by Tob Ernack
Last Updated January 12, 2018 22:20 PM

If $X$ is a totally ordered set and we give it the order topology, it is not true that any nonempty closed and bounded subset of $X$ has a smallest element.

For example if $X = \mathbb{Q}$ we can have the following set:

$$\bigcap\limits_{n=1}^{\infty}\left[\left(1 + \frac{1}{n}\right)^n, 3\right]$$

This is an intersection of closed sets and so it is closed, and it is bounded below by $2$ and above by $3$. Over $\mathbb{R}$ the set can be found to be $[e, 3]$. But there is no smallest rational number in this interval.

Now what happens if we additionally require that the subset is open in $X$? Can we now guarantee that the subset has a smallest element?

Note that in $\mathbb{Q}$ the only clopen subsets are $\emptyset$ and $\mathbb{Q}$ and none of these satisfy the requirements of the question so the claim is vacuously true here.

An example of a nontrivial clopen subset would be $\{0\}$ in the topological space $X = [-2, -1] \cup \{0\} \cup [1, 2]$ with order topology.

It is open because it is an open interval $\{x \in X: -1 \lt x \lt 1\}$, and it is closed because its complement is a union of open rays $[-2, -1] \cup [1, 2] = \{x \in X: x \lt 0\} \cup \{x \in X: x \gt 0\}$.

The set $\{0\}$ is bounded and has $0$ as a smallest element.

Pick $X = [-2, -1] \cup \{0\} \cup (1, 2)$ with the order topology. Then the set $(1,2)$ is clopen but has no smallest element.

It's not true that $\Bbb Q$ has only two clopen subsets. That would say $\Bbb Q$ was connected, which it's not.

Let $S=\{r\in\Bbb Q: r^2<2\}$. Then $S$ is a clopen subset of $\Bbb Q$, since it's open and its complement is $\{r\in\Bbb Q:r^2>2\}$, which is also open. And $S$ is bounded but has no smallest element.

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