Sequence of continuous function converging pointwise to continuous function is equicontinuous?

I've proven the following "theorem":

Let $$I \subset \mathbb{R}$$ be an interval, $$(f_n: I \rightarrow \mathbb{R})_{n \in \mathbb{N}}$$ be a family of continuous functions converging pointwise to a continuous function $$f: I \rightarrow \mathbb{R}$$ on $$I$$. Then: $$(f_n)_{n \in \mathbb{N}}$$ is equicontinuous on I.

Now my problem is, that here Equicontinuity of a pointwise convergent sequence of monotone functions with continuous limit additionally the $$f_n$$ have to be monotonic. So is my proof a generalization, or am I just missing something? Here is my proof:

Proof: Let $$\epsilon > 0$$. Observe first: $$\begin{equation} | f_n(x) - f_n(y) | \leq |f_n(x) - f(x)| + |f_n(y) - f(y)| + |f(x)- f(y)| \end{equation}$$ Now there is by pointwise convergence of $$(f_n)_{n \in \mathbb{N}}$$ a $$N \in \mathbb{N}$$ such that for all $$n \geq N$$ we have $$|f_n(x) - f(x)|<\frac{\epsilon}{3}$$ and $$|f_n(y) - f(y)| < \frac{\epsilon}{3}$$. Further there is a $$\delta > 0$$ such that $$|f(x) - f(y)| < \frac{\epsilon}{3}$$ for $$|x-y| < \delta$$ by continuity of $$f$$. Hence we have shown, that there is a $$N \in \mathbb{N}$$ and a $$\delta > 0$$ such that for all $$n \geq N$$ $$\begin{equation} |f_n(x) - f_n(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{equation}$$ holds. Now let $$n < N$$. Then, by continuity of $$f_n$$ there is a $$\delta_n$$ such that $$(x-y) < \delta_n$$ implies $$|f_n(x) - f_n(y)| < \epsilon$$. Setting $$\begin{equation} \tilde{\delta} = \min_{n < N} \delta_n \end{equation}$$ (which exists and is greater than $$0$$) we obtain, that for all $$n < N$$ the following holds: $$\begin{equation} |x - y| < \tilde{\delta} \Rightarrow |f_n(x) - f_n(y) | < \epsilon \end{equation}$$ Setting now $$\hat{\delta} = \min \{\delta, \tilde{\delta} \}$$ we have, that for all $$n \in \mathbb{N}$$ the following holds: $$\begin{equation} |x-y| < \hat{\delta} \Rightarrow |f_n(x)- f_n(y) | < \epsilon \end{equation}$$ Hence we have shown, that for all $$\epsilon > 0$$ there is a $$\hat{\delta} > 0$$ such that forall $$n \in \mathbb{N}$$ we have, that $$|x-y| < \delta$$ implies $$|f_n(x) - f_n(y)| < \epsilon$$.

• Can your interval be open? – kimchi lover Dec 15 '18 at 16:24
• Yes, i stated no assumptions regarding the interval. – warpfel Dec 15 '18 at 16:27
• The functions $f_n(x)=x^n$ converge to 0 pointwise on $(0,1)$ but not uniformly. – kimchi lover Dec 15 '18 at 16:28

Your proof is wrong, because it basically impmies that every pointwise convergence of continuous functions to a continuous function is uniform. I think the flaw is that your $$N$$ depends both on $$x$$ (not important, it is fixed) but in $$y$$ as well!