# $p$ is in $\operatorname{closure}(S)$ $\Leftrightarrow$ any ball centred at $p$ contains some point of $S$

I want to prove that $p$ is in $\mathrm{closure}(S)$ if and only if any ball centered at $p$ contains some point(s) of $S$. Where S is some subset of underlying set E of metric space (E,d). The closure of $S$ is defined the intersection of all closed subsets of E (that is, closed sets) in metric space (E,d) containing $S$.

I am really not sure how to even start on this. I am studying analysis on my own and don't really have anyone to ask for help. I think this ought to be provable without involving boundary points because in the book I am working from, boundary points have not yet been defined (interior points have been defined at this point FYI).

The book i am using is Introduction to Analysis by Maxwell Rosenlicht, Dover 1985. the question is 16 c) from chapter 3.

• $\Rightarrow\quad$ by contradiction If $p\in S$ the result is clear. Now assume $p\not\in S$ and that all ball $B$ centred at $p$ doesn't contain any point of $S$ then $B^c$ is closed containing $S$ and not $p$ so their intersection doesn't contain $p$ as well. Contradiction.
• $\Leftarrow\quad$ by contraposition Assume $p$ isn't in the closure of $S$ then there's a closed subset $C$ containing $S$ and $p\not\in C$ then $C^c$ is an open subset contains $p$ and doesn't contain any point of $C$, so take a ball subset of $C^c$ and centred at $p$.
• Spot on, dear Sami! – Namaste Mar 5 '14 at 13:17

To prove that P is in the closure of S, use the fact that any ball centered at p contains some point in S. Use this to build a sequence of points as you shrink the radius of the ball. Now you have a sequence of points in S who's limit is P. That should line up with your definition of a closed set. Something similar should get you the other direction.

If $\varepsilon>0$ and $B=\left\{ x\mid d\left(p,x\right)<\varepsilon\right\}$ with $S\cap B=\emptyset$ then $B^{c}$ is a closed set with $S\subset B^{c}$ so that $\overline{S}\subset B^{c}$. Consequently $p\notin\overline{S}$.

If conversely $p\notin\overline{S}$ then $p\in\overline{S}^{c}$ wich is an open set. Then also $\left\{ x\mid d\left(p,x\right)<\varepsilon\right\} \subset\overline{S}^{c}$ if $\varepsilon>0$ is taken small enough.

Here $\overline{S}$ denotes the closure of $S$.

It's easier to reason with balls in relation with open sets, since the definition of an open set is that it is one that contains an open ball around each of its points. So let us prove the contrapositive equivalence: $p$ is in the complement of the closure of $S$ iff there exists a ball centered at $p$ which contains no point of $S$.

I use the notation $\bar S$ for the closure of $S$. Since $\bar S$ is an open set, its complement $U$ is an open set. We need to prove that $p \in U$ iff there exists a ball centered at $p$ which contains no point of $S$.

If $p \in U$, then there exists an open ball centered at $p$ which is wholly included in $U$. This ball contains no point of $\bar S$, so it contains no point of $S$.

Conversely, suppose that there is a ball $B$ centered at $p$ which contains no point of $S$. Since $\bar S$ is the smallest closed set that includes $S$, $U$ is the largest open set that does not intersect $S$ (in other words, the complement of the closure is the interior of the complement). In particular, the open ball with the same center and radius as $B$ is a subset of $U$. In particular, $p \in U$.