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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.

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

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