interior and closure in a different topology

Question

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.

Answer 1

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.

Answer 2

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 .