How do I draw a diagram for a function space? If one considers a single function, then one can just draw its diagram as a Cartesian product. So it's relatively easy to contribute one's intuition to an argument.
However, when it is a function space, I completely lose my intuition. And I am sure this way of studying is bad since I cannot confirm myself why such theorems have to be true.
Ascoli's (classical) theorem is one example.
Firstly, I learned this theorem when I studied Rudin-Analysis.
There he attacks the theorem directly. And I had no idea why it has to be.
Then, I have encountered this theorem again in Munkres-Topology.
Here, he attacks the theorem via the uniform topology on ${\mathbb{R}^n}^X$.
Still, I have no idea why it has to be.
How do I literally draw a diagram for, or imagine a function space?
 A: I am not sure it is possible to visualize function spaces in the way that you can visualize $\mathbb{R}^2$, but here is one attempt to see Arzela-Ascoli :


*

*Consider the usual $\epsilon-\delta$ picture of a limit (See this, for instance).Now, equicontinuity of a family $S = \{f_{\alpha}\}$ says that, for any such horizonal $\epsilon-$strip, there is a vertical $\delta$-strip such that, for any $x$ within that $\delta$-strip, the corresponding values $\{f_{\alpha}(x)\}$ are all in that horizontal $\epsilon$-strip.

*Suppose $f:[0,1] \to [0,1]$ is continuous. Visualize an open ball $B(f,r)$ around a function $f\in C[0,1]$ as a band of width $r/2$ on either side of the curve (ie. It would be bounded by the curves $x \mapsto f(x) + r/2$ and $x \mapsto f(x) - r/2$) (See this, for instance)

*Now suppose $S := \{f_{\alpha} :[0,1]\to [0,1]\}$ is equicontinuous. You want to know why it is compact. So start with infinitely many bands $B(f_{\alpha}, \epsilon_{\alpha})$. Now fix $x \in [0,1]$ and look at all the bands "above" it. Since the bands above it do not go beyond the largest $Y$-value (ie. 1), there are only finitely many bands that are needed to cover $\{f_{\alpha}(x)\}$. Each of these finitely many bands contribute an $\epsilon_{\alpha}$, of which you can take the minimum - call that $\epsilon_x$.
Now there are finitely many horizontal strips of radius $\epsilon_x$ that together cover all the $Y-$values, $\{f_{\alpha}(x)\}$. Choose a vertical $\delta_x$-strip as in (1). And note that, for any $z \in (x-\delta_x,x+\delta_x)$, the corresponding $Y-$values $\{f_{\alpha}(z)\}$ are all in one of those finitely many horizontal $\epsilon_x$-strips.
Now is where the compactness of the domain comes in. There are only finitely many such $\delta_x$ that are needed to cover all of $[0,1]$. Each $\delta_x$ needs only finitely many $B(f_{\alpha},\epsilon_{\alpha})$; which gives compactness.
I have a pretty good picture in my mind right now, and I hope this explanation was able to translate that into words. Unfortunately, I am not sure how to represent this graphically. If someone can suggest a nice way to do this, then that would help greatly.
