Let $\{ f_{k} \}$ be a sequence of continuous functions on $[a, b]$. Show that if $\{ f_{k} \}$ converges uniformly on $(a, b)$, then it converges uniformly on $[a, b]$.
I attempted to use $\delta$ and $\epsilon$ definitions of uniform Cauchy and continuity, but I am really not sure about how to proceed. Below is what I have so far:
Because $\{ f_{k} \}$ converges uniformly on $(a, b)$, $\{ f_{k} \}$ is uniformly Cauchy on $(a, b)$. For every $\epsilon > 0$, there exists a large $K$ s.t. $$|f_{j} (x) - f_{k} (x)| < \frac{\epsilon}{3} \ \textrm{whenever} \ j, k > K \ \textrm{and} \ x \in (a, b)$$
Because $f_{j} (x)$ is continuous on $[a, b]$, there exists $\delta' > 0$ such that $|f_{j} (x) - f_{j} (a)| < \frac{\epsilon}{3}$ whenever $|x - a| < \delta'$. Similarly, since $f_{k} (x)$ is continuous on $[a, b]$, there exists $\delta'' > 0$ such that $|f_{k} (x) - f_{k}(a)| < \frac{\epsilon}{3}$ whenever $|x - a| < \delta''$.
Once you have $|f_{j} (x) - f_{k} (x)| < \frac{\epsilon}{3} \ \textrm{whenever} \ j, k > K \ \textrm{and} \ x \in (a, b)$ you can take limits as $x \to a$ and $x \to b$ to get $|f_{j} (x) - f_{k} (x)| \leq \frac{\epsilon}{3} \ \textrm{whenever} \ j, k > K \ \textrm{and} \ x \in [a, b]$ so we see that the sequence is Cauchy in the closed interval $[a,b]$.