First of all, go read metric space. Each metric space
is a set of points `X` combined with a distance rule `d:
X `x `X -> R` following
particular rules.

Let `P` be the set of real numbers greater than `0`. For an `x ∈ X` and `i ∈> P`, `N`_{i}(x) (the "`i`-neighborhood of `x`") is the set of points `y` such that `d (x, y) < i.` (Imagine all of the possible spheres in space, minus their surfaces).

Suppose **i > 0**, **y ∈ N**_{i}(x) and **y != x**. Notice that **d (x, y) > 0**, otherwise **d** would not be a metric. And **d (x, y) < i** from the definition of **N**_{i}(x). Thus, **0 < d (x, y) < i**.

So, given a point `y ∈ N`_{i}(x) such that **y != x**, some` i'` must exist, `0 < i' < i`, for which `N`_{i'}(y) ⊂ N_{i}(x).

Because of that, for all` x`_{1},
x_{2} ∈ X and `i`_{1}, i_{2} ∈ P, the intersection of `N`_{i1}(x_{1}) and `N`_{i2}(x_{2})
(if not empty) must be the union of several `N`_{i'}(y). (This part isn't necessary for the proof, and doesn't appear in actual descriptions, but I included it because it was the link I needed to understand it all). Briefly put, it's turtles all the way down.

Because of that, the collection of all `N`_{i}(x)
for every `x `_{∈} X and `i `_{∈} P can be used as a base for a topology of **X**.

This topology is the called the space's "metric topology". It
allows us to describe the metric space in terms that apply to topological
spaces.