Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 1

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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

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.

$\Box$


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 $X$.

$\blacksquare$