Show that the set $A=\{x\in l_2:|x_n|\le \dfrac{1}{n}\}$ is compact in $l_2$.

Question

Show that the set $A=\{x\in l_2:|x_n|\le \dfrac{1}{n}\}$ is compact in $l_2$.

A set is compact iff it is complete and totally bounded.

Step 1: To show that $A$ is complete. Let $(x_k)$ be a Cauchy sequence in $A$ such that $(x_k)\to x$.

Taking $x_k=(x_1^{(k)},x_2^{(k)},\ldots ,x_n^{(k)})\to (x_1,x_2,\ldots,x_n)$.

To show that $\sum x_i^2<\infty$ which can be shown using Minkowski and Triangle Inequality.

Step 2: To show that $A$ is totally bounded.Let $r>0$. We have to find a finite subset $A_r$ of $A$ such that $A=\cup_{a\in A_r} B(a,r)$.

I am stuck here.Will you please say how to find that finite subset.

Answer 1

The idea here is to reduce the problem to a finite dimensional one.

Choose $\epsilon>0$, then we need to find a finite $\epsilon $ net.

Choose $N$ such that $\sqrt{\sum_{k=N+1}^\infty {1 \over k^2}} < {1 \over 2} \epsilon$.

Consider the space $A'=\{ x \in A | x_k = 0, k > N \}$, and note that it is compact (can identify with a closed and bounded subset of $\mathbb{R}^{N}$), hence there is a finite ${1 \over 2} \epsilon $ net for $A'$, let this set be $F$.

Now choose $x \in A$, and let $x'=(x_1,....,x_n,0,0,...)$. Note that $\|x-x'\| \le \sqrt{\sum_{k=N+1}^\infty |x_k|^2} \le \sqrt{\sum_{k=N+1}^\infty {1 \over k^2}} < {1 \over 2} \epsilon$.

Now choose $x'' \in F$ such that $\|x'-x''\| < {1 \over 2} \epsilon$, then we have $\|x-x''\| < \epsilon$.

Hence $F$ is a finite $\epsilon$ net for $A$.