4
$\begingroup$

For example, consider the set: $$\{2,\ 3,\ 5\} \subset \mathbb R.$$

This set has no limit points.

A closed set (also this) is a set which contains all of its limit points.

The set described above contains all of its $0$ limit points, therefore it is closed.

Is this reasoning correct? Can it be made rigorous more other than just re-writing it with quantifiers?

Thank you.

$\endgroup$
12
  • 6
    $\begingroup$ Yup, that's correct - and perfectly rigorous. $\endgroup$
    – Noah Schweber
    Commented Dec 18, 2017 at 22:11
  • 1
    $\begingroup$ It does not not contain all it's limit points ;) $\endgroup$
    – Pete Caradonna
    Commented Dec 18, 2017 at 22:12
  • 2
    $\begingroup$ Seen in a different way, the set would not be closed if there is a limit point not belonging to it. Since there is no limit point at all, this can't happen. Your argument is perfectly fine. $\endgroup$
    – egreg
    Commented Dec 18, 2017 at 22:13
  • 1
    $\begingroup$ Yes. And no, it can not be made more rigorous. (Unless you want to prove it has no limit points which ... is always a fun activity (for masochists)... but is not really required). $\endgroup$
    – fleablood
    Commented Dec 18, 2017 at 22:14
  • 1
    $\begingroup$ Pet Caradonna is winking but... to not be closed means it does not contain all its limit points. Which means there would have to be a limit point that is not contained in it. That is not the case as there are no limit points whatsoever. So it is impossible for it not to be closed. No winking. That's irrefutable, isn't it? $\endgroup$
    – fleablood
    Commented Dec 18, 2017 at 22:19

1 Answer 1

Reset to default
4
$\begingroup$

Right, the set is closed, for the exact reason you described.

$$\forall x\in \emptyset: P(x)$$ is vacuously true regardless of $P(x)$. This is perfectly rigorous reasoning (in the context of real analysis).

$\endgroup$
1
  • $\begingroup$ Thank you for that. Now I feel less guilty. $\endgroup$
    – ThisIsNotAnId
    Commented Dec 18, 2017 at 22:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .