Metric spaces are metrizable?

I know a topological space need not be a metric space and every metric space can be considered a topological space (which is the one induced by a metric defined on it).

But, I've come across this question whether it is safe to say that a metric space is 'metrizable'. For ex: The uniform topology(one that is induced by the uniform metric). Does it make sense if the uniform topology is called 'metrizable' since it's a metric space?

I hope someone can clarify this for me.

• Usually one talks about whether a topological space is metrizable, but if it was induced by a metric, it must be metrizable – J. W. Tanner Dec 11 '19 at 10:49

If $$\langle X,d\rangle$$ is a metric space - i.e. If $$X$$ is a set and $$d$$ is a metric on $$X$$ - then $$d$$ induces a topology $$\tau_d$$ on $$X$$.

So starting with metric space $$\langle X,d\rangle$$ the topological space $$\langle X,\tau_d\rangle$$ is induced.

If we start with a topological space $$\langle X,\tau\rangle$$ then the natural question arises: "does there exist a metric on $$X$$ such that $$\tau=\tau_d$$?"

If the answer is "yes" then topological space $$\langle X,\tau\rangle$$ is by definition metrizable.

Further the answer will always be "yes" for topological spaces that are induced by means of a metric.

In short and with an abuse of language: "a metric space is metrizable".

A metric space is not a topological space and a metrizable space is a topological space, so the labeling is formally not correct.

A topological space $$(X, \mathcal{T})$$ is called metrizable if there exists a metric

$$d: X \times X \to \mathbb{R}^+$$

such that

$$\mathcal{T}= \{A \subseteq X \mid \forall a \in A: \exists \epsilon > 0: B_d(a, \epsilon) \subseteq A\}$$

I.e, the topology on $$X$$ is induced by a metric.

So, if we consider a metric space as a topological space (by the topology induced by the metric), it is trivially a metrizable topological space.

Note that not every topological space is a metrizable space. Indeed, the space

$$(\{0,1\}, \mathcal{S}= \{\emptyset, \{0\}, \{0,1\}\})$$ is easily seen to be a topological space but the topology does not contain $$\{1\}$$, so its complement $$\{0\}$$ is not closed. In a metrizable space, every singelton is closed so the above topological space can not be a metrizable space.

A topological space is said to be metrizable if there is a metric ("if the space is homeomorphic to a metric space"). The term is useful when talking about sufficient conditions for a topological space to be metrizable.

As others pointed out, a metric space has an induced topology that is (by definition) metrisable. But a metric space comes with a metric and we can talk about Cauchy sequences and total boundedness (which are defined in terms of the metric) and in a metrisable topological space there can be many compatible metrics that induce the same topology and so there is no notion of a Cauchy sequence etc. So what is pre-giveen (a metric or a topology ) determines what type we have and what notions are defined for it. They belong to different categories of objects.