# Compact sets in the lower topology on $\mathbb{R}$ have a minimum

Let $S$ be a subset of $\mathbb{R}$ be nonempty. Show that in the lower topology, $S$ is compact iff $S$ has minimum.

Note that the lower topology is not the lower limit topology.

I tried to prove that if $S$ is compact, then $S$ has minimum, but I didn't get anywhere.

My try:

$S$ is compact so for all open covers of $S$ there is a finite subcover of $S$. Consider the open cover $C=\{{]}a,+\infty{[}: a \in \mathbb{R} \}$ of $S$. So there are $a_1,\ldots, a_p$ such that $S \subseteq \bigcup_{i=1}^p {]}a_i,+\infty{[}$, which is equal to ${]}{\min} (a_1,\ldots,a_p),+\infty{[}$.

But this does not tell me that $S$ has a minimum. It just tells that there is a minorant.

Can anyone help? What am I doing wrong?

• That looks OK. Pick the smallest of the $a_i$; then $a_i$ is in $S$ and if $x$ is in $S$ then for some $j$ $x \in \bar{a_j}$, i.e. $x \geq a_j \geq a_i$. – HTFB Dec 14 '15 at 12:27

Your proof that $S$ is bounded below can be modified to show that it has a minimum.

Assume that your nonempty compact set $S$ does not have a minimum. Consider the family $\mathcal{U} = \{ ( x , + \infty ) : x \in S \}$ of open subsets of $\mathbb{R}$ with respect to the lower topology.

One can show that $\mathcal{U}$ covers $S$.

• For $y \in S$, since $y$ is not the minimum element of $S$ (as $S$ has no minimum) there is an $x \in S$ with $x < y$. But then $y \in ( x , + \infty ) \subseteq \bigcup \mathcal{U}$.

Since $S$ is compact there are $x_1 > \cdots > x_n$ in $S$ such that $$S \subseteq ( x_1 , + \infty ) \cup \cdots \cup ( x_n , + \infty ).$$ Since $x_1 > \cdots > x_n$ we have that $\bigcup_{i=1}^n ( x_i , + \infty ) = ( x_n , + \infty )$. But then $x_n \notin ( x_n , + \infty ) \supseteq S$, which contradicts the fact that $x_n \in S$!

Therefore the assumption that $S$ does not have a minimum cannot be true.

To show that every subset $S$ of $\mathbb{R}$ with a minimum is compact, note that if $\alpha = \min(S)$, then $S \subseteq U$ for every open set $U$ containing $\alpha$, and every open cover of $S$ must contain a set which contains $\alpha$.