On this post:

compact set always contains its supremum and infimum

People ask if compact set always contains its supremum and infimum. I know that it is true on $\mathbb{R}$ with the usual topology.

Pf: Let $K\subseteq \mathbb{R}$ be a nonempty compact set. Suppose not, by Heine Borel $K$ is closed, so the $\sup(K)$ lies in the open set. Take an open ball around the $\sup(K)$, then $\sup(K) -\epsilon$ is the new supremum, contradiction.

However, this requires Heine Borel. And in that post, people are basically saying, that since compactness = closed and bounded, the above proof always works, so compact set always contains supremum and infimum.

Does a compact set in general topological space necessarily contain sup and inf?

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    $\begingroup$ What do you mean by "sup and inf" in a general topological space? $\endgroup$ – Eric Wofsey Jul 30 '16 at 23:50
  • $\begingroup$ Might want to restrict your question to the order topology. $\endgroup$ – Hayden Jul 30 '16 at 23:51
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    $\begingroup$ @Hayden Most order topologies (other than those similar to $\mathbb R$) don't have many interesting compact sets though. The whole concept seems pretty useless outside of $\mathbb R$. $\endgroup$ – mathguy Jul 30 '16 at 23:58
  • $\begingroup$ @mathguy Oh, I don't know. The order topologies of well-ordered sets (ordinals) have compact sets which I think are kind of interesting. $\endgroup$ – bof Jul 31 '16 at 0:09
  • $\begingroup$ Let $X$ be3 a topological space, and let $\lt$ be a linear ordering of $X.$ If $\{x\in X:x\lt a\}$ is open for every $a\in X,$ then every compact nonempty subset of $X$ has a greatest element. Likewise, if $\{x\in X:x\gt a\}$ is open for every $a\in X,$ then every compact nonempty subset of $X$ has a least element. $\endgroup$ – bof Jul 31 '16 at 0:13

In a general topological space there is no order relation - and therefore no meaningful definition of sup and inf.

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    $\begingroup$ Really better as a comment. $\endgroup$ – Thomas Andrews Jul 31 '16 at 0:05
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    $\begingroup$ I initially posted it as a comment, but then I thought this is in fact "the" answer - pointing out that the question really doesn't make sense. What other answer is possible; that is, what WOULD constitute an answer rather than a comment? $\endgroup$ – mathguy Jul 31 '16 at 0:21

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