# Arzela-Ascoli Counter Example?

I am taking my second undergraduate analysis course, our text book is N.L. Carothers "Real Analysis". This question is my own after reading the chapter "The Space of Continuous Functions" and the section "Equicontinuity"

First some preliminaries:

Arzela-Ascoli Theorem. Let $X$ be any compact set. For $A\subset C(X)$, $A$ is compact iff $A$ is closed, bounded, and equicontinuous.

The set $A$ is closed if for any convergent sequence $(f_n)_{n\in\mathbb{N}}$ we have that the limit of $f_n$ is an element of A.

The set $A$ is bounded if for all $f$ in $A$ we have $||f||=sup_{x\in X}{f(x)} = M\in \mathbb{R}$ exists for all $x$ in $X$.

The set $A$ is equicontinuous if for every $\epsilon >0$ there is a $\delta>0$ such that for all $f$ in the set $A$ and for all $x, y$ in $X$, $d(x,y)<\delta$ implies that $|f(x)-f(y)| < \epsilon$.

The set A is compact if every sequence in $A$ has a subsequence that converges in $A$.

Here is my question: There is this famous problem called the 3-body problem. I don't know if you have heard of it but there is a simulation here. Each of the functions that generate a path given some starting position are in the space $C([a,b] \times [c,d])$. Let $A$ be the set of all possible functions in the three body problem. It seems to me that $A$ is closed, bounded and equicontinuous but not compact since $f$ depends so heavily on the initial position of the planet. Does this contradict the theorem?

For any physical model involving force, it only makes sense to think of compact sets, especially for gravitational problem (as the force decreases radially, and eventually becomes negligible).

Now to get more technical. Arzela-Ascoli is certainly correct, and has been verified and used for some time now. In addition, the formulation in the textbook is not the best. Arzela-Ascoli tells us when subspaces of general metric spaces (typically function spaces) are compact, since metric spaces are Hausdorff, compact subsets are closed. Knowing this means you have less work to do, you don't need to show the space is closed. Arzela-Ascoli would tell us $A$ is compact and that guarantees that $A$ is closed.

The conditions for a subspace of a metric space to be compact is that it is uniformly bounded and equicontinuous.

Let $A$ be the set of functions which satisfy your requirements.

We know that $f \in C([a,b] \times [c,d])$ is bounded because of the Extreme Value Theorem, so every function in $A$ is bounded, but nothing would indicate a priori that the functions are uniformly bounded. In fact, without an explicit description of $A$ we cannot say much about what functions live in the space. Equicontinuity can fail as well.

To summarize, without having an explicit description of the functions that make up $A\subset C([a,b] \times [c,d])$, we cannot verify equicontinuity or uniform boundedness. In addition, the functions that we choose are purely because they fit a model, there is nothing in the physical world that forces $A$ to be compact, and whatever $A$ is, it will not violate Arzela-Ascoli.

Finally, if $A$ is open (not closed), then by contrapositive, you right away know that $A$ is not compact, and the space $C(X)$ where $X$ is a compact set is itself, not compact.

• I was thinking about this recently and you are right, I didn't provide an explicit description of the functions that make up $A$. I don't know if this makes sense, but can you think of the set of all functions that are a solution to a set of differential equations as a set $A \subset C[a,b]$? – user197950 Dec 7 '16 at 2:45
• Unfortunately, that is not generally possible! For starters, the solution to the differential equation itself need not be continuous, and hence not in the set, such solutions are called weak solutions. Even if we restrict ourselves to strong solutions (i.e. smooth functions), consider the following: since $A$ is a subspace, it has a vector space basis (Schauder basis, since $A$ has uncountable dimension), this basis will allow you to express all solutions as a linear combination, but in general you cannot write down an explicit basis for any arbitrary vector space. – Kernel_Dirichlet Dec 7 '16 at 3:20
• Oh interesting. I wasn't expecting that. For example the differential equation $\frac{dv}{dt}=9.8-0.196v$ has solutions of the form $v(t)=50 + ce^{-0.196t}$. So take $a_n=1/n$ and let $A={0}\cup {a_n}$. Let $c \in A$ Then $v_n(t)=50+ce^{-0.196t}$. Let t vary in $[0,1]$. The sequence of functions is a set, is that set compact? Yes I think so. – user197950 Dec 7 '16 at 16:19
• $A=0 \cup a_n$ is compact, because it is the union of two compact sets. However, as you defined it, $A$ is a sequence of real numbers, not a sequence of functions. $c$ is just either $0$ or $1/n$ for some $n$. in addition, for $v_n(t)$ you have defined one function, there is no $n$ on the RHS, however, if you defined it as $v_n(t):[0,1]\rightarrow \mathbb{R}$, with $v_n(t)=50+c_n e^{-0.196t}$ the sequence is uniformly bounded by 50, but not equicontinuous, (take $c_0$ and another function and see there is no way to choose an appropriate $\delta$ – Kernel_Dirichlet Dec 7 '16 at 23:07