metrizing a topology

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June 27th 2006, 01:35 PM
nweissma

metrizing a topology

why does one want to metrize a topology?
June 27th 2006, 07:04 PM
ThePerfectHacker

I was finally able to find this on PlanetMath, (Source taken from this website below).
A topological space $(X,\mathcal{T})$ is said to be metrizable if there is a metric $d:X^2\to \mathbb{R}^+$ such that the topology induced by $d$ is $\mathcal{T}$ .
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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 $X$ in that case, how can we view $(X,\mathcal{T})$ as a graph!
Furthermore, $\{\} \in \mathcal{T}$ then for certainly the topology on $X$ does not contain two element subsets. That means that $G=(X,\mathcal{T})$ is not a graph and the entire concept of $\mathcal{T}= G[d]$ makes no sense.
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Forgive, me for not answersing your question. It just bothered me too.
June 28th 2006, 12:05 AM
JakeD

Quote:

Originally Posted by ThePerfectHacker

I was finally able to find this on PlanetMath, (Source taken from this website below).
A topological space $(X,\mathcal{T})$ is said to be metrizable if there is a metric $d:X^2\to \mathbb{R}^+$ such that the topology induced by $d$ is $\mathcal{T}$ .
---
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 $X$ in that case, how can we view $(X,\mathcal{T})$ as a graph!
Furthermore, $\{\} \in \mathcal{T}$ then for certainly the topology on $X$ does not contain two element subsets. That means that $G=(X,\mathcal{T})$ is not a graph and the entire concept of $\mathcal{T}= G[d]$ makes no sense.
---
Forgive, me for not answersing your question. It just bothered me too.

The link for "induced" in the PlanetMath definition is in error. You're right that graph theory has nothing to do with it.

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.
June 28th 2006, 01:27 AM
JakeD

Quote:

Originally Posted by nweissma

why does one want to metrize a topology?

Metric spaces are easier to work with than general topological spaces. One important instance of this is that in metric spaces one can work with sequences.

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

$f:X \to Y$ is continuous if $\langle x_n \rangle \to x$ implies $\langle f(x_n ) \rangle \to f(x)$

uses sequences. It applies when $X$ 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.