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Let $X$ be a nonempty totally ordered set which doesn't have a minimum or maximum and let $T$ be order topology on $X$. Let $ K \subset X $ be a nonempty subset of $X$. If $K$ is compact, prove that $K$ has a minimum and maximum.

I tried by contradiciton. Suppose that $K$ has no maximum. Then $ <- \infty, k> k \in K$ is an open cover of $K$ which has no finite subcover. So K cannit be compact. Is this correct?

user15269
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  • @InterstellarProbe Of course it doesn't work in a space with a maximal element. In the setup of the OP, we assume there is no maximal element, then for each $x\in K$, there exists $k\in K$ with $k>x$, hence with $x\in(-\infty,k)$. – Hagen von Eitzen Jun 07 '19 at 19:41

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Yes, your proof is correct, assuming that $\langle -\infty,k\rangle=\{x\in K:x<k\}$.

Indeed, if $x\in K$, $x$ is not the maximum, so there exists $k\in K$ with $x<k$; thus $x\in\langle -\infty,k\rangle$ and therefore $$ \{\langle -\infty,k\rangle:k\in K\} $$ is an open cover of $K$. It has no finite subcover, because $$ \langle -\infty,k_1\rangle\cup\langle -\infty,k_2\rangle\cup\dots\cup \langle -\infty,k_n\rangle=\langle -\infty,\max\{k_1,\dots,k_n\}\rangle $$ which cannot be the whole of $K$, as it doesn't contain the upper bound.

egreg
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