# How does one show a topological space is metrizable? Using text Intro. to Topo. by Mendelson

I'm self studying Intro. to Topology by Mendelson.

The problem I'm looking at is,

Prove that for each set $X$, the topological space $(X,2^X)$ is metrizable.

I'm not having so much trouble with this problem per se, but with the idea of metrizable spaces. From what I've understood from the reading is that a metric space $(X,d)$ satisfies a theorem which is exactly the definition of a topological space using only open sets from $X$ and so these metric spaces are known as metrizable spaces. Is my thought process sound?

Going back to this problem, would I need to find a metric $d$ that can induce/create the topology on $X$? Would I need to look at the open sets of $X$? Am I even approaching this problem correctly?

Thanks for any feedback or hints.

• The topology being the entire power set of $X$, it means that $X$ is discrete. Tried considering $d(x,y)=\begin{cases}1&x\ne y\\0&x=y\end{cases}$? This is checkably a metric. Commented Jul 20, 2013 at 10:40
• So would I need to build up the power set using this metric or is there something else to do? I'm unsure as to what the thought process is.
– S.D.
Commented Jul 20, 2013 at 10:44
• The topological space $(X,2^X)$, I think, means that the topology $\tau$ on it is $2^X$, which is the entire power set. This means that every set is open. Commented Jul 20, 2013 at 10:45
• I agree with what you said, but I'm still unsure on what I need to do exactly to show a topological space is metrizable. Do I some how use the metric you gave to create the power set?
– S.D.
Commented Jul 20, 2013 at 10:49

It is better to think of topological metrizable spaces as "metric spaces with the metric forgotten". You see, a metric space has an additional structure, namely the metric, but also "happen" to induce a topology via its open sets. So, the reason for having the two names, "metrizable space" and "metric space", is to differentiate between what we're looking at: If we say "a metric space", we mean a set with a metric function which defines distance betweeen points, satisfies the triangle inequality and so forth. If we say "a metrizable space", We mean a topological space, with open sets as its entire structure, but we also know that this topological space came from a metric, and hence satisfies many "good" qualities such as separation axioms.

Metric spaces are trivially topological spaces, but proving that a topological space is metrizable is generally hard. One way to do it is "recall" the metric. If you could realize what the metric that induced the topological space was, and show that the topology induced by it is exactly the topology of the given topological space, then you've shown the space is metrizable.

I was going to let you figure out what the metric is in your case, but Aneesh already told you.

• Thank you for your response, it did clear up some things and I will definitely work on this more tomorrow morning. I'm new to topology, but are the separation axioms similar to Hausdroff axioms or am I completely off?
– S.D.
Commented Jul 20, 2013 at 10:55
• Hausdorff is one of the separation axioms. You'll get to that later.
– Idan
Commented Jul 20, 2013 at 10:57
• Thank you Idan for the help and insight. Once I figure this out I'll make sure to repost my question/answer linking to this.
– S.D.
Commented Jul 20, 2013 at 10:59
• @Shant: Wikipedia has an extensive list of separation axioms. Commented Jul 20, 2013 at 23:40
• Thanks Brian, I'll definitely check that out.
– S.D.
Commented Jul 21, 2013 at 2:33