prove or disprove

Let $B$ be a subset of $\Bbb R$. In the topological space $(\Bbb R,\mathscr{T})$, if $B$ is not closed, then $B$ is dense.

$\mathscr{T}$ is defined as the countable complement topology, which is defined as the collection of all subsets $U$ of $X$ such that either $U = X$, $U=\varnothing$, or $X - U$ is countable.

I think it is true statement but I do not know how can I prove it ,could you please help me with that


Suppose $B$ is not closed.

Then $B$ must be uncountable, else $B$ would be closed.

Since $B$ is uncountable, the closure of $B$ is also uncountable.

But the closure of $B$ is closed, and the only uncountable closed subset of $\mathbb{R}$ is $\mathbb{R}$ itself.

It follows that the closure of $B$ is $\mathbb{R}$, hence $B$ is dense.


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