Metric spaces are metrizable?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Just to give a clear example of a non-metrizable space. Any set with the indiscrete topology, where the only open sets are the empty set and the space itself. How do we know it's not metrizable? Because it's not Hausdorff, and all metric spaces are. Consider any two different points p,q in a metric space. Then, by the axioms of metric spaces, d(p,q):=r>0. Then the balls centered at p,q respectively, both with radius r/3, are neighborhoods that separate p from q, showing them to be Hausdorff.