Is a topological space pseudo-metrizable if and only if it is regular and paracompact ?

I have a link to a reference , "General Topology" , Kelley , page 127 , theorem 18 . Also note a paracompact space has a locally finite base, and so as required there, a sigma-locally finite base.

On page 128, the proof of lemma 20 shows, if the assumption that the space is T1 is dropped, that a regular paracompact space is pseudo-metrizable.

  • $\begingroup$ ia800501.us.archive.org/16/items/GeneralTopology/… was the link i forgot to include $\endgroup$
    – kinidkinid
    Sep 16, 2018 at 20:34
  • $\begingroup$ A paracompact space does not always have such a base. A space with such a base is paracompact, that much is true. $\endgroup$ Sep 16, 2018 at 21:15
  • $\begingroup$ Henno Brandsma thanks I see my mistake, I was thinking an open refinement of a base is also a base but obviously that isn't true. However, it is true that a regular space which has a sigma locally finite base is pseudo-metrizable. $\endgroup$
    – kinidkinid
    Sep 16, 2018 at 21:25
  • $\begingroup$ Indeed, the $T_1$-ness only "bridges the gap" between a metric and a pseudometric (where we can have $d(x,y) = 0$ for distinct points). This already follows from the quotient space argument. $\endgroup$ Sep 16, 2018 at 21:32
  • $\begingroup$ Henno Brandsma are you saying that the theorem that a T1 regular space with sigma locally finite base is metrizable implies that if its just regular with signa locally finite base its pseudometrizable ? I dont see how one would take a quotient of the given space to get another that satisfies the hypotheses of that theorem $\endgroup$
    – kinidkinid
    Sep 16, 2018 at 21:39

1 Answer 1


No, paracompact is not enough. Consider the Sorgenfrey Line or the Double arrow space.

We really need $\sigma$-locally finite base or a $\sigma$-discrete base.


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