Can any Hilbert space be expressed as countable union of unit balls

compact-operatorsfunctional-analysishilbert-spaces

I was going through functional analysis text by J.Conway, and have encountered with next claim (2.4.6) :

Let $T\in \mathcal{B}_0(\mathcal{H},\mathcal{K})$ for two Hilbert spaces $\mathcal{H},\mathcal{K}$. Since $\text{cl}[T(\text{ball } \mathcal{H})]$ is compact, it is separable. Therefore $\text{cl}(\text{ran} T)$ is separable subspace of $\mathcal{K}$. Here, ball $\mathcal{H}$ is closed unit ball in $\mathcal{H}$.

However, I am not convinced since this presumes that any Hilbert space can be expressed as a countable union of unit balls, which is not very trivial for me.

  1. Is this claim well-known fact?
  2. If so, how do you prove it?

Thank you in advance.

Best Answer

It is not being claimed that $\mathcal{H}$ is a countable union of unit balls (that is not true if $\mathcal{H}$ is not separable!). Rather, what is being used here is that $\mathcal{H}$ is the union of the balls of radius $n$ around $0$. Since $T$ is linear, the images of those balls will just be scaled versions of the images of the unit balls, and therefore their images will also be separable.

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