Heine-Borel Theorem/Dedekind Complete Space

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Theorem

Let $T = \left({X, \preceq, \tau}\right)$ 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, $\left({Y, \preceq {\restriction_Y} }\right)$ is a complete lattice.



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 $\sup S, \inf S \in Y$.

Suppose for the sake of contradiction that $\sup S \notin Y$.

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.

$\blacksquare$