Identification of the infinitely dimensional unit sphere

Question

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 $S^\infty$?

The statement does seem false as stated, by the reasoning presented. As a side note, you could also use a zenos paradox sequence like $(0, 1/\sqrt{2}, 1/\sqrt{4}, 1/\sqrt{8}, \dots)$ instead of the Basel sequence. Also, it looks like there are some square roots missing in your Basel sequence. — Nick Alger, Oct 29 '19 at 21:13
oh yes, thank you. I started with the Basel sequence and then forgot about the square roots^^ — mathology, Oct 29 '19 at 21:20
Perhaps it's asking for identifying the closure of $S^\infty$ with $B^\infty$. — eyeballfrog, Oct 29 '19 at 21:23
@mathology I'm curious, were you ever able to determine what they meant by this question? Did you mention your counterexample, and what was their response? — Nick Alger, Nov 07 '19 at 19:32

score 2 · Accepted Answer · answered Nov 12 '19 at 13:34

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 $S^\infty = B^\infty$.

Identification of the infinitely dimensional unit sphere

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