User:Dfeuer/Compact Subspace of Linearly Ordered Space/Converse Proof 2

Theorem
Let $(X, \preceq, \tau)$ be a totally ordered space.

Let $Y$ be a nonempty subset of $X$.

Suppose that for each nonempty subset, $S$, of $Y$:
 * $S$ has a supremum and an infimum in $X$.
 * $\sup_X S, \inf_X S \in Y$.

Then $Y$ is a compact subspace of $X$.

Proof
All interval notation is taken in $Y$ unless subscripted with an $X$. E.g., $[a,\to)=\{y\in Y:a\le y\}$, while $[a,\to)_X=\{x\in X:a\le x\}$.

By Equivalence of Definitions of Generalized Ordered Space/Two implies One, $Y$ has a basis consisting of sets that are convex in $Y$.

By Compactness from Basis, to show $Y$ compact we need only show that a cover of $Y$ by open convex sets in $Y$ has a finite subcover.

Let $\mathcal U$ be an open cover of $Y$ all of whose elements are convex and nonempty.

Define a relation $\mathcal R$ on $\mathcal U$ by letting $U \mathrel{\mathcal R} V$ iff $U \cap V \ne \varnothing$.

Define a relation $\sim$ on $\mathcal U$ as the transitive closure of $\mathcal R$.

By Recursive Construction of Transitive Closure, $U \sim V$ iff there is a finite family $\left\{{ U_0, \dots, U_n }\right\} \subseteq \mathcal U$ such that:
 * $U=U_0$, $V=U_n$, and $U_k\cap U_{k+1} \ne \varnothing$ for $k = 0, \dots, n-1$.

$\sim$ is an equivalence relation on $\mathcal U$ by Transitive Closure of Reflexive Symmetric Relation is Equivalence.

For $U \in \mathcal{U}$, let $[U]$ be the $\sim$-equivalence class of $U$.

By Union of Overlapping Convex Sets in Toset is Convex/Infinite Union, $\bigcup [U]$ is convex for each $U \in \mathcal U$.

Then $\mathcal{C}=\left\{\bigcup[U]:U\in\mathcal U \right\}$ is a partition of $Y$ into order-convex open sets.

Since the complement of each $\bigcup [U]$ is a union of open sets, each $\bigcup [U]$ is in fact clopen.

Note that since its elements are order-convex, $\mathcal{C}$ inherits a natural total order from $Y$, which we will denote by $\preceq'$.

For each $C \in \mathcal C$, $\varnothing \subsetneqq C \subseteq Y$.

Thus by the premise, each element $C$ of $\mathcal C$ has a supremum and infimum in $X$.

$Y$ is closed in $X$ by Closed Set in Linearly Ordered Space.

$C$ is closed in $Y$.

Thus $C$ is closed in $X$.

Therefore, by Closed Set in Linearly Ordered Space, $\sup C, \inf C \in C$.

Let $a_C=\inf C$ and $b_C=\sup C$.

Since $C$ is convex in $Y$, $C = Y \cap [a_C,b_C]$.

Let $C\in\mathcal{C}$, and suppose that $a_C\ne\inf Y$. Since $[a_C,\to)$ is open in $Y$ there is an $x\in X$ such that $x<a_C$ and for all $y\in(\leftarrow,a_C)$ we have $y\le x$. Let $u=\sup\big(Y\cap(\leftarrow,x)_X\big)$; then $u\in Y$, so $u=b_D$ for some $D\in\mathcal{C}$. Clearly $D$ is the immediate predecessor of $C$ in $\mathcal{C}$. A similar argument shows that if $b_C\ne\sup Y$, then $C$ has an immediate successor in $\mathcal{C}$.

Fix $C_0\in\mathcal{C}$. Given $C_n\in\mathcal{C}$ for some $n\in\Bbb N$, either $C_n$ is the maximum element of $\mathcal{C}$, in which case the construction stops, or $C_n$ has an immediate successor in $\mathcal{C}$, and we define $C_{n+1}$ to be that successor. Suppose that the construction does not stop at any finite stage. Let $a=\sup\{a_n:n\in\Bbb N\}$. Then $a\notin\bigcup_{n\in\Bbb N}C_n$, so $a=a_D$ for some $D\in\mathcal{C}$, and this $D$ has no immediate predecessor in $\mathcal{C}$. This is impossible, so the construction must stop at some finite stage. A similar argument working to the left shows that $\mathcal{C}$ must in fact be finite.

Fix $C\in\mathcal{C}$. There are $U,V\in\mathcal{U}$ such that $a_C\in U$ and $b_C\in V$. Then $U,V\subseteq C$, so $U\sim V$, and there is a finite family $\{U_0,\dots,U_n\}\subseteq\mathcal{U}$ such that $U=U_0$, $V=U_n$, and $U_k\cap U_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$; clearly $\{U_k:k=0,\dots,n\}$ is a finite subfamily of $\mathcal{U}$ covering $C$, and it follows that $\mathcal{U}$ has a finite subcover.