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Let $X$ be an uncountable set. We define the cocountable topology $\tau$ as the set of all subsets $U\subseteq X$ such that either $U=\emptyset$ or $X\setminus U$ is countable.

I am interested in the conditions under which a set $E\subseteq X$ will be dense in $X$, and I honestly don't know where to start.

I have already shown that $(X,\tau)$ defines a topological space. I know that by definition, a set $E$ is dense in $X$ with respect to $\tau$ if for every $x\in X$ and every open set $U \subseteq X$ with $x\in U$, there exists an element $y\in E$ such that $y\in U$, or equivalently that $E$ is dense in $X$ with respect to $\tau$ if for every nonempty open set $U\subseteq X$, $E\cap U \neq \emptyset$.

I would like to see a complete proof if possible.

EDIT: I first want to show that: If $E\subseteq X$ is dense, then $X\setminus E$ is countable.

I attempted to prove this by the contrapositive (i.e. show that if $X\setminus E$ is uncountable, then $E$ is not dense in $X$). Suppose that $X\setminus E$ is uncountable. Then, $X \setminus E$ is open with respect to $\tau$, and so by definition $X\setminus (X\setminus E)=E$ is closed in $X$ with respect to $\tau$. Then, $E=\overline {E}$, where $\overline {E}$ denotes the closure of $E$. Since $E$ is countable and $X$ is uncountable, $E\neq X$. I am struggling to jump from here to the fact that $E$ is not dense in $X$.

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  • $\begingroup$ What happens if you take countably many elements from an uncountable. $\endgroup$ – William Elliot Mar 8 '18 at 4:36
  • $\begingroup$ Demanding complete proofs is like insisting on pablum. $\endgroup$ – William Elliot Mar 8 '18 at 4:40
  • $\begingroup$ Well, since all we know about is cardinality, it must be a condition on the cardinality of $E$ or perhaps its complement, right? $\endgroup$ – saulspatz Mar 8 '18 at 4:42
  • $\begingroup$ @WilliamElliot taking countably many elements from an uncountable set still leaves you with an uncountable set, but I'm not seeing how that helps $\endgroup$ – mathqueen459 Mar 8 '18 at 4:44
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    $\begingroup$ If $E$ is countable, its complement is co-countable, hence open. Thus, E is closed. $\endgroup$ – saulspatz Mar 8 '18 at 5:03
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A subset $A$ of $X$ (having the co-countable topology) is closed iff $A = X$ or $A$ is at most countable (so including finite). This is straight from the definition of the co-countable topology.

The closure $\overline{A}$ of a set $A$ is the smallest closed set $C$ such that $A \subseteq C$.

If $A$ is at most countable, it is already closed, so its closure is $A$ again (It cannot be smaller than that) and as $A \neq X$ (recall that $X$ is assumed to be uncountable), $A$ is not dense.

If $A$ is uncountable, then the only closed set of $X$ that can contain $A$ is $X$ (if $C$ is at most countable (the only other type of closed set), we cannot have $A \subseteq C$, as any subset of a set that is at most countable is again at most countable, which $A$ is not), so $\overline{A} = X$ for $A$ uncountable (there is only one candidate closed set) and thus $A$ is dense.

So the situation is atypical (for a general space): a subset is either "small" (at most countable) and closed, or "big" (uncountable) and dense.

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  • $\begingroup$ I'm having difficulty following the second part of the proof. You state: If $A$ is uncountable, then $X$ is the only closed set such that $A\subseteq X$ since for any countable closed set $C$, if $A \subseteq C$, then $A$ is countable, which is a contradiction. Then, since $X$ is the only closed set that contains $A$, $X=\overline {A}$. I understand this part. Where I get confused is last part. How does this tell us that $A$ is dense? $\endgroup$ – mathqueen459 Mar 9 '18 at 4:59
  • $\begingroup$ @mathqueen459 the definition of dense is that the closure equals $X$. $\endgroup$ – Henno Brandsma Mar 9 '18 at 5:34
  • $\begingroup$ Hmm. I am using an alternate definition (the one in my original post). Can you explain why they are equivalent? $\endgroup$ – mathqueen459 Mar 9 '18 at 5:40
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    $\begingroup$ @mathqueen459 If $\overline{A}$ equals $X$: take $U$ open and non-empty in $X$, which means that $X\setminus U$ is closed and smaller than $X$. So we cannot have that $A \subseteq X\setminus U$, so $A \cap U \neq \emptyset$. So $U$ intersects $A$. The reverse direction is similar: if $A$ intersects all open non-empty sets, it can never intersect $X \setminus \overline{A}$, which is open, so that last set must be empty and thus $\overline{A} = X$. $\endgroup$ – Henno Brandsma Mar 9 '18 at 10:23

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