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Could you give me a hint on how to solve the following problem?

Every second-countable, uncountable Hausdorff space contains a non-empty countable subspace with no isolated points.

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migrated from mathoverflow.net May 16 at 5:15

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Let $X$ be an arbitrary second countable topological space. Let $(V_n)$ be a countable basis of open subsets. Let $U$ be the union of all countable $V_n$ and $S$ its complement. Clearly $U$ is an open countable subset. Then $S$ is the set of points all of whose neighborhoods are uncountable, and has no isolated point. (Essentially clear using definition of a basis; the second assertion follows from the first: if $s$ were isolated in $S$, $\{s\}\cup U$ would be an open countable neighborhood of $s$.

Then $S$ has a dense countable subset $D$ (for each $n$ such that $V_n\cap S$ is non-empty choose a point in this intersection). Since $S$ has no isolated point, its dense subset $D$ also doesn't (this uses that $X$ is Hausdorff; $T_1$ would be enough).

Finally, if $X$ is uncountable, then $S$ is not empty, and hence $D$ is also not empty.


Edit: here's how to deal with the (possibly) non-$T_1$ case. Recall that $T_0$ means that for any two $\neq$ points there's an open subset containing exactly one of them (which one can't choose!); $T_1$ means that that for any two $\neq$ points there's an open subset containing the first and not the second, and this also means that singletons are closed. Also beware that in general, $x$ isolated means that $\{x\}$ is open, but beyond the $T_1$ case it can fail to be clopen (so "isolated" can be misleading: it could be a dense singleton for instance).

Plainly $T_0$ means that every 2-element subset is $T_0$. So the negation of $T_0$ means that some 2-element subset has no isolated point.

Finally assume that the space is $T_0$. Then every minimal nonempty open subset is a singleton, and hence there's none since there no isolated point. Hence each $V_n$ is infinite. Choose a countable subset $D$ meeting each $V_n$ in at least two points. Then $D$ has no isolated point, since it meets each element of the basis in at least two points. Hence $D$ is a dense subset with no isolated point.

Note: in the $T_1$ case, to have no isolated points passes to dense subsets but this fails in the $T_0$ case (see my comment here).

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    $\begingroup$ I posted an answer so as to emphasize where Hausdorff is used. So far unable to figure out if the result remains true without assuming $X$ to be $T_1$. ($T_1$: singletons are closed.) $\endgroup$ – YCor May 16 at 5:09
  • $\begingroup$ In fact, $T_0$ suffices. If every nonempty countable subspace of $X$ has an isolated point, $X$ is $T_0 $. Thus the result remains true without assuming any separation axiom. $\endgroup$ – YuiTo Cheng May 17 at 4:54
  • $\begingroup$ @YuiToCheng I'm not sure to follow how you reach this conclusion $\endgroup$ – YCor May 17 at 5:21
  • $\begingroup$ If $x\neq y$, $\{x,y\}$ has an isolated point. Thus there is a neighborhood separating $x$ from $y$, implying $X$ is $T_0$. $\endgroup$ – YuiTo Cheng May 17 at 5:35
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    $\begingroup$ @YuiToCheng thanks, but the linked answer seems not to be correct (I added a counterexample as a comment there). $\endgroup$ – YCor May 18 at 8:22

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