MHB Is the Uniform Limit of Continuous Functions on a Subset also Continuous?

[Chris L T521]

Sorry for posting this late. Here's this week's problem.

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Problem: Suppose that $f_n$ is a sequence of continuous functions on a subset $E\subseteq\mathbb{R}$ and that $f_n$ uniformly converges to a function $f$ on $E$. Show that $f$ is also continuous on $E$.

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[Chris L T521]

No one answered this week's question. Here's my solution.

Spoiler

Proof: Suppose that $\{f_n\}$ is a sequence of continuous functions over $E$ and $f_n\rightarrow f$ uniformly. Since $f_n$ is continuous, then $\forall\,\epsilon>0,\,\exists\,\delta>0:0<|x-y|<\delta\implies\left|f_n(x)-f_n(y)\right|<\frac{\epsilon}{3}$. Since $f_n\rightarrow f$ uniformly, then $\forall\epsilon>0,\,\exists\, N:\forall n\geq N$, $\left|f_n(x)-f(x)\right|<\frac{\epsilon}{3}$. Then for any $x,y\in E$, \[\begin{aligned}\left|f(x)-f(y)\right| &= \left|f(x)-f_n(x)+f_n(x)-f(y)\right|\\ &\leq \left|f_n(x)-f(x)\right|+\left|f_n(x)-f(y)\right|\\ &\leq \left|f_n(x)-f(x)\right|+\left|f_n(x)-f_n(y)\right|+\left|f_n(y)-f(y)\right|\\ &\leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+ \frac{\epsilon}{3} \\ &=\epsilon.
\end{aligned}\]
Q.E.D.