# Proving that the closure of a subset is the intersection of the closed subsets containing it

I'm having trouble with this exercise (from Gamelin and Greene Introduction to Topology):

Prove that the closure $\bar{Y}$ of a subset $Y$ of a metric space $X$ coincides with the intersection of all closed subsets of $X$ that contain $Y$.

My thoughts so far: if we let $V$ be the intersection of all closed subsets of $X$ containing $Y$, then we want to show that $V \subseteq \bar{Y}$ and $\bar{Y} \subseteq V$. Clearly $V$ is closed (because it's the intersection of closed sets), and $Y \subseteq V$. We know from our definition that the closure of $Y$ consists of [edit: the following is wrong, as Brian and Levon pointed out] all points for which there exists a positive $r$ such that $B(x;r) \cap Y \neq \emptyset$. So effectively we're trying to show that $x \in V \Leftrightarrow \exists r B(x;r) \cap Y \neq \emptyset$.

And yet somehow I find myself at a loss as to how to proceed. Could someone please offer a hint? Or better, ideas on what key skills or knowledge I might be lacking if I'm finding such simple exercises unduly difficult? Thank you.

ADDENDUM--- Oh, dear, it was awfully careless of me to have confused the quantifier in the definition of closure; sorry about that.

SECOND ADDENDUM--- Okay, I think I have it now. The set $\bar{Y}$ is closed and contains $Y$, therefore it was one of the sets that we intersected to get $V$, and therefore $V \subseteq \bar{Y}$. It remains to show that $\bar{Y} \subseteq V$: that is, if a point belongs to $\bar{Y}$, then it therefore belongs to $V$. But by contraposition, this is the same as saying that if a point doesn't belong to $V$, then it doesn't belong to $\bar{Y}$. If a point $x$ doesn't belong to $V$, then there must be at least one closed set (call it $W$) containing $Y$ that does not contain $x$. Because $W$ is closed, its complement $X \backslash W$ (which contains $x$) is open. But that means there's an open ball $B(x;r)$ in $X \backslash W$. Then because $Y$ is not in $X \backslash W$, we can say that there exists an $r$ such that $B(x;r) \cap Y = \emptyset$. But by the De Morgan law, this is just what it means for $x$ not to belong to $\bar{Y}$, which is quod erat demonstrandum.

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I am fairly certain to recall Arturo Magidin to have proved that in one of his answers, even somewhat recently. And I am even more certain that this was given at one point or another on the site. –  Asaf Karagila Aug 4 '11 at 20:32
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## 3 Answers

$\overline Y$ is one of the closed sets that you intersected to get $V$; use this to get one of the two inclusions ($V\subseteq\overline{Y}$ and $\overline{Y}\subseteq V$) that you need. To get the other inclusion, use the fact that if a point is not in the intersection of a family of sets, then there’s at least one set in that family that doesn’t contain it. You’ll also need to correct your definition of the closure: $x \in \overline{Y}$ if and only if $B(x,r)\cap Y \ne \varnothing$ for all $r>0$.

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Great hint; thank you! –  Zack M. Davis Aug 4 '11 at 22:52
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You are lacking the correct definition of closure. A point $x$ is a closure point of $Y$ if $\forall r$ we have $B(x;r) \cap Y \neq \emptyset$ (pay attention to $\forall$).

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I don't see where you are using the hypothesis that X is a metric space; an argument can be made for general topological spaces:

If you define the closure of a set B to be the smallest closed set C containing B (meaning that if $B\subset K\subset C$, K is not the closure of B), then the intersection $\cap V_i$, where $V_i$ contains B and $V_i$ is closed in your space, does it: assume there is a smaller (in the above sense) set B' containing B, then the intersection would equal B'.

In a metric space (X,d) , the closure of a set S consists of all points p in X with d(p,S)=0, i.e., the set of all points p such that for any r, B(p,r)$\cap$S is non-empty.

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