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
is the collection of all sets of the form V
, 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 :)