Alternate uniform convergence proof

To ensure that I understand the material well and to work on proof techniques, I like to try to create alternate proofs for the nice proofs my advanced calc text provides. (I'm not trying to create the slickest proof, but trying to better learn the material.) I believe the following works as a proof by contraposition that if a sequence of continuous functions $(f_n)$ converges uniformly to $f$, then $f$ is continuous. I'd appreciate having one more set of eyes to see if I've made any missteps.

Claim: Let $(f_n)$ be a sequence of functions on $I$ that converges uniformly on $I$ to $f$. If $f$ is discontinuous at $c\in I$, then $(f_n)$ is eventually discontinuous at $c$.

Proof: If $f$ is discontinuous at $c$, then there exists $\epsilon_0>0$ and a sequence $(x_k)$ in $I$ such that $$\lim_{k\to\infty}x_k=c, \hspace{3mm} \text{but} \hspace{3mm} |f(x_k)-f(c)|\geq \epsilon_0 \hspace{3mm} \text{for all} \hspace{3mm} k \in \mathbb{N}.$$

Since $(f_n)$ converges uniformly, there exists $N\in\mathbb{N}$ such that for all $n\geq N$, $\Vert f_n-f \Vert_{I}\leq\frac{\epsilon_0}{4}$. For all $n\geq N$, if $x_k$ is any term of the sequence, two applications of the reverse triangle inequality gives \begin{align*} |f_n(x_k)-f_n(c)|&=\big|\big(f(x_k)-f(c) \big)-\big(f(x_k)-f_n(x_k) \big)-\big( f_n(c)-f(c)\big)\big|\\ &\geq \big|\big(f(x_k)-f(c) \big)-\big(f(x_k)-f_n(x_k) \big)\big|-\big|\big( f_n(c)-f(c)\big)\big|\\ &\geq |f(x_k)-f(c)|-|f_n(x_k)-f(x_k)|-|f_n(c)-f(c)|\\ &\geq \epsilon_0-\frac{\epsilon_0}{4}-\frac{\epsilon_0}{4}\\ &=\frac{\epsilon_0}{2}. \end{align*}

Thus, eventually (for all $n\geq N$) the terms of $(f_n)$ have the property that although $$\lim_{k\to\infty}x_k=c,$$ the terms of the sequence $\{f_n(x_k)\}_{k=1}^{\infty}$ are bound away from $f_n(c)$, so that $$\lim_{k\to\infty}f_n(x_k)\neq f_n(c).$$ Hence, for all $n\geq N$, $f_n$ fails to be continuous at $x=c$.

We can now argue by contraposition that if a sequence of continuous (or eventually continuous) functions $(f_n)$, $f_n:I\longrightarrow\mathbb{R}$ for all $n\in\mathbb{N}$, converges uniformly to a function $f:I\longrightarrow\mathbb{R}$, then $f$ must be continuous to $I$. For, if $f$ was discontinuous at some point $c\in I$, then $(f_n)$ would eventually be discontinuous at $c$.

• I liked your drawing. I have never really seen the image of functions from a subset of $\mathbb{R}$ to $\mathbb{R}$ be drawn like on the real line. That is sometimes what is done to visualize graphs of complex functions, since although $\mathbb{R}^2$ can be visualized easily, $\mathbb{C}^2$ puts forward a much bigger visualization problem. Mar 9, 2018 at 2:25