# What does induced (by a metric) topology mean?

I know that this is a question that has already been discussed here, but none of them provides me an answer that I understand, unfortunately.

I know the properties that should be satisfied by a metric, a topology and what a ball is; however, I fail to establish the link between metric spaces and topological spaces. My main problem is, in fact, I do not understand what is meant by metrizable topology.

• fully faithful forgetful functor (good answer if you are avidly attracted to alliterations) – CopyPasteIt Oct 22 '17 at 10:24

Let $X$ be a set and let $\tau$ be a topology in $X$. We say that the topological space $(X,\tau)$ is metrizable if there is a distance $d\colon X\times X\longrightarrow\mathbb R$ such that the elements of $\tau$ are the open sets of the metric space $(X,d)$. In other words, if$$\tau=\left\{A\subset X\,\middle|\,(\forall x\in A)(\exists r>0):B_r(x)\subset A\right\},$$where $B_r(x)=\{y\in X\,|\,d(x,y)<r\}$.

• Can I simply use any and different epsilons to generate the open sets in the topology in terms of balls? – Larx Oct 22 '17 at 9:08
• @Larx Yes, you can. – José Carlos Santos Oct 22 '17 at 9:09
• Sorry to bother you again, but I think I do not understand the role of "p" in the definition of the topology. Can you restate the expressions after "In other words, if" in words, please? – Larx Oct 22 '17 at 9:48
• @Larx There is no $p$ in my answer. – José Carlos Santos Oct 22 '17 at 9:50
• @K.M The question makes sense indeed. I am sorry if this is going to disappoint you, but it is just a matter of habit: we say that $d$ is a map from $X\times X$ into $\mathbb R$ and right after we impose that$$(\forall x,y\in X):d(x,y)\geqslant0.$$Of course, we could just say that $d$ is a map from $X\times X$ into $[0,+\infty)$ from the start. – José Carlos Santos Oct 12 '18 at 20:37

Let $X$ be a topological space. The space is metrizable if there exists a metric $d:X\times X\to \mathbb{R}$ such that the open sets of $X$ are generated by the metric $d$.

In general, a space is not metrizable, but there are conditions that make a topological space metrizable. (See Urysohn's metrization theorem)

Early on in a course about topology, you realize the importance of the concept of basis.

In fact, if you want to study a given topology, trying to look at all of the open sets can be impossible, because they can be so different from each other. So what we do is we design a subset of the topology, which is "good enough", meaning that it doesn't necessarily contain all open sets, but by studying its elements, you can say a lot of things about the whole topology.

I'm sure you've seen that basic propositions about stuff like continuity, closure, density, ecc... can be shown to be equivalent to similar statements when instead of taking any open set you take a basis element.

Once you have this clear in mind, you realize that it works also the other way. If you want to define a topology which has certain properties, then you can choose a set of subsets which satisfy the axioms of the basis and which have some nice properties you're looking for, and create a topology from there, by taking all the possible unions of such sets.

This is exactly what happens with the metrizable topology. You have a metric on your set, and you like working with metrics because it allows you to talk about stuff like convergence, Cauchy sequences ecc... in a way very analogous with what you do in calculus. So let's define a topology which allows us to work with it the same way we would work with $\mathbb{R}^n$.

However it's very hard to describe all open sets of an abstract metrizable space. On the other hand, you know what type of sets MUST be there: open balls. Now you look at the axioms of a basis and realize that balls satisfy them. So that's it, you've just taken a potentially hideous set, and defined a topology which allows you to think of it in a way analogous to a Euclidian space, much easier to deal with and visualize.