Suppose we have a topological space, X, its

topology, T, and a subset of its points,

`A`. Recall that the topology of a

topological space is the collection of its

open subsets, then the

**subspace topology** on

`A` is the collection of all sets of the form

`V` intersect

`A`, where

`V` is one of the open subsets of X in T.

If we consider `A` as the set of points in its own topological space, then `A` is trivially open for that space. Let `W` be the intersection of `V` and `A` for some `V` open in X. By the openness of `V` in X, for any point `v` in `V` there is some `d` such that for all `x` in X where |`x` - `v`| < `d`, `x` is in `V`. Consider a point `w` in `W` and, given it's also in `V`, its corresponding `d` establishing the openness of `V` in X. Then a point `x` in X with distance less than `d` from `w`, if it's in `A`, must also be in `W`, since by the openness of `V` it is in `V` and `W` is `A` intersect `V`. Hence all such sets are open in the subspace topology on `A`, and the collection of these sets is a fully qualified topological space 'on `A`.'

Actually, the above paragraph only really works if X is a metric space. Hopefully someone will add a more general writeup later :)