# Is this set closed with respect to these topologies?

Let $\ \large \tau_1, \ \tau_2, \ \tau_3 \$ be the usual topology, half-open interval topology , countable compliment topology respectively. Consider the set $\ X=\left\{2+\frac{1}{2^n} : n \in \mathbb{N} \right\}$.

Then determine whether the set $\ X \$ is closed with respect to the above $\ 3 \$ topologies.

Case 1:

Consider the topology $\ \tau_1 =usual \ \ topology \$ ,

The limit point of the set $\ X \$ is $\ 2 \$ but $\ 2 \notin X \$.

Hence $\ X \$ is not closed with respect to usual topology.

But how can attempt for the rest two topology $\ \tau_2 \ \ and \ \ \tau_3 \$

Help me out.

By "Half open interval" topology I will assume you mean the topology generated by sets of the form $[a,b)$ where $a<b$.
Regarding $\tau_{2}$ we will claim that $X$ is not closed. Precisely for the same reason that it is not closed in $\tau_{1}$. Let $U\subseteq\mathbb{R}$ be any open set containing $2$. Then $U$ contains some half open interval of the form $[2,b)$ where $2<b$. Clearly $[2,b)\cap X\neq\emptyset$. Thus $X$ is not closed in $\tau_{2}$.
For the countable complement topology we need only note that $X$ is countable. Then the complement of $X$ has a countable complement and is therefore open, thus $X$ is closed in $\tau_{3}$.