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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.

Answer:

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.

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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 $\tau_{3}$.

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