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I am trying to solve the following problem.

Let $E \subset \mathbb{R}^2$ be open. Prove that $\mathring{\overline{E}} = E$ does not hold in general.

I haven't been able to think of a counterexample in $\mathbb{R}^2$. In $\mathbb{R}$, I can take $E = (1,2) \cup (2,3)$. As $E$ is a union of open sets, it is open. It's closure is then $[1,3]$, but the interior of $[1,3]$ is $(1,3) \neq E$ .

I can't figure out how to generalize this argument to $\mathbb{R}^2$.

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  • $\begingroup$ Take two open squares that "touch" themselves along an edge. The edge will act like $\{2\}$ in your 1-dimensionnal example. $\endgroup$
    – zwim
    Nov 8, 2021 at 2:38
  • $\begingroup$ @TheoBendit I understand why that's the case, but I'm struggling to apply this hint. $\endgroup$
    – Brad G.
    Nov 8, 2021 at 2:38
  • $\begingroup$ @BradG. Sorry, I missed the assumption that $E$ is open. My hint was indeed difficult (to impossible) to apply! $\endgroup$ Nov 8, 2021 at 2:44
  • $\begingroup$ Take $E = \Bbb R^2\setminus A$ where $A$ is any closed set with empty interior. $\endgroup$ Nov 8, 2021 at 6:43

2 Answers 2

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An alternate approach which you might find easier, if you can't convince yourself that Nightflight's $E$ is open.

Hint: Take $E = \Bbb{R}^2\setminus\{(0,0)\}.$ Convince yourself $E$ is open. What is $\overline{E}$?

Subhint for proving $E$ is open:

What is the complement of $E$? What do you know about the complement of open/closed sets?

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  • $\begingroup$ This example makes sense to me. $\{(0,0)\}$ is closed, so $E$ is open. Then $\overline{E} = \mathbb{R}^2$, the interior of which is $\mathbb{R}^2$. The only remaining question I have is, how would I prove that $\{(0,0)\}$ is closed? It's easy in $\mathbb{R}$ to show that $\{0\}$ is not open by taking complements, but I don't know how to prove it in $\mathbb{R}^2$. $\endgroup$
    – Brad G.
    Nov 8, 2021 at 2:59
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    $\begingroup$ If you know the sequential version of closedness, you can just observe that every (convergent) sequence taken from $\{(0, 0)\}$ is constantly $(0, 0)$, and hence converges to $(0, 0) \in \{(0, 0)\}$. Otherwise, you need to show $E$ is open directly; for $(x, y) \in E$, take the open ball whose radius is $\|(x, y)\|$, and prove that it is contained in $E$. $\endgroup$ Nov 8, 2021 at 3:52
  • $\begingroup$ @BradG. Well, what is your definition of closed? $\endgroup$
    – Stahl
    Nov 8, 2021 at 5:13
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    $\begingroup$ @BradG. Any set of the form $\{x\}$ in a metric space is closed as its complement is the open set $\bigcup_{y \neq x} B(y, d(x,y))$, a union of open balls, hence open. $\endgroup$ Nov 8, 2021 at 8:47
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You said $E$ is open, and you gave an example of a union of two intervals.

Generalization of interval may be $I_1\times I_2$ forms: $I_1, I_2$ is both interval.

So the specific answer is $E=(-1,0)\times(-1,1)\cup(0,1)\times(-1,1)$.

$\overline E=[-1,1]\times[-1,1]$, so $\mathring{\overline{E}}=(-1,1)\times(-1,1)\ne E$.

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  • $\begingroup$ This is very helpful, though I don't know how to prove that $I_1 \times I_2$ is open in general. Do you have any tips for showing that? $\endgroup$
    – Brad G.
    Nov 8, 2021 at 2:41
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    $\begingroup$ @BradG. , it is basically derived by defenition of product topology. If $A, B$ is both open set, then $A\times B$ is not just open set, but also element of base of product topology. $\endgroup$
    – MH.Lee
    Nov 8, 2021 at 2:43

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