From what I understand, the point of first countability is to ensure sequences capture all topological information. It's not too difficult to prove that a set function from a first countable space to any space is continuous if and only if it's sequentially continuous.

My question is: does this equivalence characterize first countable spaces? I'm guessing it does but have no idea how to prove so.

  • $\begingroup$ It does not: it characterizes sequential spaces, as is noted in this answer. $\endgroup$ – Brian M. Scott Oct 25 '16 at 21:17
  • $\begingroup$ @BrianM.Scott thanks! Could you post this as an answer? $\endgroup$ – Arrow Oct 25 '16 at 21:20
  • $\begingroup$ It probably doesn't because topology is naughty like that. $\endgroup$ – Pedro Tamaroff Oct 25 '16 at 21:22
  • $\begingroup$ Sure; just give me a few minutes. $\endgroup$ – Brian M. Scott Oct 25 '16 at 21:22

It does not: it characterizes sequential spaces, as is noted in this answer. This isn’t too hard to prove; once you get the definition of sequential space clear in your mind, you might try to prove it.

Sequential spaces are precisely those in which the convergent sequences completely determine the topology, so this is not really surprising. All first countable spaces are sequential, but there are many sequential spaces that are not first countable. Dan Ma’s Topology Blog has a series of informative posts on sequential spaces, starting here.

  • $\begingroup$ Thanks for the answer. If you happen to feel like it, could you help me here? $\endgroup$ – Arrow Oct 25 '16 at 21:31
  • $\begingroup$ @Arrow: You’re welcome. I’ll take a look; I’m doing a couple of things at once, though, so I may have to come back to it later this evening. $\endgroup$ – Brian M. Scott Oct 25 '16 at 21:33
  • $\begingroup$ Many many thanks. $\endgroup$ – Arrow Oct 25 '16 at 21:56

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