# Statement about topology of subsets of a metric space.

1. Dec 10, 2013

### mahler1

The problem statement, all variables and given/known data.

Prove that a closed subset in a metric space $(X,d)$ is the boundary of an open subset if and only if it has empty interior.

The attempt at a solution.
I got stuck in both implications:
$\implies$ Suppose $F$ is a closed subspace with $F=\partial S$ for some open subset $S$ of $(X,d)$. I want to show that $F^°=\emptyset$. So, let $x \in F^°$. Then, there is $\delta$ : $B(x,\delta) \subset F$. But then, $B(x,\delta) \subset \partial S$, which means there is $y \in B(x,\delta) \cap S^c$ and there is $z \in B(x,\delta) \cap S$. I couldn't go farther than this.

For the other implication, I need some help too. If $F^°$ is empty, then, for every $x \in F$ and every $\delta>0$, $B(x,\delta) \cap F^c \neq \emptyset$. As $F$ is closed, $F^c$ is open. I want to show that $F=\partial F^c$, we've just seen that $F \subset \partial F^c$. Now, let $x \in \partial F^c$, then, for every $n \in \mathbb N$, there is $y_n \in B(x, \dfrac{1}{n}) \cap F$. But this means $x$ is a limit point of $F$, as $F$ is closed $\implies$ $x \in F$. I've proved $\partial F^c \subset F$. This proves
$F=\partial F^c$, which means $F$ is the boundary of an open subset.

Could anyone suggest me how to prove the forward implication? Is the other one correct?

2. Dec 11, 2013

### pasmith

If $S$ is open then it contains no boundary points, so $F \subset X \setminus S$. Thus $F^{\circ} \subset (X \setminus S)^{\circ}$. Is an interior point of $X \setminus S$ a limit point of $S$?

3. Dec 11, 2013

### mahler1

I don't see why an interior point of $X \setminus S$ would have to be a limit point of $S$. For example, if I consider $S=(0,1)$, then $10 \in {S^c}^{\circ}$ but $10$ is not a limit point of $S$.

Last edited: Dec 11, 2013
4. Dec 11, 2013

### pasmith

So is the answer to the question "Is an interior point of $X \setminus S$ a limit point of $S$?" going to be "sometimes" or "never"?

5. Dec 12, 2013

### mahler1

Sorry, I was getting it all wrong. I think I've understood what you were trying to say:

As you've said, $F \subset X \setminus S$. Then, $F^{\circ} \subset {(X \setminus S)}^{\circ}$. Suppose there exists $x \in F^{\circ}$. By hypothesis, $F=\partial S$, thus $x$ is a limit point of $S$. Now, $x \in {(X \setminus S)}^{\circ}$, so there is $\delta>0$: $B(x,\delta) \subset X \setminus S$. But, as $x$ is a limit point of $S$, there is $y \in S \cap B(x,\delta)$, so $y \in S \cap (X \setminus S)$, which is clearly absurd. From here it follows $F^{\circ}=\emptyset$.

The answer is going to be "never".

Thanks!