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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?

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  • $\begingroup$ If you can't imagine an infinitely dimensional space, you probably can't really handle function spaces (in general, anyway). $\endgroup$ – Asaf Karagila Nov 7 '13 at 9:43
  • $\begingroup$ @Asaf I'm sure i can imagine an infinite dimensional space roughly enough (vector space dimension, not topological). I think function space is more than just an infinite dimensional, since every element is a function hence it has some meaning. How does understanding infinite dimensional aids understanding function space? $\endgroup$ – Jj- Nov 7 '13 at 11:28
  • $\begingroup$ Can you differ in your imagination between all the eventually zero sequences of real numbers, and all the sequences of real numbers; both vector spaces of infinite dimension (although the dimension is indeed different)? Can you differ between those and all the functions from $\Bbb R$ to itself which are non-zero only on a finite subset? If you can, then you can imagine function spaces. In fact, you already do. $\endgroup$ – Asaf Karagila Nov 7 '13 at 11:30
  • $\begingroup$ @Asaf well.. I can distinguish those spaces you listed logically literally, but cannot visualize those. And just like you can imagine well ordering of $\omega$, but cannot imagine it for $\mathbb{R}$, even though you can visualize $\mathbb{R}^\omega$, i think visualizing $\mathbb{R}^\mathbb{R}$ is a completely different matter. $\endgroup$ – Jj- Nov 7 '13 at 11:45
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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 :

  1. 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.

  2. 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)

  3. 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.

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  • $\begingroup$ Thank you, this helps a lot to me. Do you consider a function, an element, as a line in a cartesian coordinate? $\endgroup$ – Jj- Nov 7 '13 at 12:04

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