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Let $(X,d)$ be a metric space and let $x \in X$. Prove that $\{x\}$ is closed.

Defintion: $F\subseteq X$ is closed (in $(X,d)$) if $\overline{F}=F.$

We have to prove that it is open to show that it is closed. How is the set $X$ closed?

Keeran Brabazon
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Ruth Gutierrez
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2 Answers2

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I guess you mean rather "we have to prove that the complement of $\{x\}$ is open to show that $\{x\}$ is closed."

That would be a good approach.

Given a point $y$ not equal to $x$, can you find an open ball around $y$ that doesn't contain $x$? That would show the complement is open.

rschwieb
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  • Can I say: Let {x}=E. The complement of E={xeX: x is not in E}. If E is closed then the complement of E is open. So let the open set equal to the complement of E, so then E equals the complement of the open set. Conversely, if E equals the complement of the open set, then the complement of E equals to the open set, which is open. Thus E or {x} is close. Can i say this? – Ruth Gutierrez Sep 19 '13 at 19:49
  • @RuthGutierrez No, you have gone around in a circle and the conclusion isn't right. By definition, $A$ is open iff its complement is closed. You can say "${x}$ is closed iff $X\setminus{x}$ is open," but then the burden is still on you to prove that $X\setminus{x}$ is open using the metric. Did you try the approach with the open ball that I outlined? – rschwieb Sep 19 '13 at 19:57
  • I'm not too sure if this is the idea you had in mind. Does this work out? For all x,y in X, we can find two open sets U, Vy such that x is in U, y is in Vy and U intersecting with Vy is empty. Let V = union over all y that is not equal to x of Vy. Then V is open since arbitrary union of open sets is open. Also, V = X{x}. Hence, {x} is closed since its complement is open. – Ruth Gutierrez Sep 19 '13 at 20:04
  • @RuthGutierrez Well almost, except you don't mean "for all $x$" since you fixed $x$ in the problem statement. It's true though that for all $y$ you can separate $x$ from $y$ with disjoint open sets, and the rest of what you said is true. – rschwieb Sep 19 '13 at 20:10
  • Great! Thank you so much :) – Ruth Gutierrez Sep 19 '13 at 20:37
  • Consider the subspace $A={x}$ of $X$. Since $d$ is continuous, the restriction of $d$ on $A^2$ is continuous, too. Convince yourself that $A$ is the preimage of the closed set ${0}=[0]$. – Michael Hoppe Sep 20 '13 at 09:41
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You can directly determine the set of all limit points of a singleton set $\{x\}$. It is empty! Thus, if $F=\{x\}$, then $\overline F=F\cup F'=F\cup\emptyset=F$, where I use $F'$ to mean the set of all limit points of F.

Karl Kroningfeld
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