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A few months ago I read here or elsewhere that a Hausdorff space X is completely regular if it satisfies an equivalent weaker separation axiom:

Every two different points x, y in X can be separated by a continuous function f on X. Is it true?

Marek Wojtowicz
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  • If the axiom is equivalent, it is not weaker (and visa versa) – FShrike Jun 23 '22 at 14:56
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    This is not true; such a space is called "completely Hausdorff". Most of the commonly considered examples of Hausdorff spaces that are not regular are in fact completely Hausdorff (for instance, many such examples are obtained by starting with a completely regular topology and making it finer, which will preserve the complete Hausdorff condition). – Eric Wofsey Jun 23 '22 at 15:39
  • Thank you, Eric. In topological linear spaces, T_1 already implies complete regularity; that confused me a bit. – Marek Wojtowicz Jun 26 '22 at 10:30

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