Heine-Borel Theorem/Dedekind Complete Space

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Theorem

Let $T = \struct {X, \preceq, \tau}$ be a Dedekind-complete linearly ordered space.

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

Then $Y$ is compact if and only if $Y$ is closed and bounded in $T$.

Proof

Sufficient Condition

Let $Y$ be a compact subspace of $T$.

From:

Order Topology is Hausdorff

Compact Subspace of Hausdorff Space is Closed

it follows that $Y$ is closed in $T$.

From Compact Subspace of Linearly Ordered Space: Lemma 1, $\struct {Y, \preceq {\restriction_Y} }$ is a complete lattice.

This needs considerable tedious hard slog to complete it. In particular: Not much to do, just need to show that a complete lattice in this context is bounded. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Hence $Y$ is closed and bounded in $T$.

Necessary Condition

Let $Y$ be a closed and bounded subspace of $T$.

Let $S$ be a non-empty subset of $Y$.

Since $Y$ is bounded and $S \subseteq Y$, $S$ is bounded.

Since $X$ is Dedekind complete, $S$ has a supremum and infimum in $X$.

We will show that $\sup S, \inf S \in Y$.

Aiming for a contradiction, suppose $\sup S \notin Y$.

By Closed Set in Linearly Ordered Space, $b$ is not a supremum of $S$, a contradiction.

Thus $\sup S \in Y$.

A similar argument shows that $\inf S \in Y$.

Thus by Compact Subspace of Linearly Ordered Space, $Y$ is compact.

$\blacksquare$