# Does every Hausdorff space admit a metric?

1. ### Stevo

112
^.............

2. ### HallsofIvy

41,256
Staff Emeritus
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

apparently a "string theory" message board.

7,521
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. ### Stevo

112
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?

33
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?

Last edited: Apr 9, 2004