I am having trouble proving a simple proposition regarding uniform convergence and pointwise convergence in Real Analysis.
Problem:
Suppose that $\left(f_{n}\right)$ is a sequence of functions $f_{n}: A \rightarrow \mathbb{R}$ such that $\left(f_{n}\right)$ converges uniformly to $f: A \rightarrow \mathbb{R}$. Prove that $\left(f_{n}\right)$ also converges pointwise to $f: A \rightarrow \mathbb{R}$
Relevant definitions/Notations:
One says that $f_{n} \rightarrow f$ uniformly on $A$ if, for every $\epsilon>0,$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$ for all $x \in A$.
Finally, we say that $f_{n} \rightarrow f$ pointwise on $A$ if, given $\epsilon>0,$ for each $x \in A$ there exists $N \in \mathbb{N}$ such that $n>N$ implies that $$\left|f_{n}(x)-f(x)\right|<\epsilon$$
Attempts:
I tried to write down the definition of uniform convergence and then arguing that, in particular, since $N \in \mathbb{N}$ from uniform convergence works for any given point in $x \in A$, then it must work for a given point and from that conclude pointwise convergence.
I also checked some proofs like the one that states that uniform continuity implies continuity and write something similar, but i just dont know how to do it.
I would highly appreciate a detailed proof regarding this fact, i am trying to become proeficient at proof writing.
Thanks in advance, Lucas
Best Answer
Your attempt looks fine. Take $x\in A$; you want to prove that $\lim_{n\to\infty}f_n(x)=f(x)$. Now, take $\varepsilon>0$; you want to prove that there is some $N\in\Bbb N$ such that$$n\geqslant N\implies\bigl|f_n(x)-f(x)\bigr|<\varepsilon.\tag1$$So, take $N\in\Bbb N$ such that$$(\forall a\in A)(\forall n\in\Bbb N):n\geqslant N\implies\bigl|f_n(a)-f(a)\bigr|<\varepsilon.$$It follows from this that, for such a $N$, $(1)$ holds.