What is the main difference between pointwise and uniform convergence as defined here? I have a little confusion here. I have seen the following several times and seem to be a bit confused as to differentiating them.
Let $E$ be a non-empty subset of $\Bbb{R}$. A sequence of functions $\{f_n\}_{n\in \Bbb{N}},$ converges pointwise to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
On the other hand $\{f_n\}_{n\in \Bbb{N}},$ converges uniformly to $f$ on $E$ if and only if \begin{align}f_n(x)\to f(x),\;\forall\,x\in E.\end{align}
QUESTION:
Why is $f_n(x)\to f(x),\;\forall\,x\in E,$ is used for both uniform and pointwise convergence or I'm I missing something important? Can't we distinguish them?
 A: $f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$
$f_n$ converges uniformly means that  for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.
In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.
A: Uniform convergence is actually $\mathcal L^\infty$ convergence, i.e. 
$$
f_n \rightrightarrows f [x \in E]\!\! \iff \!\! \sup_{x \in E} \vert f_n - f\vert(x) \to 0[n \to \infty]. 
$$
This is strictly stronger than pointwise convergence. 
Alternatively, uniform convergence implies pointwise convergence, so $f_n \to f$ in both cases. 
A: Sorry, but yes, you probably are missing something important, because the second statement in your post

On the other hand $\{f_n\}_{n\in\mathbb{N}}$, converges uniformly to $f$ on $E$ if and only if
  $$f_n(x)\to f(x),\;\forall\,x\in E.$$

is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.
Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $\varepsilon/\delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.
Your best bet is to check the original source to find out what exactly it says there.
A: What you wrote is actually ill-formed, and that is why it is completely wrong. Nowhere did you specify or quantify $n$ in "$f_n(x) → f(x)$, $∀x∈E$". Furthermore, do not ever use symbols like "$∀$" unless you use them properly (quantifiers must always be put in front of the quantified statement).

$f_n → f$ pointwise as $n→\infty$ iff ( $f_n(x)-f(x) → 0$ as $n→∞$ ) for every $x∈E$.
$f_n → f$ uniformly as $n→\infty$ iff $\sup_{x∈E} |f_n(x)-f(x)| → 0$ as $n→∞$.

In one case the convergence may proceed differently for different $x∈E$. In the other case the sup norm between $f_n$ and $f$ (over all $x∈E$) must tend to zero, which intuitively means the convergence must proceed uniformly for all $x∈E$.
In no way is it possible to correctly express uniform convergence using the expression you used, so it is most likely that you did not actually copy the definition or notation given to you.

In purely logical form:
$
\def\nn{\mathbb{N}}
\def\rr{\mathbb{R}}
$

$f_n → f$ pointwise as $n→\infty$ iff $∀x∈E\ ∀ε∈\rr_{>0}\ ∃k∈\nn\ ∀n∈\nn_{≥k}\ ( |f_n(x)-f(x)| < ε )$.
$f_n → f$ uniformly as $n→\infty$ iff $∀ε∈\rr_{>0}\ ∃k∈\nn\ ∀n∈\nn_{≥k}\ ∀x∈E\ ( |f_n(x)-f(x)| < ε )$.

This is an instance of a (restricted) quantifier swap. It is a basic logic fact that one implies the other, easily summarized as "$∃∀ ⇒ ∀∃$", but the reverse implication may not hold.
