How to prove a topological space is metrizable

by variety
Tags: metrizable, prove, space, topological
 P: 22 1. The problem statement, all variables and given/known data X is a set and P(X) is the discrete topology on X, meaning that P(X) consists of all subsets of X. I want to prove that X is metrizable. 2. Relevant equations My text says that a topological space X is metrizable if it arises from a metric space. This seems a little unclear to me, which is probably why I am slightly confused. 3. The attempt at a solution This problem seems really easy, I am just unsure of what I am supposed to prove. I want to show that X, together with the discrete topology, is metrizable. I choose the discrete metric d, which is defined by d(x,y) = 0 if x=y and d(x,y) = 1 if x=/=y. This is where I am unsure of what I am supposed to show. I can start by showing that if we have the metric space (X,d), then every subset of X is open since all the points are isolated. Then P(X) satisfies the property of a topology on X, so (X,P(X)) is a topological space. But I don't think this is correct because we already assumed (X,P(X)) is a topological space. In fact, it is ALWAYS a topological space for any X, right? Am I supposed to show that if (X,P(X)) is a topological space, (X,d) is a metric space? But this is also obvious because I already know that d is a metric. What am I supposed to be proving?
 Sci Advisor HW Helper Thanks P: 25,228 You start with a space with the discrete topology. You want to find a metric that induces the discrete topology. You already did that with the discrete metric (did you show it's a metric?) and you showed it induced the discrete topology which you did when you said "I can start by showing that if we have the metric space (X,d), then every subset of X is open since all the points are isolated." I'll admit it's not a hard proof, but it's not completely without substance.
 P: 1 counter example of topological space which is not metric space
HW Helper
P: 3,220
How to prove a topological space is metrizable

 Quote by variety But I don't think this is correct because we already assumed (X,P(X)) is a topological space. In fact, it is ALWAYS a topological space for any X, right? Am I supposed to show that if (X,P(X)) is a topological space, (X,d) is a metric space? But this is also obvious because I already know that d is a metric. What am I supposed to be proving?
Given any set X, (X, P(X)) is a topological space, this is a fact, and your starting assumption.

Of course you know that d is a metric. You found a metric which induces the discrete topology. As Dick noted, there's not much more to it.

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