Let $r_{1},r_{2},...$ a sequence that includes all rational numbers in $[0,1]$. Define $$f_n(x)=\begin{cases}1&\text{if }x=r_{1},r_{2},...r_{n}\\0&\text{otherwise}\end{cases}$$
this sequence converges($lim_{n\to \infty}f_n(x)$) to dirichlet function in $[0,1]$
$$f(x)=\begin{cases}1&\text{if }x\in\mathbb Q\\0&\text{otherwise}\end{cases}$$
but does it converges uniformly? I think that the answer is no (but I´m not sure) my attempt to prove that this sequence of functions does not converges uniformly to Dirichlet function (if this was the case):
If i need to prove that a sequence of functions $f_{n}(x)$ converges uniformly to another function $f(x)$ i need to prove that:
$\forall \epsilon\gt 0, \exists N$ (that dependes on $\epsilon$)so that $\forall n\gt N$ so that $|f_{n}(x)-f(x)|\lt \epsilon \forall x\in[a,b]$
hence in order to prove that a sequence of functions $f_{n}(x)$ does not converges uniformly to $f(x)$ I need to prove that:
$\exists \epsilon_{0}\gt 0, x\in[a,b]$ so that $\forall N\gt 0 \exists n\gt N$ so that $|f_{n}(x)-f(x)|\ge \epsilon_{0}$
let $\epsilon_{0}=1$ and $x=r_{n+1}$ we have that
$f_{n}(x)=0 \forall n\in \mathbb N $ but $f(r_{n+1})=1$ (because $r_{n+1}$ is an element of the sequence of rational numbers in $[0,1]$) hence
$|f_{n}(r_{n+1})-f(r_{n+1})|=1=\epsilon_{0}$
I think that my idea is correct(if this sequence does not converges uniformly) but I don´t know how to formaly prove it ; I don´t know how to put $N$ in my proof
So I would like you to tell me if this sequence converges uniformly or not, and if this was the case how can I formaly write the proof
I would really appreciate your help :)
You basically have it right and this does not converge uniformly. The thing that creates you issues is that you did not really negate the condition properly.
Actually you get for non uniform: there exists $\varepsilon$ such that for all $N$ there exists $n > N$ such that there exists $x$ such that $|f_n(x) -f(x)| \ge \varepsilon$.
Since you made a good start by yourself I am conveinced with this you will be able to fill in the details.