# Compactness of a subset of $\ell^2$

Let $$K \subset \ell^2(\mathbb{N})$$ be a set defined as follows: $$K := \left\{x = (x_1, x_2, \dots) \in \ell^2(\mathbb{N}) \,|\, |x_n| \le \frac{1}{n}\right\}.$$ Since $$\ell^2(\mathbb{N})$$ is a reflexive space, we know that any sequence of $$K$$, being bounded, admits a converging subsequence by the Eberlein-Smulian theorem. Moreover, since $$K$$ is closed, the weakly converging subsequence has a limit in $$K$$, and thus $$K$$ is weakly sequentially compact.

Since $$(\ell^2(\mathbb{N}))^*$$ is separable, we know that $$\overline{B(0,1)}$$ is metrizable for the weak topology and hence $$K$$ is not only weakly sequentially compact but compact for the weak topology.

Is it compact for the strong topology as well?

• If the "strong" topology is the norm topology then yes. (Given a sequence in $K$, a diagonal gives a pointwise-convergent subsequence, which then converges in norm by DCT.) Commented Dec 21, 2020 at 13:47
• $K$ is homeomorphic to $\prod_{n=1}^\infty [0,1/n]$, which is compact. Commented Dec 21, 2020 at 14:15

Yes it is. It is a standard (I mean "very common") argument in functional analysis, you can pick a sequence of points $$(x_n)_n \in K^{\mathbb{N}}$$ and proceed a Cantor's diagonal argument to extract a convergence subsequence. Then, it is just a matter of applying Lebesgue Dominated Convergence theorem to conclude convergece in $$||.||_{\ell_2}$$ as well.
• If I am not mistaken, for a given sequence, the sequence given by the first component is in $[0,1]$ and thus admits a converging subsequence, same for the second component in $[0,1/2]$, and so on... Now, we take recursively subsequence of subsequence for each component and we are left we a converging subsequence in each component in addition to convergence in $\ell^2$? Commented Dec 21, 2020 at 16:47