# Proof that sequence of uniformly continuous functions which converges to a function is uniformly continuous

Part 1) Let $g_n:[0,1]\to R$ be a sequence of uniformly continuous functions which converges uniformly to a function $g:[0,1]\to R$. Prove that $g$ is uniformly continuous.

Part 2) Let $g_n:(0,1)\to R$ be a sequence of uniformly continuous functions which converges uniformly to a function $g:(0,1)\to R$. Prove that $g$ is uniformly continuous.

The second part is just a variant of the first, except with an open interval instead of a closed one. Does anyone know of a way prove both of these rigorously?

Hint: Apply $\epsilon/3$-argument to the inequality $|g(x)-g(y)|\leq|g(x)-g_{n}(x)|+|g_{n}(x)-g_{n}(y)|+|g_{n}(y)-g(y)|$.
• Actually this argument does not depend on whether open or closed interval because you have assumed that $g_{n}$ are uniformly continuous on $(0,1)$. – user284331 Nov 19 '17 at 4:15
• As I said because you have assumed that $g_{n}$ are uniformly continuous on both cases, if it were only continuous on $(0,1)$, of course there is a counterexample. – user284331 Nov 19 '17 at 4:20