# interior and closure in a different topology

Let $$\tau = \{ U ∈ \tau_e |\sin x > 0, ∀x ∈ U\} ∪ \{\mathbb{R}\}$$ where $$\tau_e$$ is the euclidean topology on $$\mathbb{R}$$, find the interior and closure of: $$[0,1]$$ and $$[0,2\pi]$$

Another part of the exercise is verifying if $$\tau$$ iss actually a topology on $$\mathbb{R}$$ and I've not encountered any trouble with that. Knowing that the interior is the biggest open set contained and the closure the smallest closed set which contain the set, I was thinking that $$\mathring{[0,1]} = (0,1)$$ Since in $$\{0\}$$, $$sin(x) = 0$$ and also $$\{1\}$$ is not contained in any open interval inside $$[0,1]$$. $$\overline{[0,1]} = \mathbb{R}$$ Since every other closed set can't contain the interval $$(0,1]$$, where $$sin(x)>0$$. Following same reasoning $$\mathring{[0,2\pi]} = (0,\pi)$$ $$\overline{[0,2\pi]} = \mathbb{R}$$ Is my solution wrong? thank you very much.

The interior of $$[0,1]$$ is here $$(0,1)$$ since this is Euclidean-open and $$\sin(0,1)\subseteq(0,\infty)$$ as desired, and the interior of $$[0,2\pi]$$ will be $$(0,\pi)$$ for the same reasons.

Please note that the hard cup symbol, $$\sqcup$$, is used to denote disjoint union as opposed to $$\cup$$.

A set $$F$$ is closed in this topology if the complement is open (definition) which means $$\Bbb R\setminus F$$ is Euclidean-open and $$\sin(\Bbb R\setminus F)\subseteq(0,\infty)$$ (which forces $$F$$ to be closed in $$\Bbb R$$ as well). Equivalently, $$\Bbb R\setminus F\subseteq\sin^{-1}((0,\infty))$$ and $$F\supseteq\sin^{-1}(-\infty,0]=\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]$$. Then a closure of a set $$A$$, the intersection of all closed containers, must contain $$\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]$$ (which is Euclidean-closed) and the smallest way to achieve this and the containment property is to take the Euclidean closure of $$A$$ and take the union of this with $$\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]$$.

For example, the closure of $$[0,1]$$ will be $$\overline{[0,1]}\cup\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]=(0,1]\sqcup\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]$$. The closure of $$[0,2\pi]$$ will be $$(0,\pi)\sqcup\bigsqcup_{n\in\Bbb Z}[(2n-1)\pi,2n\pi]$$ since $$0,[\pi,2\pi]$$ are already present in the big union.

To understand the "set of all limit points" idea in this instance, every Euclidean-open neighbourhood of, e.g. $$2n\pi$$, will contain elements $$x$$ for which $$\sin(x)\le0$$ (such as $$x=2n\pi$$ itself). It is thus impossible for any set $$U$$ of the form $$\sin(x)\gt0:\forall x\in U$$ to cover $$2n\pi$$, so the only open neighbourhood of this point is $$\Bbb R$$ itself. So, in a very unintuitive way, $$2n\pi$$ is a limit point, not only of $$[0,1]$$ but of every single point in $$\Bbb R$$ in this topology. This demonstrates dramatically why limits aren't unique in non-Hausdorff space.

• Thank you, but it sounds so counterintuitive to me. I'll explain my worst doubt, in my mind the closure of a set $A$ is the intersection of all the closed set that contained $A$, in this particular case I don't know how a closed set of $\tau$ could contain the interval $(0,1)$ because there $sin(x) > 0$ and since in my mind each closed set is contained at least in $(-\infty, 0] \cup [1, \infty)$, $(0,1)$ could not be in the intersection of all the closed set. Could you please clarify this part because maybe I'm picturing things the wrong way. Thank you very much! Jun 23, 2022 at 16:02
• @TurquoiseTilt Do you understand why $F\supseteq\sin^{-1}(-\infty,0]$ holds? This should clear your doubt. And yes, it is a very weird topology. A closed set can contain $(0,1)$ because then the complement would not contain $(0,1)$, which is ok! The complement can still be open even if it doesn't contain $(0,1)$ (e.g. it could contain $(2\pi,3\pi)$) Jun 23, 2022 at 16:08
• I was thinking a little bit about the definition of $\tau$ and it says $\forall x$ so correct me if I'm wrong but the interval $(0,3/2\pi)$ is actually in $\tau$ because is not open right? Jun 23, 2022 at 16:10
• That interval is not open since $\sin(3\pi/2)=-1\lt0$ @TurquoiseTilt Jun 23, 2022 at 16:11
• That interval is not closed since, in particular, it is not Euclidean closed. Its complement also is not open since its complement contains $7\pi/2$ and $\sin(7\pi/2)=-1\lt0$ @TurquoiseTilt This is why we must take a very large union to get closed sets Jun 23, 2022 at 16:15

Interior of $$[0,1]$$ is $$(0,1)$$ as it is the largest open set contained in $$[0,1]$$.

Closure of $$[0,1]$$ is not $$\Bbb{R}$$ because for example points in $$(2\pi,3\pi)$$ are not accumulation points of this set because $$(2\pi,3\pi)\cap[0,1]$$ is empty.

The closure is $$\displaystyle[0,1]\cup\bigg(\bigcup_{n=1}^{\infty}\bigg[(2n+1)\pi,(2n+2)\pi\bigg]\bigg)\cup\bigg(\bigcup_{n=1}^{\infty}\bigg[-2n\pi,(-2n+1)\pi\bigg]\bigg)$$ .

Similarly you can work out for $$[0,2\pi]$$ using accumulation points arguments .

• Your closure should not contain $(1,\pi)$ as this is open in the topology but disjoint from $[0,1]$ Jun 23, 2022 at 15:37
• Ahh my bad. I wanted to wrtie $[0,1]$. I'll edit it Jun 23, 2022 at 15:37