Is every finite subset of Rn\mathbb{R}^n closed?

I have a problem asking me to prove that every finite subset of Rn\mathbb{R}^n is closed. However, consider a punctured disk D:={x∈Rn:‖x‖≤1}∖→0D:=\{x\in\mathbb{R}^n:\lVert x\rVert\le 1\}\backslash\vec{0}. Then we can show that DD is not open by considering an open ball on its boundary. But can’t we also show that DD is not closed by considering the sequence \{(1/n, 1/n, …, 1/n|)\}\{(1/n, 1/n, …, 1/n|)\}, which converges to \vec{0}\vec{0}, thus proving that \vec{0}\vec{0} is a limit point of DD? Then DD would neither be open nor closed (“clopen”). Isn’t it true that clopen sets exist in \mathbb{R}^n\mathbb{R}^n?

UPD: Oh, I see, I got it. A finite subset is a subset containing finitely many points, not a bounded one…

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Yes, “finite” and “bounded” are quite different things 🙂
– arkeet
Oct 20 at 22:26

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You seem to have got it. Just one small note: there is no common word for a set that is neither open nor closed. The word “clopen” refers to a set that is both open and closed. In \mathbb R^n\mathbb R^n these are only \emptyset\emptyset and \mathbb R^n\mathbb R^n.
– Mees de Vries
Oct 20 at 22:27

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2 Answers
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It is true that every finite subset of \mathbb{R}^n\mathbb{R}^n is closed. (If a sequence in a finite set converges, then the limit must be one of the points, because you can choose \epsilon\epsilon to be smaller than half the distance between any two points.) This holds for any metric space.

But that fact seems unrelated to DD. It is true that DD is not closed and not open. (But clopen is a different thing: a set is clopen iff it is both closed and open, so it is not clopen if either it is not closed or not open. The only two clopen subsets of \mathbb{R}^n\mathbb{R}^n are \varnothing\varnothing and \mathbb{R}^n\mathbb{R}^n, because \mathbb{R}^n\mathbb{R}^n is connected.)

If you are considering \mathbb{R}^n\mathbb{R}^n with the topology induced by the Euclidean metric then all singletons are closed i.e any finite set is a finite union of closed sets (singletons). Also, there is nothing special about \mathbb{R}^n\mathbb{R}^n for this problem. Every metric space has this property.