Heine-Borel Theorem/Dedekind Complete Space
If you are fairly certain that there are none, please remove {{proofread}} from the code.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can remove this instance of {{MissingLinks}} from the code.
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:
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.
Not much to do, just need to show that a complete lattice in this context is bounded.
To discuss it in more detail, feel free to use the talk page.
When this has been done, {{Finish}} should be removed 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 $\sup S, \inf S \in Y$.
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.
$\blacksquare$