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Problem: Let $X$ be a regular space. If $Y$ is a subspace that is dense and second countable (with respect to the subspace topology), then $X$ is second countable.

I have this problem and I'm having trouble figuring it out. I have a hint, it says: consider B a countable basis for Y and show that $B' = \{\text{int}(\overline{U}) \, : \, U \in B\}$ is a countable basis for X, where $\overline{U}$ stands for the closure in $X$.

Now, it's obvious that $B'$ is countable. The problem is to show that $B'$ is a basis for $X$. Let me show what I tried.

Let $p \in X$ a point and $V$ a neighborhood of $p$. We know that $Y$ is dense, so $V \cap Y$ is a nonempty open set in $Y$. Because $B$ is a basis, there is a $U_V$ such that $U_V \subseteq V\cap Y \Rightarrow U_V \subseteq V$.

I know that if I show $\overline{U_V} \subseteq V$ and $p \in \text{int}(\overline{U_V})$ then we're done: $p \in \text{int}(\overline{U_V}) \subseteq V$ and, therefore, $B'$ is a basis.

I'm having trouble to prove both things in the above paragraph. I can't see where to use regularity or the closure of $U_V$ in $X$.

Any help would be appreciated. Any solution not using the hint is welcome too. Thanks in advance.

(P.S.: Sorry for my grammar mistakes. Englsh isn't my mother's tongue.)

user233397
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  • Your English is excellent. The only small foreignism that I noticed is having trouble to prove: the normal idiom is having trouble proving. – Brian M. Scott Apr 22 '15 at 01:59
  • I think the hint is misleading, since I think for some countable bases $B$ of $Y$ the resulting construction will not yield a basis for $X$. Consider $X = \Bbb R, Y = \Bbb Q$. Then you can choose a countable basis for $\Bbb Q$ by choosing those intervals $\Bbb Q \cap (a,b)$ such that $a,b$ are rational and $\pi \not\in (a,b)$. Then the resulting construction will give open sets none of which contain $\pi$. – Rolf Hoyer Apr 22 '15 at 03:42
  • @Rolf: Thanks for waking me up earlier. In fact the result is false. – Brian M. Scott Apr 22 '15 at 04:17

2 Answers2

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The result is false. Here’s one of the more elementary counterexamples.

Subsets $A$ and $B$ of $\Bbb N$ are almost disjoint if their intersection is finite. As is shown in the answers to this question and this question, there is a family $\mathscr{A}$ of $2^{\omega}=\mathfrak{c}$ almost disjoint infinite subsets of $\Bbb N$. Let $X=\Bbb N\cup\mathscr{A}$. Points of $\Bbb N$ are isolated. For each $A\in\mathscr{A}$, the basic open nbhds of $A$ are the sets of the form $\{A\}\cup(A\setminus F)$ such that $F$ is a finite subset of $\Bbb N$.

Because the members of $\mathscr{A}$ are almost disjoint, $X$ is Hausdorff. Moreover, the basic open sets are also closed, so $X$ is zero-dimensional and therefore Tikhonov, and in particular regular. $\Bbb N$ is a dense subset, and $\big\{\{n\}:n\in\Bbb N\big\}$ is a countable base for $\Bbb N$. However, $X$ is not second countable, because $\mathscr{A}$ is an uncountable discrete subset of $X$: each $A\in\mathscr{A}$ has a nbhd that contains no other point of $\mathscr{A}$ (e.g., $\{A\}\cup A$).

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Brian M. Scott
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As Brian M. Scott says, this is false. Another basic example is the lower-limit topology on $\mathbb R$ (that is, the topology generated by the half-open intervals $[a,b)$ for $a<b$). Let's denote this space by $\mathbb R_\ell$. It is not difficult to see that $\mathbb R_\ell$ is regular, and one can show that it is even perfectly normal.

The set $\mathbb Q$ of rationals is a countable dense set, and as $\mathbb R_\ell$ is first-countable it follows that $\mathbb Q$ is a second-countable subspace. However $\mathbb R_\ell$ is not second-countable, as shown here.

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