# When does a set of real numbers have a minimal element?

I Believe the answer is when the set is closed in the sense of standard topology (we exclude $$\mathbb{R}$$) itself.

Examples:

1. Point sets have a minimum, and they are closed.

2. Closed interval also have a minimum, which is simply the lower bound $$[a,b]->a$$

If my guess is right, how do I prove it, else what is the correct characterization?

I'm guessing this has something to do with well orderedness. What I got so far from MSE:

• $(-\infty, 0]$ is closed, but it does not have a minimal element... Commented Jun 16 at 20:10
• $[0,1)$ is not closed, but it does have a minimum element. Commented Jun 16 at 20:13
• Suppose that $X\subseteq (0,\infty)$, then $\{0\}\cup X$ has a minimum and you can choose $X$ to be as complicated as you want it to be. Commented Jun 16 at 20:17
• also, being well ordered is not the same as having a minimum element. Commented Jun 16 at 20:17
• There is no topological characterization: $[0, \infty)$ has a minimum element, $(-\infty), 0]$ doesn't, but $-$ is a homeomorphism of $\mathbb R$, which maps one to the other. Similar for well-ordered.
– Ulli
Commented Jun 16 at 20:25

Here is the nicest description I could come up with : $$\{\{x\} \cup Y_x | x\in \mathbb{R}, Y_x \subset (x,\infty)\}$$ is the collection of subsets of $$\mathbb{R}$$ with a least element.
• In other words, a subset of $\mathbb R$ has a least element if and only if it has a least element. Commented Jun 16 at 21:33