Is it possible to define metric spaces from pure topological concepts without the need to define a distance function? Reading a book this question comes to my mind: would it be possible to define metric spaces using only topological concepts, without an explicit definition of a metric?
For example, can we say that all Hausdorff spaces are metric spaces?
 A: It is possible to define the class of metrizable spaces in purely topological terms. Indeed, there are many so-called metrization theorems that characterize the class of metrizable spaces topologically; here is a bundle of four of them.

Theorem. The following are equivalent for a topological space $\langle X,\tau\rangle$:
  
  
*
  
*$X$ is $T_3$ and has a $\sigma$-locally finite base.  
  
*$X$ is $T_3$ and has a $\sigma$-discrete base.  
  
*$X$ has open covers $\mathscr{G}_n$ for $n\in\Bbb N$ such that
  
  
*
  
*if $G_0,G_1\in\mathscr{G}_{n+1}$ for some $n\in\Bbb N$, and $G_0\cap G_1\ne\varnothing$, then there is a $G_2\in\mathscr{G}_n$ such that $G_0\cup G_1\subseteq G_2$, and  
  
*whenever $x\in U\in\tau$, there is an $m\in\Bbb N$ such that $\bigcup\{G\in\mathscr{G}_n:x\in G\}\subseteq U$ for each $n\ge m$.
  
  
*$X$ is a paracompact Moore space.  
  
*$X$ is metrizable.
  

However, a metric space by definition comes equipped with a metric, and a metric space with more than one point always has more than one metric that generates the same topology. Thus, we don’t actually have a metric space until we specify the metric, even when we know that there is one that generates the desired topology. If a topological space $\langle X,\tau\rangle$ satisfies one of the four conditions above, there is a way to construct a metric on $X$ that generates the topology $\tau$, but there are many other metrics on $X$ that generate the same topology.
