The book I'm reading doesn't explicitly give a definition of separable metric spaces. The only type of separability definition I know that a separable topological space is one that has a countable dense subset.

Could someone give me a definition of a separable metric space? I'm assuming it would have something to do with the metric that induces the topology, but I'm unsure as to how to write this.


Any metric space is a topological space. So topological terms generally have the same meaning as in a general topological space.

In particular a metric space is separable if it has a countable dense set.

  • 2
    $\begingroup$ Technically a metric space is not a topological space. However, every metric induces a topology, and we speak of the metric space as having each of the topological properties of the resulting topological space. $\endgroup$ – Brian M. Scott May 23 '16 at 17:42
  • $\begingroup$ @BrianM.Scott Thanks. I was aware that formally speaking a metric space is not a topological space. Actually gave some thought to how to say what I said, decided that this was one of those times when a lie would make things clearer than telling the literal truth. $\endgroup$ – David C. Ullrich May 23 '16 at 17:44
  • $\begingroup$ As a teacher I don’t mind occasionally telling lies for that reason, but I think that one should always make it clear that one is doing so. Here I’d have started with the lie, perhaps modified by for all practical purposes or the like, and then gone on to explain the technicality. (In this case it seems especially reasonable to do so, since the OP appears to realize that they actually are different animals.) $\endgroup$ – Brian M. Scott May 23 '16 at 17:50
  • $\begingroup$ @BrianM.Scott Good thing the comments are visible then... $\endgroup$ – David C. Ullrich May 23 '16 at 17:53
  • 1
    $\begingroup$ @Oliver: That’s exactly right. A metric space is second countable, compact, connected, etc. iff the induced topological space is second countable, compact, connected, etc. $\endgroup$ – Brian M. Scott May 23 '16 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.