Does every Hausdorff space admit a metric?

Stevo

^.............

HallsofIvy

Every metric space is Hausdorff but not every Hausdorff space is metrizable!

Googling on "Hausdorff" and "metrizable", I found
"Metrizable requires, in addition to Hausdorf, separability and existance of at least one countable locally finite cover. Those three are independent requirements; if you could do without any one of them you would have a much stronger theorem, and be famous among topologists (nobody else would notice or care)." attributed to a "DickT" on

http://superstringtheory.com/forum/g...ages3/143.html

apparently a "string theory" message board.

selfAdjoint

That was me, and I stand behind it. I should, because I got it straight out of one of my old textbooks!

Stevo

Yeah, it's pretty easy to show that every metric space is Hausdorff... I wasn't sure if the converse was true. Thanks for that.

Does anybody have a proof, a link to a proof, or a reference to a proof that metrisation requires Hausdorff, separability, and existence of a countable locally finite cover?

maddy

Try these:-
1) Manifolds at and beyond the limit of metrisability at arXiv:math.GT/9911249
2) Metrisability of manifolds - paper in preparation (the file is labelled metrisability.pdf)
both by David Gauld at University of Auckland Department of Mathematics.

A mathematical physics prof taught me that paracompactness must also be one of the criteria of metrisability.
Can there really be a proof that doesn't include this criteria?

selfAdjoint

Paracompactness is a generalization from the countable locally finite cover. If a space is paracompact then every open cover of it has a countable locally finite refinement. So you get a little narrower theorem by specifying the CLF cover specifically, but in many instances, you would use the given paracompactness of the space to prove the CLF cover exists.

The theorem is called Urysohn's theorem. here is a sketch of the proof.