How to prove f = g?

Question

Let X and Y be topological spaces. Suppose Y is Haursdorff and let f and g be continuous functions from X to Y. Suppose f and g agree on a subset D of X, dense in X. Show that f(x) = g(x) for every x in x, that is f = g.

I could prove this when X and Y are metric spaces, but here we have a general case.

Answer 1

Assume x is an element of X such that f(x) != g(x).

Then let U and V be disjoint neighborhoods of f(x) and g(x) in Y. (Possible because f(x) and g(x) are different and Y is Hausdorff.)

f^-1(U) and g^-1(V) are both open sets of X containing x.

So their interestection is an open set containing x.

Since D is dense in X, their intersection must contain a point d in D.

So d is in f^-1(U) and g^-1(V) and f(d) = g(d). But then f(d) is in U and g(d) is in V, so U and V are not disjoint.

Answer 2

I was about to answer this question, then I read Thomas one. Mine would be almost exactly like his proof, so I'd rather not give it.

Thomas proof is perfect.

Edit:

Since you mentioned you proved this for metric spaces, you probably based your proof on sequences. Then, a different proof came to my mind, based on the concept of net, which is an extension of the concept of sequences. The proof below is probably the same you gave for metric spaces except that the word sequence is replaced by net.

Let x be in X. Since D is dense in X, there's a net N in D that converges to x. Since f and g are continuous, the image nets f(N) and g(N) converge to f(x) and g(x), respectively. Since f and g agree on D, f(N) and g(N) are the same net and its limit is unique, because Y is Hausdorff. So, f(x) = g(x) for every x in X.