Proving a necessary and sufficient condition for compactness of a subset of $\ell^p$

Solution 1:

The properties 1) and 2) are quite obvious when $A$ is a finite set. So we see that a compact set "almost behaves as a finite set". Here is a formal proof.

• As $\ell^p$ is complete, $A$ is complete whenever it's closed. So we just have to show pre-compactness. Fix $\varepsilon>0$, and apply 2) with $\varepsilon/2$. This gives an integer $N$ such that for all $x\in A$, $\sum_{j\geq N}|x_j|^p<\varepsilon/2$. As $A$ is bounded, we can find $M>0$ such that $\lVert x\rVert\leq M$ for all $x\in A$. Since $[-M,M]^N$ is pre-compact, we can find an integer $K$ and sequences $x^{(0)},\dots,x^{(K)}$ such that for all $v\in [-M,M]^n$, there exists $i\in \{0,\dots,J\}$ such that $\sum_{j=0}^{N-1}|v_j-x^{(i)}_j|^p\leq \frac{\varepsilon^p}{2^p}$. Define $y^{(j)}:=(y^{(j)}_0,\dots,y^{(j)}_{N_1},0,\ldots,0)\in \ell^p$ to see that $A$ is pre-compact.

• Conversely, we assume $A$ compact. A compact subset of a Hausdorff space is closed. It's bounded, as we can extract from the open cover $\{B(x,1)\}_{x\in A}$ a finite subcover $\{B(x,1)\}_{x\in F}$, where $F$ is finite. Then for all $x\in A$, $\lVert x\rVert\leq 1+\max_{y\in F}\lVert y\rVert$. Fix $\varepsilon>0$, then by pre-compactness, we can find an integer $K$ and $x^{(1)},\dots,x^{(K)}$ such that $\bigcup_{j=1}^KB(x^{(j)},\varepsilon/2)\supset A$. For each $j\leq K$, take $N_j$ such that $\sum_{i\geq N_j}|x_i^{(j)}|^p<\frac{\varepsilon^p}{2^p}$. Then take $N:=\max_{1\leq j\leq K}N_k$.