Does every Hausdorff space admit a metric?

1. Apr 8, 2004

Stevo

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

2. Apr 8, 2004

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/geomboard/messages3/143.html [Broken]

apparently a "string theory" message board.

Last edited by a moderator: May 1, 2017
3. Apr 8, 2004

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

4. Apr 8, 2004

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?

Last edited by a moderator: May 1, 2017
5. Apr 9, 2004

Try these:-
1) Manifolds at and beyond the limit of metrisability at arXiv:math.GT/9911249
2) http://www.math.auckland.ac.nz/~gauld/research/ (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?

Last edited by a moderator: Apr 20, 2017
6. Apr 9, 2004