Is every set metrisable?

Question

Can any set be turned into a metric space? Let $X$ be any set, let $x,y\in X$ define a metric $d:X\times X\rightarrow \mathbb{R}$ by: $$ d(x,y) = \left\{\begin{array}{cc} 1 & x\neq y \\ 0 & x= y \end{array}\right. $$ Obviously $d$ is a valid metric.

If any set can become a metric space why are they so special? Is the only interesting part of a metric space the distance function?

EDIT

Some context:

I was trying to prove a result about metric spaces namely,

Theorem: If $X$ is a metric space with distance function $d:X\times X\rightarrow \Bbb{R}$ and $\varphi:X\rightarrow Y$ is a bijection then $Y$ is metrisable with a distance function $d'(y_1,y_2) = d(\varphi^{-1}(y_1),\varphi^{-1}(y_2))$

I remember feeling all accomplished after proving this result, but it seems rather trivial in light of this. I later found out that this is an actual result presented in some textbooks. Is there any real use for this theorem?

Answer 1

The metric spaces are special because the metric is special. Your question is really vague. So I cannot add much more than that. I, however, mention that the Euclidean space $\mathbf{R}^d$ is not the same metric space (not even the same topological space) as the space $\mathrm{X}$ whose "underlying set" is $\mathbf{R}^d$ and whose metric is the one you give.

ANSWER TO EDIT: The result is useful because it allows you to transport, so to speak, a metric (which you regard as given and special) from the metric space $\mathrm{X}$ onto the set $\mathrm{Y}.$

Answer 2

To answer the question if “is the interesting part of a metric space the distance function?” I would say yes, and that the interesting part of a topological space is the topology, and the interesting part of a group is the operation (rather than the underlying set).

This is because

1. given a set, you can often impose many different (metric space, group, etc) structures on it
2. given a set with structure, you can find many different sets with the “same” (isomorphic) structures

This is what is going on in your example. The interesting part is not that we make $X$ a metric space, but that we make it one whose metric is “the same as” $Y$, showing that the underlying set of a metric space doesn’t really matter (up to cardinality).