e/3 argument? Uniform convergence from a topoloical space to a metric space

Question

I am trying to prove that if we take a sequence (say fn) in C(X,Y) that converges uniformly to to a function f: X to Y, then f must be an element of the space C(X,Y). Where f moves from a topology to a metric. Is this along the lines of an e/3 argument?

Answer 1

Let $(f_n)$ be a sequence in $C(X,Y)$ converging uniformly to $f$. We aim to show that $f$ is continuous at $x_0\in X$. Let $\varepsilon>0$.

Choose $n_0\in\mathbb N$ large enough so that $d(f_n(x),f(x))<\varepsilon/3$ for all $x\in X$ and all $n\geq n_0$. (We can do this because $f_n\to f$ uniformly as $n\to\infty$.) Fix $n\in\mathbb N$ with $n\geq n_0$.

Since $f_n$ is continuous at $x_0$, there exists an open neighbourhood $U$ of $x_0$ such that $d(f_n(x),f_n(x_0))<\varepsilon/3$ for all $x\in U$. Now, for each $x\in U$, we have $d(f(x),f(x_0))<\varepsilon$ (these details need filling in). Thus $f$ is continuous at $x_0$. Since $x_0$ was arbitrary, it follows that $f$ is continuous.