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

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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## 1 Answer

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).

• 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.) – YCor May 16 at 5:09
• 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. – YuiTo Cheng May 17 at 4:54
• @YuiToCheng I'm not sure to follow how you reach this conclusion – YCor May 17 at 5:21
• If $x\neq y$, $\{x,y\}$ has an isolated point. Thus there is a neighborhood separating $x$ from $y$, implying $X$ is $T_0$. – YuiTo Cheng May 17 at 5:35
• @YuiToCheng thanks, but the linked answer seems not to be correct (I added a counterexample as a comment there). – YCor May 18 at 8:22