# Does compact set always contain supremum and infimum without Heine Borel?

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?

• What do you mean by "sup and inf" in a general topological space? Jul 30 '16 at 23:50
• Might want to restrict your question to the order topology. Jul 30 '16 at 23:51
• @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 $\mathbb R$. Jul 30 '16 at 23:58
• @mathguy Oh, I don't know. The order topologies of well-ordered sets (ordinals) have compact sets which I think are kind of interesting.
– bof
Jul 31 '16 at 0:09
• 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.
– bof
Jul 31 '16 at 0:13