Pointwise vs. Uniform Convergence 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.
 A: 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 $[0,1]$. 
A: Comparison
Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).
Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$

Uniform convergence means there is an overall speed of convergence.
In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.
Another Approach
One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.
A: $f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$. Notice this is a pointwise (local) criterion.
On the other hand, $f_n\to f$ uniformly on $(a,b)$ if $\sup_{a< x<b}|f_n(x)-f(x)|\to 0$ as $n\to\infty$. This is a "global" criterion in that is requires the maximum of all the pointwise errors on $[a,b]$ to tend to zero.
As an example, $f_n(x)=x^n$, $0\le x\le 1$ converges pointwise to $f(x)=\begin{cases} 0, &0\le x<1,\\ 1, &x=1.\end{cases}$, because the first condition above holds, but the convergence is not uniform since the second condition does not hold.
A: It may help if you unfold the definition of limit in pointwise convergence.
Then pointwise convergence means that for each $x$ and $\epsilon$ you can find an $N$ such that (bla bla bla). Here the $N$ is allowed to depend both on $x$ and $\epsilon$.
In uniform convergence the requirement is strengthened. Here for each $\epsilon$ you need to be able to find an $N$ such that (bla bla bla) for all $x$ in the domain of the function. In other words $N$ can depend on $\epsilon$ but not on $x$.
The latter is a stronger condition, because if you have only pointwise convergence, it may be that some $\epsilon$ will require arbitrarily large $N$ for some $x$s.
For example, the functions $f_n(x)=\frac{x}{n}$ converge pointwise to the zero function on $\mathbb R$, but do not converge uniformly. For example, if we choose $\epsilon=1$, then the convergence condition boils down to $N>|x|$. For each $x\in\mathbb R$ we can find such an $N$ easily, but there's no $N$ that works simultaneously for every $x$.
A: Let $(f_n)$ be a series of functions defined on a subset $E$ of $\mathbb R$, and suppose that $f:E\to\mathbb R$ has the property that for all $x\in E$,
$$
f(x)=\lim_{n\to\infty}f_n(x) \, .
$$
Then, $f$ is called the limit function of the sequence $(f_n)$, and the functions $(f_n)$ are said to converge pointwise to the function $f$.
It is natural to ask the question: "to what extent do the functions $f_n$ end up resembling the function $f$"? If no further hypotheses are added, then the resemblance may be weak. For instance, the limit of a sequence of continuous functions might be discontinuous, the limit of a sequence of integrable functions might be non-integrable, and so on.
Why does this alarming state of affairs occur? Well, if we unfold the meaning of the statement "for all $x\in E$, $f(x)=\lim_{n\to\infty}f_n(x)$", it becomes
$$
(\forall x\in E)(\forall\varepsilon>0)(\exists N\in\mathbb N)(\forall n\in\mathbb N)(n\ge N\implies |f_n(x)-f(x)|<\varepsilon) \, .
$$
This formidable looking statement holds the answer to our question. Note that the clause "$(\forall x\in E)$" appears before the clause "$(\exists N\in\mathbb N)$", and so the value of $N$ is allowed to depend on the value of $x$. Therefore, there is a possibility that no value of $N$ works for every value of $x$.
It is easiest to understand this with an example. If $f_n(x)=x^n$ on $[0,1]$, then $(f_n)$ converges pointwise to the function $f(x)$ that equals $0$ when $0\le x<1$ and equals $1$ when $x=1$. This is trivial to see for $x=1$, and for $x\neq1$ we can proceed as follows: given an $\varepsilon>0$, we can choose an $N\in\mathbb N$ satisfying $N>\log(\varepsilon)/\log(x)$. Then, if $n\ge N$, it follows that $x^n\ge x^N>x^{\log(\varepsilon)/\log(x)}=\varepsilon$.
On the other hand, in the above example, it is impossible to choose an $N\in\mathbb N$ which does not depend on the value of $x$. In other words, it is not the case that
$$
(\forall\varepsilon>0)(\exists N\in\mathbb N)(\forall n\in\mathbb N)(\forall x\in E)(n\ge N\implies|f_n(x)-f(x)|<\varepsilon)\tag{*}\label{*}
$$
To prove this, we can have to demonstrate the negation of $\eqref{*}$, which is
$$
(\exists\varepsilon>0)(\forall N\in\mathbb N)(\exists n\in\mathbb N)(\exists x\in E)(n\ge N\text{ and }|f_n(x)-f(x)|\ge\varepsilon) \, .
$$
Take $\varepsilon=1/2$, and consider that if $x=\sqrt[N]{1/2}$ then $x^N=1/2$.
What this shows is that no matter how big $N$ is, there is a value of $x$ such that $|f_N(x)-f(x)|=1/2$, and consequently the function $f_N$ looks noticeably different to $f$ at this point.
If, on the other hand, the condition $\eqref{*}$ holds, then for any $\varepsilon>0$ thrown at you, there is an $N\in\mathbb N$ such that the functions $f_N,f_{N+1},f_{N+2},\dots$ are "within $\varepsilon$" of $f$: at no point does the value of $f$ differ from one of these functions by a margin greater than $\varepsilon$. Consequently, the functions $f_n$ really do end up visually resembling the function $f$, and many of the properties that the functions $f_n$ have will end up being shared by $f$.
The condition $\eqref{*}$ is precisely what uniform convergence means.
