Motivation for the Arzela-Ascoli's theorem I'm studying by myself Arzela-Ascoli's theorem and I'm reading this chapter of lecture notes. Firstly, I would like to be clear that I know that the motivation of Arzela-Ascoli's theorem is to characterize when $E \subset \mathcal{C}(K,N)$ is a compact set in $\mathcal{C}(K,N)$ (I'm considering the space of continuous functions on a compact set $K$ in a metric space $M$ and $N$ is another metric space), but there are some points in the chapter that I linked that it are not clear to me.
I didn't understand when the author talked about the "directions" of an infinite-dimensional vector space in the paragraph below:

Intuitively, this example shows that the unit ball in $\mathbb{R}^n$ is much ”smaller” than the unit ball in $\mathcal{C}[0,1]$. We will see in later sections that the difference can be understood in terms of the vector space structure of the two sets. $\mathbb{R}^n$ is a finite-dimensional vector space, and so bounded sequences intuitively only have a few directions to go in, and thus they must cluster up in at least one direction. In contrast, $\mathcal{C}[0,1]$ is infinite-dimensional, and so sequences, even in bounded sets, have an infinite number of ”directions” in which they can go without clustering. This intuitive idea of size based on dimension will be formalized later. For now, we merely confirm that $\mathbb{R}^n$ is in fact finite-dimensional, while $\mathcal{C}[0,1]$ is infinite-dimensional.

Unfortunately, I don't have the complete lecture notes, so I am not sure about this "intuitive idea of size based on dimension" that the author comments about. I would like to know what he meant to say with this. Also, on page $4$, the author states:

We will see that the additional requirement for a subset of $\mathcal{C}[0,1]$ to be convergent is somehow related to requiring all of the elements of the subset to be close to each other. To make this precise, we will shortly be introducing the definition of a property called equicontinuity, which is meant to deal with continuity of the entire set at once. Before this, we will motivate the idea of equicontinuity.

I think it is reasonable that you need equicontinuity of the set $E \subset \mathcal{C}([0,1])$ by the argument given in section $5$, but how is this related to "requiring all of the elements of the subset to be close to each other"? I think to give a notion of proximity of a function $f$ to another function $g$ in a set $E$ we need to talk about the distance between $f$ and $g$, but the equicontinuity just tells us about the continuity of the functions in a family $E$. I really don't see how equicontinuity gives us the proximity of functions in $E$.
Finally, I would like to know if there is a link between the "directions" commented on previously and the relation of the proximity between functions in $E$.
Thanks in advance!
$\textbf{EDIT:}$
I found this lecture notes on function spaces by Terence Tao, which is possibly related with the "intuitive idea of size based on dimension" commented by the author. As far as I understand, "size" depends about whatever I want, but I don't know what "size" I want in the context of the Arzela-Ascoli's theorem and how this "size" is related to equicontinuity.
 A: By direction is meant a subspace of dimension 1. By independant direction is meant subspaces of dimension 1 which are in direct sum. In this sense, in a finite space, you just have a finite number of independant directions. 
In finite dimension, if you take a sequence of non zero vectors $(v_n)$, you can say in some sense that you can extract a subsequence $(v_{\phi(n)})$ such that the directions converges : "the directions of the $v_n$ tends to cluster at some places". More precisely, define $w_n = \frac{v_n}{||v_n||}$ a unitary vector generating the same subspace as $v_n$. Then $w_{\phi(n)}$ converges to some $w$ ; so in some sense $D_n = Vect(v_n) = Vect(w_n)$ converges to $D = Vect(w)$. All of this is possible to make this because in finite dimension, the balls (and hence, the spheres) are compact. So in that sense, there is few directions.
In infinite dimension, this does not longer work : see Riesz theorem, the unit ball is compact (or equivalently, all the balls are compact) iff the space if of finite dimension : https://en.wikipedia.org/wiki/Riesz%27s_lemma. Let $E$ a vector space which has a countable basis, $e_n$, $n\in \mathbb{N}$, and suppose $||e_n|| = 1$. It is possible to build a base $f_n$ such that $||f_n|| = 1$ and $||f_n - f_m|| > \frac{1}{2}$ for $n\neq m$. It is then easy to check that $(f_n)$ cannot have any convergent subsequence. In this sense, a sequence of directions needs not to accumulate in some place. We are losing compacity : in that sense, the set of directions is much more bigger. In fact, it is possible to caracterize the relative compacity of a subset $K$ of a Banach space $E$ : to say it somewhat vaguely, the width of $K$ in some direction must converge to $0$ when the direction gets far away from finite subspaces.  For instance, in $R^\mathbb{N}$ with the uniform norm, $\prod_{n\in \mathbb{N}} B(a_n, r_n)$ is compact iff $\lim_{n\rightarrow \mathbb{N}} r_n = 0$
Let us translate this in the Arzela Ascoli theorem's setting.
Take $K = [|1, n|]$ a finite set, and $X = C(K, \mathbb{R})$. Then it is easy to check that $X = \mathbb{R}^n$, and a natural basis is constitued by the characteristic functions $e_k = \chi_{\{k\}}$. 
Say that a sequence $f_n = \sum f_{n,k} e_k$ converges is equivalent to say that all the coordinates $f_n,k$ in $e_k$ converges when $n\rightarrow \infty$ (this is automatically uniform in $k$ since there is only a finite number of coordinates).
Say that $F\subset X$ is relatively compact is equivalent to say that all coordinates of the $f$ in $e_k$ are bounded uniformly on $f$.
Take : 
$K = \{0\} \cup \{\frac{1}{n}, n \in \mathbb{N}^*\}$ and $X = C(K, \mathbb{R})$.
Here, there are a lot of directions for $X$. Intuitively, the independant directions should be something like this : the caracteristic functions $e_n = \chi_{ \{\frac{1}{n} \} }$, $e_0 = \chi_{\{0\}}$ : these vectors give a free family. In a algebraic sense, there are a lot of other directions. Take for example for all subset $A$ of $K$, $\chi_A$ : if $A$ is infinite, $\chi_A$ is not generated by the $e_n$. Via a diagonal argument, it is possible to check that $X$ cannot have a countable basis. It not obvious at all to exhibit a base of this vector space. Even with this simple example, it is impossible to exhibit a base without using the axiom of choice. See http://math.uchicago.edu/~may/REU2014/REUPapers/Barnum.pdf , "applications of the axiom of choice" for the demonstration that the axiom of choice is equivalent to find a base of all vector spaces.
Nevertheless, in a topological sense, these other algebraic directions are not determinant ;  the $e_n$ constitutes a Schauder basis : https://en.wikipedia.org/wiki/Schauder_basis . Here, an element $f$ of $X$ can be written uniquely $f = f_0 e_0 + \sum_{n\in \mathbb{N}^*} f_n e_n$ where $lim_{n\rightarrow \infty} f_n = f_0$. Furthermore, $||f|| = max_{n\in \mathbb{N}} |f_n|$.
Say that a sequence $f_n = \sum f_{n,k} e_k$ converges is equivalent to say that all the coordinates $f_n,k$ in $e_k$ converges **uniformly in $k$ ** when $n\rightarrow \infty$
The problem is that the directions $e_n$ are not close, even when $n$ is big : $||e_n - e_m|| = 1$ for all $n\neq m$. In particular, a set $F$ which contains infinitely many $e_n$ cannot be compact. But a compact can contain all the $f = \alpha_k e_k$ if $\alpha_n \rightarrow 0$ for instance. Here, the speed of convergence of all the sequences of coefficients of $f$ (i.e $0, 0, .., \alpha_k, 0, ..$) when $n\rightarrow \infty$ is majorated by $|\alpha_n|$, uniformly on $f$.
More generally, let $F$ a subset of $X$. I will denote $f = \sum f_n e_n$ the elements of $F$. Then the set $F$ is relatively compact iff (boundedness) for all integer $n$, the $f_n$ are bounded uniformly on $f$ and (equicontinuity) the $f_n$ converges with a bound on the speed independant of $f$ when $n\rightarrow \infty$.
This kind of phenomena corresponds to the intuition that the width of $K$ should be small when we are far from spaces for finite dimension. Here, the $f_0$ are bounded and the $f_n$ when $n$ big belongs to the "translations by $(f_0, 0, ..)$ of a fixed cone pointed at $(0, 0, ... )$" given by the uniform bound of the rate of convergence.
When $K$ is more complicated, it is the same kind of ideas (but $0$ is no more the only accumulation point ! ). 
EDIT : 
For $K = [0,1]$ you don't have any natural Schauder basis anymore (or maybe no Schauder basis at all ! But I'm not 100% sure of it). I talked about this tool before to make things a little more concrete ; in the general case, it is not necessary to use it to show the theorem. Still, the configuration is essentially the same. Intuitively, you may think "$f = \sum_{x\in [0,1]} f(x) e_x$" where $e_x = \chi_{x}$... But this infinite sum haven't a topological meaning linked to the uniform convergence norm we are using ! So you cannot use this kind of notation to make rigorous reasoning. Still, intuitively, the f(x) should be uniformly (on $f\in F$) close to $f(x_0)$ when $x$ is close to $x_0$ ($x_0$ plays the role of $0$, and $x$ the role of $\frac{1}{n}$ in the previous case $K = \{0\} \cup \{\frac{1}{n}, n\in \mathbb{N}\}$). That is exactly the hypothesis of equicontinuity.
