Seperable + What Implies Second Countable?

[lugita15]

A topological space is second countable if it has a countable basis. A space is seperable if it has a countable dense subset. Now being second countable implies being seperable, but the converse doesn't hold in general. My question is, what is the weakest condition you need to add to seperability in order to imply second countability? It's not enough for a space to be first countable, because ℝ with the lower limit topology is first countable and seperable but not second countable. On the other hand, a seperable metric space is always second countable. So is there a condition, stronger than being first countable but weaker than being metrizable, which is sufficient to ensure that being seperable implies being second countable? Wow, that was a tongue twister!

Any help would be greatly appreciated.

Thank You in Advance.

 

[micromass]

In compact Hausdorff spaces, it turns out that second countable is equivalent to metrizable. So in compact Hausdorff spaces, the answer to your question is that metrizable is the weakest possible condition. So this gives the intuition that you need a rather strong condition to ensure second countable.

In other spaces, I don't know the answer.

Is there perhaps some concrete problem that you're trying to solve?

Maybe you should also look at the Nagate-Smirnov metrization theorem. That says that seperable + regular + has a $\sigma$-locally finite basis implies second countable (and actually metrizable). So the answer to this question will be a condition much in the spirit of $\sigma$-locally finiteness.

 

[lugita15]

micromass, it looks like you were right. It seems that the condition required for seperable to imply second countable is first countable + having a $\sigma$-locally finite basis. (I haven't proven this, but the excellent database of topological counterexamples here seems to confirm it.) And this is the weakest possible condition, because it is implied by second countable.

 

[micromass]

OK, so you conjecture that separable + first countable + $\sigma$-locally finite basis => second countable. Hmm, let me think about a proof.

 

[lugita15]

micromass, did you have any success proving or disproving my conjecture?

 

[micromass]

micromass, did you have any success proving or disproving my conjecture?

No success here My gut feeling says it's false, but I can't seem to find a decent counterexample.

 

[lugita15]

micromass, I finally got a proof of my conjecture, after asking my question on Usenet. But first of all, I should note that having a $\sigma$-locally finite basis implies first countable. That's because any point x can only belong to finitely many elements of a locally finite collection, so it can only belong to countably many elements of a $\sigma$-locally finite collection. The reason I thought otherwise was because Counterexamples in Topology apparently has errors in it; it lists two spurious examples of spaces that have a $\sigma$-locally finite basis but are not first countable.

So the statement is just that separable + $\sigma$-locally finite basis implies second countable.

First let me prove that if X is seperable, then any locally finite collection C of nonempty open sets must be countable. Let T be a countable dense subset of X. Then for each t in T, let C_t be the collection of all sets in C which contain t. Since C is locally finite, t can only belong to finitely many elements of C, so each C_t is finite. But T is dense, so every set in C must contain an element of T, so C is equal to the union of C_t for all t. Thus C is the union of countably many finite sets, so it is countable.

So any locally finite collection of nonempty open sets in a seperable space is countable, and thus any $\sigma$-locally finite collection of nonempty open sets is countable. In particular any $\sigma$-locally finite basis must be countable.

So to sum up, a space is second countable if and only if it is seperable and has $\sigma$-locally finite basis, a result I find really surprising.