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I'm guessing here that $U$ would have to be closed, especially since for example the proof of the theorem that the union of two closed sets is closed is also valid if one of the sets is $U$. Still, I'd like to make sure my approach is correct:

Since p is the single point in U, that means $\exists r>0, B_r(p)\subset U$ can never be true, so $p$ is not an interior point, which means it has to be a boundary point, which means since $p$ is the only point in $U$, $U$ contains all its boundary points which means $U$ is closed.

Asaf Karagila
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  • The arbitrary intersection of closed sets is closed. Can you write $p$ as such an intersection? – Listing Dec 07 '13 at 10:06
  • The singleton ${x}$ can be an open set though. In fact, this happens if, and only if, the metric space itself consists of just one point. – Ittay Weiss Dec 07 '13 at 10:39
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    @IttayWeiss it is not necessary that the metric space consists of a single point. Every isolated point has that property. – Emanuele Paolini Dec 07 '13 at 11:33
  • @EmanuelePaolini yes, of course. The lack of context in the question made me think of continua only. – Ittay Weiss Dec 07 '13 at 19:34

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I assume you're in a general metric space $(X,d)$. Consider {x} as a subset of $X$, and consider a point $y \neq x $ in $X$ . What can you say about d(x,y) (is it larger than 0)? If so, can you construct a ball $B(y,r)$, for all $y \neq x $ that does not intersect the set {$x $}? (Remember that the complement of a closed set is open.)

user99680
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let (X,d) be a metric space. we show X-{p} is open. let x belong to X-{p} and define r=d(x,p) then B(x, r/2) is a subset of X-{p}. in actually we show that each point in X-{p} is interior and complement of {p} is open.this means {p} is close.

Sajjad Beyranvand
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