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We equip an infinite and uncountable set $X$ with the topology $$\mathcal{T}=\left\lbrace U \subset X : U = \emptyset \text{ or } X \setminus U \text{ is countable}\right\rbrace.$$ Prove that $A \subset X$ is dense in $X$ if and only if $A$ is an uncountable set.

I have found a solution for $(\Leftarrow)$ by proving its contradiction is wrong. I get an arbitrary open set $U \neq \emptyset$, then I have its contradiction $U \cap A=\emptyset$, that means $A \subset X \setminus U (!)$, because $A$ is uncountable but $X \setminus U$ is countable.

I have tried to prove $(\Rightarrow)$ but I didn't manage to prove it. Is there any hint for it? Thank you!

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    $\begingroup$ Closed sets are exactly $X$ and the countable subsets... $\endgroup$ – Gae. S. Jul 19 '19 at 14:07
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Suppose that $A$ is dense, and for the sake of contradiction suppose that $A$ is countable. So $A=X\setminus(X\setminus A)$ is countable, then $(X\setminus A)$ is an open set. Note that $X\setminus A$ is not empty, because $X$ is uncountable. As $A$ is dense, it must happen that $A\cap(X\setminus A))$ is not empty, which is a contradiction. So $A$ is uncountable.

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  • $\begingroup$ The OP was asking for a hint, not for a solution. $\endgroup$ – Mars Plastic Jul 19 '19 at 14:27
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Hint: By the definition of $\mathcal{T}$, the full space $X$ is the only uncountable closed set.

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