Hi I want to see if this I am doing this right.
Suppose $X$ is Hausdorff, $Y$ is a metric space, and $\langle f_n\rangle$ is equicontinuous.
Show that $$C=\{x\in X: \langle f_n(x)\rangle \text{ is a cauchy sequence in $Y$}\}$$ is closed.
Proof: Suppose that $\langle x_\alpha\rangle\subset C$ is a net that converges to $x$ and let $d$ be the metric on $Y$. Then $$d(f_n(x),f_m(x))\leq d(f_n(x),f_n(x_\alpha))+d(f_n(x_\alpha),f_m(x_\alpha))+d(f_m(x_\alpha),f_m(x))$$ for any $\alpha,n,m$.
So if I fix an $\epsilon$, the first and last terms can be controlled by equicontinuity: I can produce a $U$ such that each is less that $\epsilon/2$. Then pick an element in the net that satisfies this condition. Now, since $\alpha$ is fixed I can use the cauchy condition on $x_\alpha$.
Is this correct? This is a prelim problem and so this argument seems suspiciously too easy.