Show Uniform Convergence of fn to f in Compact Metric Space

[Enjoicube]

Homework Statement

Alright, here is the problem. Given a compact metric space X, and a sequence of functions fn which are continuous and f_{n}:X->R (reals), also f_n->f (where f is an arbitrary function f:X->R). Also, given any convergent sequence in X x_{n}->x, f_{n}(x_{n})->f(x). The problem is to show that fn converges uniformly to f.

The Attempt at a Solution

Alright, I can prove this relatively easily if I can prove that f (the limit function) is continuous. However, I don't know if this is possible, does anyone see a way to do this? Only little hints if you see a way.

 

[Enjoicube]

Nevermind continuity, I don't think the continuity of f directly matters in this situation. However, maybe a start on the problem would be to assume that f is not uniformly continuous, and then, picking an epsilon e>0, for each n, pick one x in X such that |fn(x)-f(x)|>=e, and so create a sequence of points in X, which then must have a convergent subsequence due to the compactness of X. From there on, I am stumped, even if this is the correct way to solve the problem.

 

[Office_Shredder]

I would think proof by contradiction is a good way to go. If the f_n do not converge to f uniformly, then for all \epsilon, there exists an infinite subsequence of functions f_{n_k} such that for each such function, there is a point x_{n_k} with |f(x_{n_k})-f_{n_k}(x_{n_k})|>\epsilon

Relabel these sequences as f_k and x_k for convenience. x_k has a convergent subsequence which we will label y_k. Now let f_k just be the functions corresponding to the points y_k

What can you say about |f(y_k)-f_k(y_k)|

 

[Enjoicube]

Haha, took a walk and that's exactly what came to me. I think I had a proof of a similar theorem stuck in my head and wanted to follow that technique.