fréchet product space on $\R^\omega$'''.
Then $\struct {\R^\omega, d}$ is a complete metric space.

Proof:
From Fréchet Product Space is Metric Space, $\struct {\R^\omega, d}$ is a metric space.
It remains to be demonstrated that $\struct {\R^\omega, d}$ is complete.
Let $\sequence {x^{\paren n} }_{n \mathop \in \N}$ be an arbitrary Cauchy sequence in $\R^\omega$, where as an exception we denote the index of the Cauchy sequence as a superscript.
Recall that each $\forall n \in \N: x^{\paren n} \in \R^\omega$ itself is a sequence in $\R$ by the definition of $\R^\omega$.
More precisely, each $x^{\paren n} \in \R^\omega$ can be written as
:$x^{\paren n} := \sequence { {x_i}^{\paren n} }_{i \mathop \in \N}$
where ${x_i}^{\paren n} \in \R$ for all $i \in \N$.
Theorem: Separable Metric Space is Homeomorphic to Subspace of Fréchet Metric Space

Let $M = \struct {A, d}$ be a metric space whose induced topology is separable.
Then $M$ is homeomorphic to a subspace of the Fréchet product space $\struct {\R^\omega, d}$ on the countable-dimensional real Cartesian space $\R^\omega$.

Proof:
Let $f: M \to \R^\omega$ be the mapping defined as:
:$\forall x \in M: \map f x = \sequence {\map d {x, x_i} }$
where $\set {x_i}$ is a countable dense subset of $A$.
It remains to be shown that $f$ is a homeomorphism.
{{ProofWanted}}
Theorem: Hilbert Cube is Metric Space

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a metric space.

Proof:
As defined, $M$ is a subspace of the Hilbert sequence space $\ell^2$.
We have that Hilbert Sequence Space is Metric Space.
The result follows from Subspace of Metric Space is Metric Space.
{{qed}}
Theorem: Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals

Let $M_1 = \struct {I^\omega, d_2}$ be the Hilbert cube:
:$M_1 = \ds \prod_{k \mathop \in \N} \closedint 0 {\dfrac 1 k}$
under the same metric as that of the Hilbert sequence space:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$
Let $M_2$ be the metric space defined as:
:$M_2 = \ds \prod_{k \mathop \in \N} \closedint 0 1$
under the product topology.
Then $M_1$ and $M_2$ are homeomorphic.

Proof:
Let $x = \sequence {x_i}$ be an arbitrary element of $M_1$.
Let $f: M_1 \to M_2$ be the mapping defined as:
:$\forall x \in M_1: \map f x = \tuple {x_1, 2 x_2, 3 x_3, \ldots}$
Then $f$ is seen to be a bijection.
It remains to be shown that an open set in $M_1$ is mapped to an open set in $M_2$ by $f$.
{{ProofWanted}}
Theorem: Hilbert Cube is Completely Normal

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a completely normal space.

Proof:
We have that Hilbert Cube is Metric Space.
The result follows from Metric Space is Completely Normal.
{{qed}}
Theorem: Hilbert Cube is Separable

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a separable space.

Proof:
Consider the set $H$ of all points of $M$ which have finitely many rational coordinates and all the rest zero.
Then $H$ forms a countable subset of $A$ which is (everywhere) dense.
{{finish|Demonstrate that it is (everywhere) dense.}}
The result follows by definition of separable space.
{{qed}}
Theorem: Hilbert Cube is Second-Countable

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a second-countable space.

Proof:
From Hilbert Cube is Separable, $M$ is a separable space.
We also have that Hilbert Cube is Metric Space.
The result follows from Separable Metric Space is Second-Countable.
{{qed}}
Theorem: Hilbert Cube is Compact

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a compact space.

Proof:
Let $M'$ be the metric space defined as:
:$M' = \ds \prod_{k \mathop \in \N} \closedint 0 1$
under the product topology.
By definition, $\closedint 0 1$ is the closed unit interval under the usual (Euclidean) topology.
From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals, $M$ is homeomorphic to $M'$.
From Closed Real Interval is Compact in Metric Space, $\closedint 0 1$ is a compact space.
But $M'$ is also a compact space by Tychonoff's Theorem.
From Compactness is Preserved under Continuous Surjection, if $M'$ is compact then so is $M$.
{{qed}}
{{expand|Add the direct first-principles proof from S&S as Proof 2.}}
Theorem: Hilbert Cube is Injectively Path-Connected

Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is an injectively path-connected space.

Proof:
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$.
Consider the mapping $f: \closedint 0 1 \to I^\omega$ defined as:
:$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop \ge 0} \paren {t x_i + \paren {1 - t} y_i}^2
      | r = \sum_{i \mathop \ge 0} \paren {t^2 x_i^2 + \paren {1 - t}^2 y_i^2 + 2 t \paren {1 - t} x_i y_i}
      | c = 
}}
{{eqn | r = t^2 \sum_{i \mathop \ge 0} x_i^2 + \paren {1 - t}^2 \sum_{i \mathop \ge 0} y_i^2 + 2 t \paren {1 - t} \sum_{i \mathop \ge 0} x_i y_i
      | c = 
}}
{{end-eqn}}
which is convergent.
{{explain|Find the appropriate rules to prove the above statement.}}
Then $f$ is an injective path joining $x$ to $y$.
Hence the result.
{{qed}}
Theorem: Interval of Totally Ordered Set is Order-Convex

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $I \subseteq S$ be an interval in $S$.
Then $I$ is order-convex.

Proof:
There are a number of cases to investigate.
=== Open Interval ===
Let $I = \openint a b$ be an open interval:
:$I = \set {x \in S: a \prec x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \prec s \prec x$
and:
:$x \prec t \prec b$
and so:
:$a \prec x \prec b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is order-convex by definition.
{{qed|lemma}}
=== Left Half-Open Interval ===
Let $I = \hointl a b$ be a left half-open interval:
:$I = \set {x \in S: a \prec x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \prec s \prec x$
and:
:$x \prec t \preceq b$
and so:
:$a \prec x \preceq b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is order-convex by definition.
{{qed|lemma}}
=== Right Half-Open Interval ===
Let $I = \hointr a b$ be a right half-open interval:
:$I = \set {x \in S: a \preceq x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \preceq s \prec x$
and:
:$x \prec t \prec b$
and so:
:$a \preceq x \prec b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is order-convex by definition.
{{qed|lemma}}
=== Closed Interval ===
Let $I = \closedint a b$ be a closed interval:
:$I = \set {x \in S: a \preceq x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \preceq s \prec x$
and:
:$x \prec t \preceq b$
and so:
:$a \preceq x \preceq b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is order-convex by definition.
{{qed}}
Theorem: Interval of Totally Ordered Set is Order-Convex

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $I \subseteq S$ be an interval in $S$.
Then $I$ is order-convex.

Proof:
An interval can be represented as the intersection of two rays.
{{explain|Obvious though it is, the above needs to be stated as a theorem in its own right.}}
Thus by Ray is Order-Convex and Intersection of Order-Convex Sets is Order-Convex, $I$ is order-convex.
{{qed}}
Theorem: Order-Convex Subset of Ordered Set is not necessarily Interval

Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $C$ be an order-convex subset of $S$.
Then it is not necessarily the case that $C$ is an interval of $S$.

Proof:
Consider the open ray of $S$:
:$R = \set {x \in S: a \prec x}$
for some $a \in S$.
From Ray is Order-Convex, $R$ is a order-convex subset of $S$.
But $R$ is not an interval of $S$.
{{qed}}
Theorem: Union of Non-Disjoint Order-Convex Sets is Order-Convex

Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\CC$ be a set of order-convex sets of $S$ such that their intersection is non-empty:
:$\bigcap \CC \ne \O$
Then the union $\bigcup \CC$ is also order-convex.

Proof:
Let $x, y, z \in S$ be arbitrary elements of $S$ such that $x \prec y \prec z$.
Let $x, z \in \bigcup \CC$.
First let $x, z \in C$ where $C \in \CC$.
Then as $C$ is order-convex, $y \in C$.
Hence, by definition of union, $y \in \bigcup \CC$.
Now let $x \in C_1, z \in C_2$ where $C_1, C_2 \in \CC$.
We have that $\bigcap \CC \ne \O$.
Thus $C_1 \cap C_2 \ne \O$.
Then $\exists a \in C_1 \cap C_2: x < a < z$.
Hence one of the following cases holds:
{{begin-itemize}}
{{item|(1):|$x < y < a < z$, whence $y \in C_1$, by order-convexity of $C_1$}}
{{item|(2):|$x < a < y < z$, whence $y \in C_2$, by order-convexity of $C_2$}}
{{item|(3):|$y {{=}} a$, whence $y \in C_1$ and $y \in C_2$, by definition of $a$.}}
{{end-itemize}}
Thus in all cases $y \in \bigcup \CC$.
Thus $\bigcup \CC$ is order-convex by definition.
{{qed}}
Theorem: Separated Subsets of Linearly Ordered Space under Order Topology

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
{{begin-eqn}}
{{eqn | l = A^*
      | r = \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}
}}
{{eqn | l = B^*
      | r = \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}
}}
{{end-eqn}}
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Then $A^*$ and $B^*$ are themselves separated sets of $T$.

Proof:
=== Lemma ===
{{:Separated Subsets of Linearly Ordered Space under Order Topology/Lemma}}{{qed|lemma}}
So, from the {{Lemma|Separated Subsets of Linearly Ordered Space under Order Topology}}:
:$A \subseteq A^*$
:$B \subseteq B^*$
:$A^* \cap B^* = \O$
Let $p \notin A^* \cup A^-$.
Thus $p \notin A^*$ and $p \notin A^-$.
Then there exists an open interval $\openint s t$ which is disjoint from $A$ such that $p \in \openint s t$.
Now $\openint s t$ can only intersect $A^*$ only if it intersects some $\closedint a b \subseteq A^*$ where $a, b \in A$.
But we have:
:$\openint s t \cap A = \O$
and as $a, b \in A$ it follows that:
:$\openint s t \subseteq \openint a b$
That means $p \in A^*$.
But we have $p \notin A^*$.
Therefore:
:$\openint s t \cap A^* = \O$
Thus:
:$p \notin \paren {A^*}^-$
Hence:
{{begin-eqn}}
{{eqn | l = \paren {A^*}^-
      | o = \subseteq
      | r = \paren {A^* \cup A^-} \cap B^*
      | c = 
}}
{{eqn | r = \paren {A^* \cap B^*} \cup \paren {A^- \cap B^*}
      | c = 
}}
{{eqn | r = \O
      | c = 
}}
{{end-eqn}}
Hence the result.
{{explain|Fill in the justification for the above chain of reasoning}}
{{qed}}
Theorem: Generators of Special Linear Group of Order 2 over Integers

Let:
:$ S = \begin{pmatrix}
0 & - 1 \\
1 & 0
\end{pmatrix}$
and:
:$T = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}$
Then $S$ and $T$ are generators for the special linear group of order $2$ over $\Z$.

Proof:
Let:
:$ g = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$
be an element of $\SL {2, \Z}$.
Observe that:
:$T^n = \begin{pmatrix} 
1 & n \\ 
0 & 1
\end{pmatrix}$
so that:
:$T^n g = \begin{pmatrix}
a + nc & b + nd \\
c & d
\end{pmatrix}$
Also note that:
:$S^2 = -I$
and:
:$S g = \begin{pmatrix}
-c & -d \\
a & b
\end{pmatrix}$
We now describe an algorithm to reduce $g$ to the identity matrix only using multiplication by $S$ and $T$. 
This will show that $g^{-1}$ can be written as a word in $S$ and $T$. 
We can then take the inverse of this expression in $S$ and $T$ and it too will be a word in $S$ and $T$. 
Recall $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. 
If $c = 0$ then, since $\map \det g = 1$, we must have $a d = 1$. 
Thus $a = d = \pm 1$. In this case:
:$g = \pm\begin{pmatrix}
1 & b' \\
0 & 1
\end{pmatrix} = \pm T^{b'}$
where $b' = \pm b$. 
Since $S^2 = -I$ we either have $g = T^{b'}$ or $g = S^2 T^{b'}$. 
If $c \ne 0$ we proceed as follows. 
{{WLOG}}, we may suppose $\size a \ge \size c$, for if $\size a < \size c$ we could apply $S$ to swap $a$ and $c$. 
By the Division Theorem, we can write $a = c q + r$ where $q, r \in \Z$ and $0 \le r < \size c$. 
Then $T^{-q} g$ has upper left entry $a - q c = r$ which is smaller than $\size c$.
Applying $S$ switches the rows (with a sign change), which then makes our lower left entry $-r$. 
We have applied a combination of $S$ and $T$ to turn $g$ into a matrix where the absolute value of the lower left entry is strictly less than the one we started with. 
If $r = 0$ then we're in our original case, and we know we can multiply by a multiple of $T$ and $S$ to get to the identity matrix. 
Note that this process can have at most $r$ steps, since there are only $r$ natural numbers between $r$ and $0$. 
Thus the process terminates in a finite number of steps.
{{qed}}
Category: Special Linear Group
m8281ehsc3tjumf07oeqjy7cd3ll2gp
Theorem: Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
{{begin-eqn}}
{{eqn | l = A^*
      | r = \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}
}}
{{eqn | l = B^*
      | r = \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}
}}
{{end-eqn}}
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Then the set $M = \set {A_\alpha, B_\beta, C_\gamma}$ inherits a linear ordering from $S$, and so is a linearly ordered set.

Proof:
{{ProofWanted|I lack both the skill and the patience to disentangle this material.}}
Theorem: Successor Sets of Linearly Ordered Set Induced by Convex Component Partition

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
{{begin-eqn}}
{{eqn | l = A^*
      | r = \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}
}}
{{eqn | l = B^*
      | r = \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}
}}
{{end-eqn}}
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
:$M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Then each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$, the set of strict upper bounds for $A_\alpha$.
Similarly for $B_\beta$.
That immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.

Proof:
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$.
This belongs to the complement in $S$ of the closed set $\paren {B^*}^-$.
Hence there exists a neighborhood $\openint x y$ of $p$ which is disjoint from $\paren {B^*}^-$.
Then:
:$\openint x y \cap S_\alpha \ne \O$
and so:
:$\openint p y \ne \O$
But $\openint p y$ is disjoint from both $A^*$ and $B^*$.
Thus there must exist some $C_\gamma$ which contains $\openint p y$.
{{qed}}
Theorem: Linearly Ordered Space is T5

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a $T_5$ space.

Proof:
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
{{begin-eqn}}
{{eqn | l = A^*
      | r = \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}
}}
{{eqn | l = B^*
      | r = \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}
}}
{{end-eqn}}
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
:$M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Let:
:$S_\alpha$ be the set of strict upper bounds for $A_\alpha$.
:$S_\beta$ be the set of strict upper bounds for $B_\beta$.
From Successor Sets of Linearly Ordered Set Induced by Convex Component Partition:
:each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$
:each of the sets $B_\beta \in M$ has an immediate successor in $M$ if $B_\beta$ intersects the closure of $S_\beta$
:that immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
For each $\gamma$, let $k_\gamma \in C_\gamma$ be selected and fixed.
{{explain|We probably need AoC for this.}}
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then there exists a unique ${k_\alpha}^+ \in {C_\alpha}^+$.
In such cases, let $I_\alpha = \hointr p { {k_\alpha}^+}$ where $p \in A \cap {S_\alpha}^-$.
The other case is when $A_\alpha \cap {S_\alpha}^- = \O$.
In this case, let $I_\alpha = \O$.
Let $J_\alpha$ be defined similarly for the strict lower bounds of $A_\alpha$, using the same set of points $k_\gamma \in C_\gamma$.
Then:
:For each $\alpha$, let $U_\alpha = J_\alpha \cup A_\alpha \cup I_\alpha$.
:For each $\beta$, let $V_\beta = J_\beta \cup A_\beta \cup I_\beta$.
Each $U_\alpha$ and $V_\beta$ is an order-convex open set in $S$ containing $A_\alpha$ and $B_\beta$ respectively.
Thus $\ds U = \bigcup U_\alpha$ and $\ds V = \bigcup V_\beta$ are open sets in $T$ containing $A^*$ and $B^*$ respectively.
We have that:
:no $A_\alpha$ intersects no $B_\beta$
:the same $K_\gamma$ being used throughout implies that no $J_\beta$ or $I_\beta$ can intersect any $J_\alpha$ or $I_\alpha$.
Thus no $U_\alpha$ can intersect any $V_\beta$.
Thus $U \cap V = \O$.
So we have two disjoint open sets $U, V$ in $S$ containing the two separated sets $A$ and $B$ respectively.
As $A$ and $B$ are arbitrary, it follows that such $U$ and $V$ can be found whatever $A$ and $B$ are, provided they are separated.
Hence the result by definition of $T_5$ space.
{{qed}}