8

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.

Keshav Srinivasan
  • 10,804
  • 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 02:27
  • Thanks for catching that Brian! – Keshav Srinivasan Apr 10 '13 at 03:35
  • 3
    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

3 Answers3

5

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.

Brian M. Scott
  • 631,399
  • 2
    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 03:41
5

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...

Henno Brandsma
  • 250,824
1

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.

Paul
  • 21,141
  • 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 02:11
  • The space needn't $\sigma$-locally finite base. It is just sigma-locally finite. – Paul Apr 10 '13 at 02:16
  • 1
    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 02:17
  • What is meaning of "OP"? Do you let it mean open problem? – Paul Apr 10 '13 at 02:19
  • 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 02:20
  • En; maybe I would take some time to consider it. – Paul Apr 10 '13 at 02:24
  • Paul, Brian's right. I meant having a sigma-locally finite basis. – Keshav Srinivasan Apr 10 '13 at 03:36
  • If you use the Spacebook site and ask for a first countable, separable space with a sigma-locally finite base that is not second countable, you get no results. So it could be true based on the small sample of spaces from Counterexamples in Topology. – Henno Brandsma Apr 12 '13 at 09:30
  • Henno, yes, as I mentioned in my question, the lack of counterexamples on Spacebook is what led me to suspect that my guess was right. – Keshav Srinivasan Apr 19 '13 at 00:54