Uniform convergence implies pointwise convergence

Question

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

Answer 1

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.

Answer 2

To begin with, it is useful to remind that uniform convergence is equivalent to convergence with respect to the sup norm.

Having said that, let $f_{n}:(X,d_{X})\to(Y,d_{Y})$ converges uniformly to $f:(X,d_{X})\to(Y,d_{Y})$.

Thus for every $\varepsilon > 0$ and every $x\in X$ there corresponds $n_{\varepsilon} \in \mathbb{N}$ such that we have that \begin{align*} n\geq n_{\varepsilon} & \Rightarrow d_{Y}(f_{n}(x),f(x)) \leq \sup_{x\in X}d_{Y}(f_{n}(x),f(x)) \leq \varepsilon \end{align*}

Hence $f_{n}$ converges to $f$ point-wise.

Hopefully this helps!