The cardinality of the classic Hilbert space Question. The classic Hilbert space consists of all infinite sequences $(x_n)$ of real numbers, called points, for which the series $x_1^2+x_2^2+\cdots $ converges. Show that the classic Hilbert space contains just as many points as the real in $\mathbb R$.
Proof says $c\le \vert H\vert \le c^{\aleph_0} =(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0 \aleph_0}=c. $
Why is $c\le \vert H\vert \le c^{\aleph_0}$ so trivial that the book adds no explanation to it?

$H = \{(x_1,x_2,x_3,\ldots) \in \mathbb R^{\aleph_0} : x_1^2+x_2^2+x_3^2+\cdots \} \subseteq \mathbb R^{\aleph_0}$ and that last set has cardinality $c^{\aleph_0}$.
As to the first inequality, every one-dimensional subspace of $H$ has cardinality $c$.
