Suppose $L^{2}(\mathbb{N})$ is the Hilbert space of sequences
$(a_{n})_{n \in \mathbb N}$ which satisfy $\sum |a_{n}|^{2}$ with $(a,b) = \sum a_{n} \bar{b_{n}}.$
Prove the set of sequences $\{a_{n}\}_n$ that satisfy the condition $|a_{n}| \le \frac{1}{n} \forall n$ is compact in $L^2(\mathbb{N})$.
I'm rather stuck on this idea, and was curious if I could be given help.
This is $\mathbb{not}$ for homework, but self-study. My initial thought would be to use the fact that
suppose we have a compact subset $M$ of a metric space, then $M$ is bounded and complete.
Something I'm trying to use is the Bolzano-Weierstrass property. Thanks!
You can use the following criterium of compacity. A metric space is compact if it is complete and for every $\epsilon$ one can cover it by a finite number of balls of radius $\epsilon$.
For complete, your set is a closed subset in a complete space, as intersection of the sets $\vert a_n\vert \leq 1/n$
For the ball, let $\epsilon $ be given and $n_0 $ an integer such that $\sum _{n\geq n_0} 1/n^2 < {1\over 100 \epsilon ^2}$. Note that every point in your set is at the distance $\leq 1/10 \epsilon$ of a point with $a_n=0$ if $n>n_0$. Now cover the compact set$ \{ (a_n) / a_n=0 if n>n_0, \vert a_n\vert \leq 1/n \}$ by a finite set of ball of radius $\leq \epsilon /2$. The balls with the same center and radius $\epsilon$ cover your set.