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In a euclidean space $\mathbb{R}^k$, is the set consisting of a single point open, closed, neither, or both?

I would say that a set $E$ consisting of a single point $p$ doesn't have any limit points, so $E$ contains all of its limit points and is therefore closed. But it might be open, too, since a ball of radius zero around $p$ is a subset of $E$. When using balls to define interior points, do balls have to have radius greater than zero?

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    $\begingroup$ Do you have any thoughts? In particular proving it isn't open should be quite obvious $\endgroup$
    – HBeel
    Mar 29, 2015 at 20:12
  • $\begingroup$ @HBeel Proving it isn't open would not say about closed or not $\endgroup$ Jan 22, 2022 at 15:45

2 Answers 2

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One point sets are closed in $\mathbb{R}^n$. The only closed and open sets are $\emptyset,\mathbb{R}^n$.

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    $\begingroup$ why is it not vacously open? $\endgroup$ Jun 11, 2018 at 0:10
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    $\begingroup$ @CharlieParker - You can clearly see its closed, as $\cap_{n \geq 1} [x-1/n, x+1/n] = \{x\}$ is a countable intersection of closed sets. As for it not also being open, note that its complement is not closed -- $x$ is a limit point. Alternatively, if you know about connectedness, it is also not hard to prove that a topological space is connected if and only if the only sets which are both closed and open are the empty set and whole set: proofwiki.org/wiki/Connected_iff_no_Proper_Clopen_Sets. $\endgroup$
    – Batman
    Jun 11, 2018 at 1:49
  • $\begingroup$ I confused closed and open for a second. I think I see why its clearly closed (vacuously?). Since a single point doesn't have a limit point of course, then its closed. Sorry! $\endgroup$ Jun 14, 2018 at 1:01
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    $\begingroup$ In any T1 space, one point sets are closed. Metric implies T1. $\endgroup$
    – Batman
    Jun 15, 2018 at 3:01
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Balls must have a radius greater than zero or else the definition would not be very useful, since that would mean any point is an interior point. Any neighborhood of radius greater than zero would include more than one point, so it cannot be a subset of E. Hence E cannot not be open.

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