TITLE: What is the main difference between pointwise and uniform convergence as defined here?
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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.