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


  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:

  1. Well orderedness is preserved under subsets

  2. Well ordered subsets must be countable

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    $\begingroup$ $(-\infty, 0]$ is closed, but it does not have a minimal element... $\endgroup$ Commented Jun 16 at 20:10
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    $\begingroup$ $[0,1)$ is not closed, but it does have a minimum element. $\endgroup$
    – Lee Mosher
    Commented Jun 16 at 20:13
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    $\begingroup$ 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. $\endgroup$
    – dialegou
    Commented Jun 16 at 20:17
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    $\begingroup$ also, being well ordered is not the same as having a minimum element. $\endgroup$
    – Carlyle
    Commented Jun 16 at 20:17
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    $\begingroup$ 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. $\endgroup$
    – Ulli
    Commented Jun 16 at 20:25

1 Answer 1


This question unfortunately does not have an interesting answer. It also doesn’t have an answer based on the topology (as shown by @Ulli)

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.

  • 1
    $\begingroup$ In other words, a subset of $\mathbb R$ has a least element if and only if it has a least element. $\endgroup$ Commented Jun 16 at 21:33
  • $\begingroup$ @HagenvonEitzen exactly. This (hopefully) makes it clear that there is nothing of interest in this question. $\endgroup$
    – Malady
    Commented Jun 16 at 21:51

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