Pointwise and Uniform convergence intuition and examples I am having trouble understanding the intuition behind the difference between point-wise convergence of functions and  uniform convergence of functions once revisited. I am quoting the following definition from a book:
Theorem: Let$\{f_n\}$ be defined on $S$. Then 
(a) $\{f_n\}$ converges point wise to $F$ on $S$ if and only if there is, for each $\epsilon>0$ and $x\in S$ an integer $N$(which may depend on $x$ as well as $\epsilon$) such that:
$|F_n(x)-F(x)|<\epsilon\:\:\:\:\:\:\text{if}\:\:\:\:\:n\geqslant N$
(b)$F_n$ converges uniformly to $F$ on $S$ if and only if there is for each $\epsilon>0$ an integer $N$(which depends only on $\epsilon$ and not on any particular $x$ in $S$) such that:
$|F_n(x)-F(x)|<\epsilon\:\:\:\:\:\:\text{for all}\:x\in S\:\text{if}\:n\geqslant N$
So uniform convergence means that after the $N$ the $f_n$ is going to approach $f$ at the same pace $\epsilon$ for all $x$? While point wise convergence means that a function converges on a given point $x$ that is why $N$ depends on $x$, right?
Could someone provide me examples of both cases, functions that converge point wise and functions that converge uniformly?
Thanks in advance!
 A: You can at least try to see that uniform convergence implies pointwise convergence, which follows directly from the definitions.
For a function sequence converging pointwisely but not uniformly, a handy example that came to my mind is this:
$$
f_{n}: x \mapsto \frac{x}{n}, \mathbb{R} \to \mathbb{R}
$$
for all $n \in \mathbb{N}$.
For every $x \in \mathbb{R}$, the sequence $(f_{n}(x))$ of real numbers converges to $0$ apparently. However, for every $l \neq 0$, there is some  $\varepsilon > 0$, say $\varepsilon := |l|/2$, such that for every $N \in \mathbb{N}$ there are some $n \geq N$ and some $x \in \mathbb{R}$, say $n := N$ and $x := 2Nl$, such that
$$
|f_{n}(x) - l| = |\frac{x}{n} - l| = |l| \geq \varepsilon.
$$
For $l = 0 $, take $\varepsilon := 1$, for example, to see that it still holds true that for every $N \in \mathbb{N}$ there are some $n \geq N$ and some $x \in \mathbb{R}$, say $n := N$ and $x := 2N$, such that 
$$
|f_{n}(x) - l| = |\frac{x}{n}-l| = 2 \geq \varepsilon.
$$
This proves the negation of the definition of uniform convergence; so $(f_{n})$ is not uniformly convergent.
An example of uniformly convergent function sequence is even easier. Let
$$
g_{n}: x \mapsto \frac{1}{n}, \mathbb{R} \to \mathbb{R}
$$
for all $n \in \mathbb{N}$.
I assert that $g_{n} \to 0$ uniformly.
Note that for every $\varepsilon > 0$, there is some $N \in \mathbb{N}$, say $N := \lfloor \frac{1}{\varepsilon} \rfloor + 1$, such that
$$
|g_{n}(x) - 0| = \frac{1}{n} < \varepsilon
$$
for all $n \geq N$.
We are done.
A: I can provide a pretty beautiful heuristic to think about it. It has a few ideas that may not be found anywhere else , so this answer has a chance to attract more downvotes, but, I kindly request that the answer is read to the end before it is voted on.
I focus on functions uniformly convergent from $\mathbb{R} \to  \mathbb{X}$ where $X$ is any space where it makes sense to talk about uniform convergence of functions:

The idea is that on the top of the real line, we can think of like a solid cylinder standing of top of the real line, that is, to each point $x$ of the real line a disc attached . This disc is a metaphor for the codomain for the family of functions.
It can be seen that for a given $x$, we have some sort of sequence in one of the disc (Fig 2.).
But, we can see the functions also "globally" in the sense that we can think of a bundle of curves through the cylinder by collecting all the different curves (Fig 3.). If it is not clear, a single curve is made by fixing an index, and then "lighting up" all the points inside the cylinder corresponding to $f_n(x)$ for all $x$.
All the uniform convergence tells us is that these curves can be made as close to the limit function's curve as we want.
Now, what is pointwise convergence then? Well we no longer have this coming together of curves on the global level but for each individual cross section, we can find a cut off index such that distance of $f(x)$ from terms above that index is less than epsilon.
