TITLE: Identification of the infinitely dimensional unit sphere
QUESTION [6 upvotes]: I have some doubts about a statement from class:
Let $\mathbb{R}^\omega$ be the countable product of $\mathbb{R}$. Equip $\mathbb{R}^\omega$ with the scalar product $\sum_{k = 1}^\infty x_i y_i$ and the induced norm. We identify $\mathbb{R}^n$ with the space generated by the first $n$ coordinates. Let $S^\infty = \bigcup_n S^n \subset \mathbb{R}^\omega$ ($S^n$ being the unit sphere in $\mathbb{R}^n$) and let $B^\infty = \{(x_k)_{k \in \mathbb{N}} \mid \sum_{k = 1}^\infty x_k^2 = 1\}$. Show that we can identify $S^\infty$ with $B^\infty$.
What bothers me about this question is that clearly $S^\infty \subset B^\infty$. However, I could take the sequence $(0, \frac{\sqrt{6}}{\pi} \frac{1}{1},  \frac{\sqrt{6}}{\pi}  \frac{1}{2}, \frac{\sqrt{6}}{\pi}  \frac{1}{3} , \dotsc)$. This sequence has norm $1$ under the assumptions in the statement. But the sequence also has infinitely many non-zero elements. So how can it be in on of the $S^\infty$?

REPLY [2 votes]: If $\mathbb{R}^\omega$ is the countable product of $\mathbb{R}$, then $\sum_{k = 1}^\infty x_i y_i$ does not define a scalar product because the series may diverge.
Thus we have to restrict to either
$$\ell^2 = \{ (x_n) \in \mathbb{R}^\omega \mid \sum_{n = 1}^\infty x_n^2 < \infty\}$$
or to
$$\ell^2_0 = \{ (x_n) \in \mathbb{R}^\omega \mid x_n = 0 \text{ for almost all } n\} \subset \ell^2.$$
Then we get a scalar product and an induced norm.
If we work in $\ell^2$, then $S^\infty \subsetneqq B^\infty$ as you correctly stated in your question.
However, if we work in $\ell^2_0$, then trivially  $S^\infty = B^\infty$.