Prove that if $\mathcal{O}$ is open, then the interior of $\bar{\mathcal{O}} – \mathcal{O}$ is empty

general-topologymetric-spaces

I am trying to show the following:

Let $\mathcal{O}$ be an open subset of a metric space $X$. Show that
$\text{int} \, (\bar{\mathcal{O}} – \mathcal{O}) = \emptyset$.

Here, for $E \subset X$, $\text{int} \, E$ denotes the interior of a set and $\bar{E}$ denotes its closure in $X$.

My attempt:

For the sake of contradiction, suppose $\bar{\mathcal{O}} – \mathcal{O}$ has non-empty interior, so there is $x \in \text{int} \, (\bar{\mathcal{O}} – \mathcal{O})$. This means there is $r > 0$ such that $B(x, r) \subset \bar{\mathcal{O}} – \mathcal{O}$. However, since $x$ is a point of closure of $\mathcal{O}$, for every $t > 0$, $B(x, t) \cap \mathcal{O} \neq \emptyset$. This contradicts with $B(x, r) \subset \bar{\mathcal{O}} – \mathcal{O}$. Hence $\text{int} \, (\bar{\mathcal{O}} – \mathcal{O})$ is empty. $\blacksquare$

The problem I have with my attempt is that the assumption of $\mathcal{O}$ being open is not used. The argument seems to work for any subset of $X$, or in other words, it suggests that if $E$ is any subset of $X$, then $\text{int} \, (\bar{E} – E) = \emptyset$. I do not think this is true, but I have not came up with a counter-example to debunk it.

Please help by:

  1. spotting any mistakes in my attempt;
  2. answering the question " for $E \subset X$, does $\text{int} \, (\bar{E} – E) = \emptyset$ holds for all $E \subset X$ "?

Any help is appreciated.

Here are the definitions I used: For $E \subset X$, $x \in X$ and $r > 0$,

  • $B(x, r)$: open ball centered at $x$ with radius $r$.
  • $\text{int} \, E$: interior of $E$, the points in $E$ which there is an open ball centered at $x$ covered by $E$, i.e.
    $$
    \text{int} \, E = \{ x \in E : \exists r > 0 \, , B(x, r) \subset E \} \ .
    $$
  • $\bar{E}$: closure of $E$, the points in $X$ which every open set containing $x$ also contains a point in $E$, i.e.
    $$
    \bar{E} = \{ x \in X: \forall \text{ open set } O \subset X \text{ such that }x \in O \, , O \cap E \neq \emptyset\} \ .
    $$

Best Answer

Your proof is indeed correct for general $O.$ Nice! I'll provide you with a different version, because, why not.

Let $V$ be the interior of $\bar{O} \setminus O$. Then $V$ is open and a subset of $\bar{O}$ and since the interior of $\bar{O}$ is the same as that of $O,$ namely, $\mathring{O},$ and the interior of a set is the largest open subset, we get $V\subseteq \mathring{O} \subseteq O.$

On the one hand, we have $V \subseteq \bar{O} \setminus O.$ On the other, we have $V\subseteq O.$ Therefore $V = \emptyset.$

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