The question asks:

Do $E$ and $\bar E$ always have same interiors?

Here, $\bar E$ denotes the closure of $E$. That is, $\bar E = E \cup E'$, where $E'$ denotes the set of limit points of $E$. Also, let $E^\circ$ denote the interior of $E$.

My attempt at proof:

Answer. Yes

Clearly, $E^\circ \subseteq (\bar E)^\circ$. Therefore, we need to prove only that $(\bar E)^\circ \subseteq E^\circ$.

If $(\bar E)^\circ$ is empty, then $E^\circ$ is also empty. Therefore, assume that $(\bar E)^\circ$ is non-empty.

Consider a point $p \in (\overline E)^\circ$.

For some $r > 0$, we have that for any $0 < r' \le r$, $N_{r'}(p) \subseteq \bar E$. Here, $N_a(p)$ denotes the (open) neighborhood of $p$.
We need to show that $p \in E^\circ$.
That is, there exists some $\rho > 0$ such that $N_\rho(p) \subseteq E$.

Therefore, assume for contradiction that for all $\rho > 0$, there exists a point $p' \in N_\rho(p)$ such that $p' \not \in E$.
In particular, for any $0 < \rho \le r$, there exists a $p' \in N_\rho(p)$, but $p' \not \in E$.
Therefore, $N_\rho(p) \subseteq E'$. In particular, $p \in E'$, i.e. $p$ is a limit point of $E'$.

However, this is a contradiction because for every neighborhood of a limit point of $E$ must contain a point of $E$.

But the result is not true as shown by the following example.
$E = \mathbb Q \subset \mathbb R$. Here, $\mathbb Q^\circ = \emptyset$, but $\bar{\mathbb{Q}} = \mathbb R$.

So what is wrong with my proof? I thought for a while (even considering the same example!), and I can't find my mistake.

  • $\begingroup$ I didn't really your proof but I think this logic flows better: You have $\bar{E} = E \sqcup E'$ i.e the union is actually a disjoint one. Hence, if $p \in \textbf{int}(E \sqcup E')$ then $p \in \textbf{int}(E)$ since $E'$ is closed and using the fact that $\textbf{int}(X \cup Y) \subset \textbf{int}(X) \cup \textbf{int}(Y)$. $\endgroup$ May 28, 2017 at 5:41
  • 3
    $\begingroup$ $E=(0,1)\cup(1,2)$ is a simpler counterexample. $\endgroup$ May 28, 2017 at 6:24

1 Answer 1


In particular, for any $0 < \rho \le r$, there exists a $p' \in N_\rho(p)$, but $p' \not \in E$. Therefore, $N_\rho(p) \subseteq E'$.

The second sentence does not follow. You know that there exists some point $p'\in N_\rho(p)$ which is not in $E$, but that doesn't mean every point of $N_\rho(p)$ is not in $E$, which is what $N_\rho(p) \subseteq E'$ says. Maybe some points of $N_\rho(p)$ are in $E$ and other points are not. Indeed, this is exactly what happens in the case $E=\mathbb{Q}$: any open ball contains both points of $\mathbb{Q}$ and points which are not in $\mathbb{Q}$.

  • $\begingroup$ Of course! How silly of me. Thanks for the answer. :) $\endgroup$
    – taninamdar
    May 28, 2017 at 5:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .