# Uniform convergence preserves continuity Proof verification

Proposition:

Let $$f_n \rightrightarrows f$$ and each $$f_n$$ is continuous at $$x_0$$, then $$f$$ is continuous at $$x_0$$

I know there are some posts that have proof of this proposition, but I didn't find any that proves it with the sequential definition of continuity, instead, they use the $$\delta - \epsilon$$ definition.

When I saw this problem, I thought it'd be more natural to prove it with the sequential definition of continuity because we are working with sequences. However, I'm not sure if my proof is correct.

So here it is.

First I want to justify this inequality (triangle inequality with 2 points?)

$$|a - b| = |a+c-c+d-d-b| = |(a-d) + (d-c) + (c-b)| \leq |a-d| + |d-c| + |c-b| \tag{1}$$

So, suppose that $$f_n: I \to \mathbb{R} \text{ where } I \subseteq \mathbb{R}$$, let $$(x_n)$$ be a sequence in $$I$$ such that $$x_n \to x_0\text{ , } x_0 \in I$$.

Since each $$f_n$$ is continuous, we have $$f_n(x_n) \to f_n(x_0)$$. So there exists an $$N_1$$ such that $$\forall q \geq N_1 \rightarrow |f_n(x_q) - f_n(x_0)| < \epsilon /3$$

Since $$f_n \rightrightarrows f$$ there exists $$N_2$$ such that $$\forall n \geq N_2 \rightarrow |f_n(x_q) - f(x_q)| < \epsilon /3 \text{ and } |f_n(x_0) - f(x_0)| < \epsilon /3 \tag{2}$$

Let $$N = max(N_1, N_2)$$ and in the inequiality $$(1)$$, let $$a = f(x_q)$$, $$b = f(x_0)$$, $$c = f_n(x_0)$$ and $$d = f_n(x_q)$$

so $$\forall q,n \geq N$$ we have

$$|f(x_q) - f(x_0)| \leq |f_n(x_q) - f(x_q)| + |f_n(x_q) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \epsilon$$ (I also used $$|a-b| = |b-a|$$)

Thus $$f(x_q) \to f(x_0)$$ and $$f$$ is continuous at $$x_0$$.

Is it correct? I don't feel so sure about it. Also. If the sequence didn't converge uniformly, but only point-wisely, in $$(2)$$, there'd be different $$N_2, N_3$$ that would guarantee those inequalities, so If I chose an $$N$$ to be equal to the max of those two, I could guarantee those inequalities without the hypothesis of uniform continuity, and I could "prove" that point-wise convergence preserves continuity which is false, this bothers me a lot.

If the proof is wrong(I think it's wrong), how can I fix it?

Thanks for taking the time to read it.

I think that the problem with your proof is that you chose $$q$$ in a way that depends on the value of $$n$$ and in the next line you chose $$n$$ in a way that depends on $$q$$. Uniform convergence is a universal property that holds for all elements in $$I$$ and you haven't used this in your proof.
Suppose that $$x_n\rightarrow x_0$$. Given $$\epsilon>0$$, choose $$N_1\in\mathbb N$$ such that for all $$x\in I$$ and $$n>N_1$$, $$|f(x)-f_n(x)|<\frac{\epsilon}{3}.\quad (\dagger)$$ Since $$f_n$$ continuous, there is an $$N_2$$ such that for all $$m>N_2$$, $$|f_n(x_m)-f_n(x_0)|<\frac{\epsilon}{3}.\quad (\dagger\dagger)$$ Then as you stated, for all $$m>N_2$$ (Here, we're not varying $$n$$.) \begin{aligned}|f(x)-f(x_m)|&\le|f(x)-f_n(x)|+|f_n(x)-f_n(x_m)|+|f_n(x_m)+f(x_m)|\\ &\le \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}. \end{aligned} Here we used $$(\dagger)$$ for the first and the third term and $$(\dagger\dagger)$$ for the second term.
Also as an aside, the counterexample to the claim that pointwise convergence preserves continuity is \begin{aligned}f_n:[0,1]&\rightarrow \mathbb R\\ x&\mapsto x^n\end{aligned} You can easily check that $$f_n$$ converges pointwise to a function $$f(x)=\begin{cases}0&\text{if x\neq 0,}\\ 1&\text{if x=0.}\end{cases}$$