# Interior of the closure of an open set

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 $$\mathbb{R}^2$$.

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

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?

• 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 $\mathbb{R}^2$. Nov 8, 2021 at 2:59
• 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 $E$. Nov 8, 2021 at 3:52
• @BradG. Well, what is your definition of closed? Nov 8, 2021 at 5:13
• @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. Nov 8, 2021 at 8:47

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

• 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? Nov 8, 2021 at 2:41
• @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. Nov 8, 2021 at 2:43