I have a problem asking me to prove that every finite subset of $\mathbb{R}^n$ is closed. However, consider a punctured disk $D:=\{x\in\mathbb{R}^n:\lVert x\rVert\le 1\}\backslash\vec{0}$. Then we can show that $D$ is not open by considering an open ball on its boundary. But can't we also show that $D$ is not closed by considering the sequence $\{(1/n, 1/n, …, 1/n|)\}$, which converges to $\vec{0}$, thus proving that $\vec{0}$ is a limit point of $D$? Then $D$ would neither be open nor closed ("clopen"). Isn't it true that clopen sets exist in $\mathbb{R}^n$?
UPD: Oh, I see, I got it. A finite subset is a subset containing finitely many points, not a bounded one…
Best Answer
If you are considering $\mathbb{R}^n$ 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 $\mathbb{R}^n$ for this problem. Every metric space has this property.