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Stevo
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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.
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?
A Hausdorff space is a topological space where every pair of distinct points have disjoint neighborhoods. This means that for any two points in the space, there exist open sets containing each point that do not intersect.
A metric is a function that measures the distance between two points in a space. It assigns a non-negative value to the distance between any two points and satisfies certain properties such as symmetry and the triangle inequality.
No, not every Hausdorff space admits a metric. There are Hausdorff spaces that are not metrizable, meaning there is no metric that induces the topology on the space. Examples include the long line and the Sorgenfrey line.
Having a metric on a Hausdorff space allows for more precise distance measurements and can help with studying the convergence of sequences in the space. It also allows for the use of various tools and techniques from metric space theory in the study of the space.
In order to prove that a Hausdorff space admits a metric, one must construct a metric that induces the given topology on the space. This can be done by defining a distance function that satisfies the necessary properties, such as the Hausdorff property and compatibility with the open sets in the space.