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.
Best Answer
$\overline Y$ 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 $r>0$.