TITLE: Pointwise vs. Uniform Convergence
QUESTION [97 upvotes]: This is a pretty basic question. I just don't understand the definition of uniform convergence.
Here are my given definitions for pointwise and uniform convergence:
Pointwise convergence: Let $X$ be a set, and let $F$ be the real or complex numbers. Consider a sequence of functions $f_n$ where $f_n:X\to F$ is a bounded function for each $n\in \mathbb N$. $f:X\to F$ is the pointwise limit of $f_n$ if for every $x \in X$, $$\lim_{n\to \infty}f_n(x)=f(x).$$
Uniform convergence: Let $f_n$ be a sequence of functions in the set of all bounded functions from $X$ to $F$ where $F$ is the real or complex numbers. The sequence is said to converge uniformly to a bounded function $f:X \to F$ if, given $\epsilon>0$, there exists an $N\in \mathbb N$ s.t. $\sup\{|f_n(x)-f(x)| : x \in X \}<\epsilon$ for $n\ge N$
I'm sorry I don't have a more specific question. I just don't see the exact relation/difference between the two definitions. I've asked two different professors to explain this to me but neither of their explanations helped.
Edit: Attempting to show that uniform convergence implies pointwise convergence
if $f_n$ converges uniformly to f, then $\sup\{|f_n(x)-f(x)| : x\in X \}$ for $n\ge N$. Thus, $|f_n(x)-f(x)|<\epsilon$ for $n\ge N$, which is the definition of pointwise convergence.

REPLY [3 votes]: The pointwise convergence depends on each $x$, that is you need to count and fix every $x$ before passing to the limit. On the other hand, uniform convergence is independent of $x$.
Hence, the pointwise convergence is enough when your space $X$ is a countable set and not enough when you are working with an uncoutable set like $[0,1]$, in such case you need a uniform convergence to generalise your convergence on $[0,1]$.