1

It is well-known that for metric spaces, being separable, strongly Lindelöf and second-countable are equivalent. I know how to prove the equivalence between separable and second-countable, and I guess I could prove the equivalence between being second-countable and strongly Lindelöf. Moreover, I can prove that if a metric space is Lindelöf, then it is separable. But I don't know how to prove the converse without proving that it is second-countable first.

Do you know a proof like that? If there is some, do we need some kind of Choice?

Dog_69
  • 1,877
  • Yes yes, the question is about second countable. But you will find the references are about $\Bbb N$ being Lindelöf (in its standard, discrete, topology). And there is no space more second-countable than that. – Asaf Karagila Nov 05 '18 at 23:06
  • Also relevant: https://math.stackexchange.com/questions/309313/second-countable-implies-separable-axiom-countable-choice, there might be another couple of relevant threads out there too. (For example, https://math.stackexchange.com/questions/1249606/equivalence-of-countable-choice-for-subsets-of-the-reals-and-second-countable) – Asaf Karagila Nov 05 '18 at 23:08
  • Right, but that was not my question. I asked whether I can prove that every separable metric space is Lindelöf without using that every separable metric space is second-countable and hence (using Choice) strongly Lindelöf. I have already read the question you refer to (In fact I thought I cited it). And in my opinion it is completely different. Anyway, I guess that the answer to my question is 'No, you cannot prove such a equivalence without proving first that every separable metric space is second-countable too' (as HennoBransdma did in this old web site. – Dog_69 Nov 06 '18 at 09:50
  • And I pointed out that $\Bbb N$ with the discrete topology, which is the most separable space in the history of spaces, is not Lindelöf without some modicum of choice. And that this is what you can find in those duplicates. – Asaf Karagila Nov 06 '18 at 10:08
  • Ok. Got it. Thanks – Dog_69 Nov 06 '18 at 11:07

0 Answers0