why does one want to metrize a topology?

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- Jun 27th 2006, 02:35 PMnweissmametrizing a topology
why does one want to metrize a topology?

- Jun 27th 2006, 08:04 PMThePerfectHacker
I was finally able to find this on PlanetMath, (Source taken from this website below).

A topological space is said to be metrizable if there is a metric such that the topology induced by is .

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This is what bothers me,

*induced*as in graph theory.

By the problem is that the topology can contain more than two element subsets of in that case, how can we view as a graph!

Furthermore, then for certainly the topology on does not contain two element subsets. That means that is not a graph and the entire concept of makes no sense.

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Forgive, me for not answersing your question. It just bothered me too. - Jun 28th 2006, 01:05 AMJakeDQuote:

Originally Posted by**ThePerfectHacker**

Here's a definition of induced topology from*Topology*by James Dugundji.

Let Y be a set and d be a metric in Y. The topology T(d), having for basis the family { Bd(y,r) | y in Y, r > 0 } of all d-balls in Y, is called the topology in Y induced (or determined) by the metric d. - Jun 28th 2006, 02:27 AMJakeDQuote:

Originally Posted by**nweissma**

For example, the definition of continuity of a function that we learned in calculus

is continuous if implies

uses sequences. It applies when is a metric space but not a general topological space. But try teaching the general definition of continuity--the inverse image of an open set is open--to a first-year calculus student.

There are many theorems that apply to metric spaces but not general topological spaces. Thus it is important to know when a topology is metrizable so those theorems apply.

I have a book on probability measures on metric spaces. What matters there is that the topological spaces are metrizable, not the particular metric used.