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A topological space is separable if it has a countable dense subset. A space is first countable if it has a countable basis at each point. It is second countable if there is a countable basis for the whole space. A collection of subsets of a space is locally finite if each point has a neighborhood which intersects only finitely many sets in the collection. A collection of subsets of a space is sigma-locally finite (AKA countably locally finite) if it is the union of countably many locally finite collections.

My question is, if a space is separable, first countable, and has a sigma-locally finite basis, must it also be second countable? I think the answer is yes, because I haven't found any counterexample here.

Any help would be greatly appreciated.

Thank You in Advance.

EDIT: I fixed my question. I meant that the space should have a locally finite basis, not be locally finite itself, which doesn't really mean much.

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Your title is clear, but your wording of the question in the body of the post is flawed. A space can’t be $\sigma$-locally finite: that’s a property of families of sets, not of spaces. I assume that you meant to repeat the question in the title: if a space is separable and first countable and has a $\sigma$-locally finite base, must it be second countable? –  Brian M. Scott Apr 10 '13 at 2:27
Thanks for catching that Brian! –  Keshav Srinivasan Apr 10 '13 at 3:35
Having a $\sigma$-locally finite base already implies being first countable: for any $x$ in $X$, for such a base, $x$ is at most countably many base members (finitely many for each of the locally finite subfamilies, of which we have countably many), and these members form a local base at $x$ by definition. –  Henno Brandsma Apr 28 '13 at 11:49
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3 Answers

up vote 2 down vote accepted

Let $\mathcal{B} = \cup_n \mathcal{B}_n$ be a $\sigma$-locally finite base (all non-empty sets), and let $D = \{d_n: n \in \mathbb{N} \}$ be a dense subset of $X$ (as said in my comment, $X$ is automatically first countable from having such a base, so I won't use that assumption).

For every $n$, $d_n$ is in at most countably many members of $\mathcal{B}$, as in is at most finitely many members of $\mathcal{B}_k$ for every $k$, note that we only need that each of these is point-finite, not locally finite; the same holds for the first countability. Call these members that contain $d_n$: $\mathcal{B}^n$, this is a countable set.

Then, as $D$ is dense, so every member of $\mathcal{B}$ contains some $d_k$, so $\mathcal{B} = \cup_n \mathcal{B}^n$, so $\mathcal{B}$ is already a countable base for $X$.

So in short: yes, this holds, even: every space $X$ with a $\sigma$-point-finite base that is separable has a countable base. Or, a point-countable family of non-empty open subsets in a separable space is (at most) countable...

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@user14111 Indeed. It almost becomes a semi-tautology if one thinks about it: if stuff is at most countable, countably many stuff is also at most countable.... –  Henno Brandsma May 18 '13 at 12:02
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The Nagata-Smirnov metrization theorem says that a $T_3$ space with a $\sigma$-locally finite base is metrizable (and conversely). A separable metrizable space is second countable. Thus, if your space is $T_3$, the answer to your question is yes.

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Thanks Brian, but I knew that already. The whole reason I asked the question is that I was curious about whether there's a condition stronger than first countable, but weaker than metrizable, which is sufficient to ensure that separable implies second countable. The Nagata=Smirnov theorem is what inspired my guess that having a sigma-locally finite basis is relevant. –  Keshav Srinivasan Apr 10 '13 at 3:41
+1) for the Nagata-Smirnov metrization theorem –  Paul Apr 10 '13 at 5:42
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For example:

Helly Space; Right Half-Open Interval Topology; Weak Parallel Line Topology.

These space are all separable, first countable and paracompact, but not second countable. Note that a paracompact is the union of one locall finite collection.

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I don’t know the other two by name, but the Sorgenfrey line, your second example, does not have a $\sigma$-locally finite base. –  Brian M. Scott Apr 10 '13 at 2:11
The space needn't $\sigma$-locally finite base. It is just sigma-locally finite. –  Paul Apr 10 '13 at 2:16
Look at the title: it must have a $\sigma$-locally finite base. Saying that the space itself is $\sigma$-locally finite is meaningless: the term applies to collections of sets, not to spaces. Keshav was simply a bit careless in the body of the post. –  Brian M. Scott Apr 10 '13 at 2:17
What is meaning of "OP"? Do you let it mean open problem? –  Paul Apr 10 '13 at 2:19
I meant Original Poster; it’s sometimes also used for Original Post. (You’ll see it used quite often here, in both senses.) –  Brian M. Scott Apr 10 '13 at 2:20
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