I am just trying to understand the proof of the following exercise:
Define $l^2(\mathbb{N}) = \{(x_n)_n \in \mathbb{C} | \sum_n |x_n|^2 < + \infty \}$. Let $(a_n)_n \subset \mathbb{R}_+$ such that $\sum_n a_n^2 < + \infty$. Show that $B = \{(x_n) \in l^2(\mathbb{N}) | |x_n| \leq a_n\}$ is compact in $l^2(\mathbb{N})$.
It is easy to show that $B$ is closed. We want to show that $B$ is totally bounded: Let $\epsilon > 0$. Because of the convergence of $\sum_n | a_n|^2$, there exists a $N_{\epsilon} \in \mathbb{N}$ such that $\sum_{n \geq N_{\epsilon}}|a_n|^2 < \epsilon^2$.
Now we define $D = \{ (x_n)_n \in l^2(\mathbb{N}) \ | \ |x_n| \in \{ 0, \frac{\epsilon}{\sqrt{N_\epsilon}} , \frac{2\epsilon}{\sqrt{N_{\epsilon}}}, \cdots, \frac{K_n \epsilon}{\sqrt{N_{\epsilon}}} \} \text{ if } 0 \leq n \leq N_\epsilon -1 , x_n = 0 \text{ if } n \geq N_{\epsilon} + 1\}$ such that $a_n \leq K_n \frac{\epsilon}{\sqrt{N_{\epsilon}}} < a_n + \frac{\epsilon}{\sqrt{N_{\epsilon}}} $ and $K_n = \left\lfloor\dfrac{a_n \sqrt{N_{\epsilon}}}{\epsilon}\right\rfloor + 1$. $D$ is finite and with this set $D$ we can show that $B$ is totally bounded.
I don't really understand the construction of the set $D$. Why does this help us to show that $B$ is totally bounded, i.e. compact?
Best Answer
Of course I can understand that an undergraduate student might not feel comfortable using Tychonoff's theorem, so I will give another answer that seems also to be closer to OP's idea. Let's use the notation of my other answer. Again, I will work with real vector spaces but the same arguments work for the complex ones.
The functionals $\phi_n:\ell^2\to\mathbb{R}$ given by $x\mapsto x(n)$ (i.e. we map a sequence $x=\{x(n)\}$ to its $n$ term) is of course continuous, since $|x(n)|\leq(\sum_{k=1}^\infty|x(k)|^2)^{1/2}$. Now one immediately sees that $$B_a=\bigcap_{n=1}^\infty\phi_n^{-1}([-a_n,a_n])$$ Now each interval $[a_n,a_n]$ is closed, so their preimages under continuous maps are closed, so their intersection is an intersection of closed sets, thus $B_a$ is closed. In particular, the set $B_a$ is complete, as we do know that $\ell^2$ is complete. Now to show compactness all we must do is show that $B_a$ is totally bounded, as it is a general fact that for metric spaces, completeness and totally boundedness is equivalent to compactness.
By definition, a space is totally bounded when for each $\varepsilon>0$ there exists a finite collection of open balls of radius $\varepsilon$ that covers the space.
The idea is very simple. Let $\varepsilon>0$. Since $\sum_na_n^2<\infty$, find an integer $N\ge1$ so that $\sum_{n=N+1}^\infty a_n^2<\varepsilon^2$. Now have a look at the intervals $[-a_n,a_n]$ for $n=1,\dots,N$. We can find a partition of each interval $-a_n=t_{0,n}<t_{1,n}<\dots<t_{d_n,n}=a_n$ so that $|t_{i,n}-t_{i+1,n}|<\frac{\varepsilon}{\sqrt{N}}$ for all $i=0,\dots,d_n-1$ and all $n=1,\dots,N$.
Now we simply have to consider all these sequences that, in slot $1$ have a value from the partition of $[-a_1,a_1]$ we chose, in slot $2$ they have a value from the partition of $[-a_2,a_2]$ that we chose and so on up to the $N$-th slot, where we want them to have a value from the partition of $[-a_N,a_N]$ that we chose. Let them have $0$ from that point on.
Note that these sequences are finitely many: how many of them are there? well, for the first slot we have $d_1+1$ choices. For the second slot we have $d_2+1$ choices and so on until the $N$th slot, for which we have $d_N+1$ choices. So there exist precisely $D:=\prod_{n=1}^N(d_n+1)$ such seqeunces, this is a finite number.
We now claim that the balls centered at these sequences and of radius $\sqrt{2}\varepsilon$ actually cover the entire set $B_a$. Let's prove this.
Suppose that we are given a sequence $x=\{x(n)\}\in B_a$. How do we determine in which one of the $D$ balls (with center one of the $D$ sequences we constructed and radius $\sqrt{2}\varepsilon$) $x$ will lie in? Well, there's only one natural thing to do! look at its first $N$ terms. For $x(1)\in[-a_1,a_1]$, there exist some interval of our partition that contains $x(1)$. Note that our partition is so thin that each of the subintervals has length at most $\frac{\varepsilon}{\sqrt{N}}$. Let $t_1$ be one of the end points of the interval that contains $x(1)$. In particular, we have $|t_1-x(1)|<\frac{\varepsilon}{\sqrt{N}}$. Similarly for $x(2)$ and so on ..., up to $x(N)$: $x(N)\in[-a_N,a_N]$, so $x(N)$ belongs to some interval of our partition that contains $x(N)$. Let $t_N$ be one of the end points of the subinterval of the partition that contains $x(N)$. Then we have $|x(N)-t_N|<\frac{\varepsilon}{\sqrt{N}}$.
Now take the sequence $$t=(t_1,\dots,t_N,0,0,0,\dots)$$ This is actually one of the finite set (with cardinality $D$) of the sequences that we described earlier. But look:
$$\|x-t\|_2^2=\sum_{n=1}^N|x(n)-t_n|+\sum_{n\ge N+1}|x(n)|^2\leq\varepsilon^2+\sum_{n\ge N+1}a_n^2\leq2\varepsilon^2$$ so $\|x-t\|_2<\sqrt{2}\varepsilon$. This shows that the set $B_a$ is totally bounded.
A final comment: There is nothing special about intervals and partitions. What is being used here actually, is the fact that each of the intervals $[-a_N,a_N]$ is totally bounded, as a set. This is how one would work the complex case for example.