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Definitions:

Neighborhood: The neighborhood of a point $x$ is a subset of the topological space $X$ that contains an open set such that $x$ is in that open set.

Closure: The closure of a set $S$ is the intersection of all the closed subsets of the topological space $X$ which contain $S$ as a subset.

Closed: A set is closed if the complement of that set is open.

I want to prove that every open neighborhood of a point $x$ in the closure of a set $S$ contains a point in $S$.

If $x \in S$, then we're done. If $x \not\in S$, then $x \in \operatorname{cl}S \setminus S$. Since we don't know if $S$ is open, closed, both, or neither, I guess we have to do cases:

1) $S$ is closed: then $S = \operatorname{cl}S$, so we're done.

2) $S$ is open: then $S = \operatorname{Int}S$ and $\operatorname{cl}S \setminus S$ is closed. If $x \in \operatorname{cl}S \setminus S$, and there exists an open neighborhood of $x$.... I don't know how to show that a point in this neighborhood can also be in $S$.

3) $S$ is both open and closed: No idea how to start this one

4) $S$ is neither open nor closed: No idea how to start this one

mr eyeglasses
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3 Answers3

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If $S$ is open and closed, then you can use either argument $1$ or argument $2$, since argument $1$ does not say "if $S$ is closed, but not open".

That said, I think your separation into cases is not the optimal way of going about this. You are better of simply taking $x\in\mathrm{cl}(S)$ and a neighborhood $O$ of $x$ and show that $O\cap S\neq \emptyset$.

For that, you can assume that $O\cap S = \emptyset$, then notice how the complement of $O$ is a closed set which contains $S$, meaning that it also contains the closure of $S$. Can you come up with a contradiction now?

5xum
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  • I see that the complement of $O$ is a closed set (since $O$ is open, its complement is closed), but I don't see how the complement of $O$ necessarily contains $S$. For example if $x \in S \subset \mathrm{cl}(S)$, then the complement of a neighborhood of $x$ does not contain $S$. – mr eyeglasses Apr 01 '15 at 14:52
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    @ᴇʏᴇs Well, since $O\cap S=\emptyset$, this means that $O$ does not contain any elements of $S$, therefore, all elements of $S$ are contained in $O^c$. Or, more technically, knowing that $O\cap S=\emptyset$ and knowing that $S\cap X= S$ and $X=O\cup O^c$, you can see that $S\cap (O\cup O^c)=S$, and using distributivity, that means $(S\cap O)\cup (S\cap O^c)=S$. That means that $S\cap O^c=S$, or, in other words, $S\subseteq O^c$. – 5xum Apr 01 '15 at 15:30
  • Okay, so now I understand why the complement of $O$ is a a closed set that contains $S$, but why does it necessarily contain the closure of $S$ as well? Isn't it possible that an open set has a closed subset, so isn't it possible for $O$ to contain $\mathrm{cl}(S) - S$, so that the complement of $O$ only contains $S$, but not the entire closure? – mr eyeglasses Apr 01 '15 at 16:38
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    @ᴇʏᴇs By definition, the closure of $S$ is the intersection of all closed sets that include $S$, thus (because $O^c$ is closed and includes $S$), the closure of $S$ is the intersection of $O^c$ and several other closed sets, so $\mathrm{cl}(S) = O\cap C$ for some closed set $C$, meaning that $\mathrm{cl}(S)\subseteq O^c$ – 5xum Apr 01 '15 at 17:27
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We proceed by contraposition and prove: if a point $x$ has a neighbourhood (not necessarily open) that doesn't meet $S$, it is not in the closure of $S$.

Indeed, we may suppose this neighbourhood $V$ is open. Then, $F_X\smallsetminus V$ is a closed subset of $X$, and it contains $S$. Hence $x$ doesn't lie in all closed subsets of $X$, i.e. it doesn't lie in the closure of $S$.

Note. This proof is valid in any topological space, not only in metric spaces.

Bernard
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We have to show that $B(x,r)\cap S\neq \emptyset \forall r>0$

Suppose $B(x,r)\cap S= \emptyset $ for some $ r_1>0$

Then $S\subset X\setminus B(x,r_1)$ ; $X\setminus B(x,r_1)$ is a closed set not containing $x$.

But we have assumed that $x\in S_\alpha$ forall $\alpha$ where $S_\alpha$ is a closed set containing $x$

Thus arriving at a contradiction because we have got a closed set not containing $x$

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