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## Homework Statement

This is a nice one, if it's correct.

Show that if (fn) is a sequence of elements of C(X, Y) (where Y is a metric space) which converges uniformly, then the collection {fn} is equicontinuous.

## The Attempt at a Solution

Let ε > 0 be given and let x0 be a point of X. The proof is based on the following:

d(fn(x0), fn(x)) <= d(fn(x0), f(x0)) + d(f(x0), f(x)) + d(f(x), fn(x)).

Since fn converges uniformly, we can find a positive integer N such that, for all x in X, and for n >= N, we have d(f(x), fn(x)) < ε/3. Since f is continuous, for ε/3 choose a neighborhood U of x0 such that whenever x is in U, we have d(f(x0), f(x)) < ε/3. Now the upper inequality becomes d(fn(x0), fn(x)) < ε, for all x in X (and hence for all x in U), and for all n >= N, i.e. for all but finitely members of the collection {fn}. Now, for every of the remaining members of {fn], use continuity of fn at x0 to find neighborhoods of x0 such that d(fn(x0), fn(x)) < ε. Now, intersect this finite number of neighborhoods of x0 with the neighborhood U, call this intersection U'. Clearly, for any x in U, we have d(fn(x0), fn(x)) < ε, for all elements of {fn}.