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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?

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Hint: Apply $\epsilon/3$-argument to the inequality $|g(x)-g(y)|\leq|g(x)-g_{n}(x)|+|g_{n}(x)-g_{n}(y)|+|g_{n}(y)-g(y)|$.

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  • $\begingroup$ Thank you. I can see how you would apply that argument to the closed interval. How would it change for the open interval, however? $\endgroup$ – user485755 Nov 19 '17 at 4:13
  • $\begingroup$ Actually this argument does not depend on whether open or closed interval because you have assumed that $g_{n}$ are uniformly continuous on $(0,1)$. $\endgroup$ – user284331 Nov 19 '17 at 4:15
  • $\begingroup$ So the argument for both open and closed intervals would be the same? It seems counterintuitive that my textbook would list it this way then. $\endgroup$ – user485755 Nov 19 '17 at 4:19
  • $\begingroup$ 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. $\endgroup$ – user284331 Nov 19 '17 at 4:20

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