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 $\textbf{int}(X \cup Y) \subset \textbf{int}(X) \cup \textbf{int}(Y)$. $\endgroup$ – Faraad Armwood May 28 '17 at 5:41
  • 3
    $\begingroup$ $E=(0,1)\cup(1,2)$ is a simpler counterexample. $\endgroup$ – Thomas Andrews May 28 '17 at 6:24

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 '17 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.