Every Hausdorff space is $T_1$ and sober. Does the converse hold? I expect not. What's a counterexample?

I expected I should be able to look this up in Counterexamples in Topology, but unfortunately that book doesn't appear to discuss sober spaces.

  • 6
    $\begingroup$ Wikipedia has an example. "Let X be the set of real numbers, with a new point p adjoined; the open sets being all real open sets, and all cofinite sets containing p." $\endgroup$ – Nate Eldredge Sep 20 '18 at 19:18
  • $\begingroup$ @NateEldredge Thanks, I can't believe I missed that. $\endgroup$ – Tim Campion Sep 20 '18 at 19:22
  • $\begingroup$ I suppose more generally one can take any infinite $T_1$ and sober space $Y$ and adjoin a new point in a similar manner to get a $T_1$ and sober space which is not Hausdorff. $\endgroup$ – Tim Campion Sep 20 '18 at 19:26
  • $\begingroup$ I've just added references to this example to a few relevant nlab pages. $\endgroup$ – Tim Campion Sep 20 '18 at 19:42
  • $\begingroup$ I recommend making Nate's comment a CW answer, accepting it, and then this question is resolved. $\endgroup$ – David White Sep 20 '18 at 20:31

As Nate Eldredge points out in the comments, there's a counterexample on Wikipedia. See there or Nate's comment for a description.


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.