# Compact Subspace of Linearly Ordered Space/Reverse Implication

## Theorem

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.

Let $Y \subseteq X$ be a non-empty subset of $X$.

Let the following hold:

- $(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
- $(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Then $Y$ is a compact subspace of $\struct {X, \tau}$.

## Proof 1

Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.

Let $\preceq'$ be the restriction of $\preceq$ to $Y$.

### Lemma

$\struct {Y, \preceq', \tau'}$ is a linearly ordered space.

### Proof

By definition of a generalized ordered space:

- $\tau'$ has a sub-basis consisting of upper and lower sets in $Y$.

To prove that $\struct {Y, \preceq', \tau'}$ is a linearly ordered space, we need to show that each element of this sub-basis is open in the $\preceq'$ order topology.

Let $U$ be a non-empty, $\tau'$-open upper set in $Y$.

If $U = \O$ or $U = Y$, then $U$ is open in the $\preceq'$-order topology by the definition of topology.

Suppose then that $\O \subsetneqq U \subsetneqq Y$.

Let $C = Y \setminus U$.

Then $C$ is non-empty and $\tau'$-closed.

$Y$ is $\tau'$-closed by Closed Set in Linearly Ordered Space.

Thus $C$ is also $\tau'$-closed.

$C$ has a supremum in $X$ by the premise.

Let $c = \sup_X C$

Since $C$ is $\tau'$-closed, Closed Set in Linearly Ordered Space shows that $c \in C$.

Since $c \in C = Y \setminus U$ and $U$ is an upper set in $Y$:

- $c \prec u$ for all $u \in U$.

Furthermore, if $y \in Y$ and $c \prec y$, then by the definition of supremum, $y \notin C$, so $y \in U$.

Thus:

- $U = {c^\succeq}_Y$

where ${c^\succeq}_Y$ denotes the upper closure of $c$ in $Y$.

So $U$ is open in the $\preceq'$-order topology for $Y$.

The same argument shows that $\tau'$-open lower sets in $Y$ are open in the $\preceq'$-order topology.

Thus $\tau'$ is the $\preceq'$-order topology.

$\Box$

The premises immediately show that $\struct {Y, \preceq'}$ is a complete lattice.

By Complete Linearly Ordered Space is Compact, $Y$ is a compact subspace of $X$.

$\blacksquare$

## Proof 2

Let $\FF$ be an ultrafilter on $Y$.

For $S \in \FF$, let $\map f S = \inf S$.

Let $p = \sup \map f \FF$.

Then $\FF$ converges to $p$:

### Upward rays

Let $a \in X$ with $a \prec p$.

Since $\FF$ is an ultrafilter, either $Y \cap {\uparrow} a \in \FF$ or $Y \cap {\bar \downarrow} a \in \FF$.

Aiming for a contradiction, suppose that $Y \cap {\bar \downarrow} a \in \FF$.

For each $S \in \FF$:

- $S \cap {\bar \downarrow} a \in \FF$ because an ultrafilter is a filter.

- $S \cap {\bar \downarrow} a \ne \O$ because a filter on a set is proper.

By applying the definition of supremum to $p$, it follows that there exists an $S \in \FF$ such that $a \prec \inf S$.

By the definition of infimum:

- $S \cap {\bar \downarrow} a = \O$

which is a contradiction.

Thus $Y \cap {\uparrow} a \in \FF$.

### Downward rays

Let $b \in X$ with $p \prec b$.

Either $Y \cap {\downarrow} b \in \FF$ or $Y \cap {\bar \uparrow} b \in \FF$.

Aiming for a contradiction, suppose that $Y \cap {\bar \uparrow} b \in \FF$.

Let $b' = \map \inf {Y \cap {\bar \uparrow} b}$.

We have that $b$ is a lower bound of $Y \cap {\bar \uparrow} b$

So by the definition of infimum:

- $b \preceq b'$

Since $p \prec b$ and $b \preceq b'$, $p \prec b'$ by Extended Transitivity.

By the definition of $b'$ and the definition of $f$:

- $b' \in \map f \FF$

But this contradicts the fact that $p$ is the supremum, and hence an upper bound, of $\map f \FF$.

$\Box$

By the definition of the order topology, the upward and downward rays containing each point form a neighborhood sub-basis for that point.

Thus by the Neighborhood Sub-Basis Criterion for Filter Convergence, $\FF$ converges.

Since every ultrafilter on $Y$ converges, $Y$ is compact by Equivalent Definitions of Compactness.

#### Boolean Prime Ideal Theorem

This proof depends on the Boolean Prime Ideal Theorem (BPI), by way of Equivalent Definitions of Compactness.

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.