Does every Hausdorff space admit a metric?

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Every metric space is Hausdorff, but not every Hausdorff space is metrizable, as metrization requires additional conditions such as separability and the existence of a countable locally finite cover. The discussion highlights that these requirements are independent, and without them, one could develop stronger theorems in topology. Participants seek proofs or references that establish the necessity of these conditions for metrization, with suggestions for resources provided. Additionally, the role of paracompactness as a criterion for metrization is debated, emphasizing its connection to the countable locally finite cover. The conversation underscores the complexity of the relationship between Hausdorff spaces and metrization.
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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 existence of at least one countable locally finite cover. Those three are independent requirements; if you could do without anyone 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.
 
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That was me, and I stand behind it. I should, because I got it straight out of one of my old textbooks!
 
HallsofIvy said:
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 existence of at least one countable locally finite cover. Those three are independent requirements; if you could do without anyone 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.

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?
 
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Stevo said:
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?

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?
 
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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. http://www.cs.utk.edu/~mclennan/Classes/594-MNN/MNNH/MNNH-3/node20.html is a sketch of the proof.
 
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