# Two (different) infinite-dimensional spheres

Let $$\mathbb{R}^\infty$$ be the set of sequences $$x = (x_0, x_1, x_2, \ldots)$$ with $$x_i \in \mathbb{R}$$, let $$\{e_n\}$$ be the standard basis and let $$||x||_2 = \sum\limits_{i=0}^\infty x_i^2$$.
We consider a nice subset, known by many as $$\ell^2 = \{ x \in \mathbb{R}^\infty : ||x||_2 < \infty \}$$. The norm $$||\cdot ||_2$$ defines a topology on $$\tau_A$$ on $$\ell^2$$, as well as the unit disk $$D_A^\infty = \{ x \in \ell^2 : ||x||_2 \leq 1\}$$ and the unit sphere $$S_A^\infty = \{ x \in \ell^2 : ||x||_2 = 1\}$$.

And so we have obtained the disk and the sphere that make analyst happy. But there is another sect of mathematicians, who define infinite-dimensional spheres in a different way: topologists.

Let us consider the "standard" sphere $$S_B^n \subseteq \mathbb{R}^\infty$$, $$S^n_B = \{(x_0, \ldots, x_n, 0, 0, \ldots) \text{ such that }\sum\limits_{i=0}^n x_i^2 = 1 \}$$. Let $$S_B^\infty = \bigcup\limits_{n=0}^\infty S_B^n$$ the union of all finite-dimensional spheres. The disks $$D_B^n$$ and $$D_B^\infty$$ are defined in a similar way, substituting $$||x||_2 = 1$$ with $$||x||_2 \leq 1$$.

In other words $$S_B^\infty$$ is the set of all sequences that are eventually null, with norm equal to 1, while $$S_A^\infty$$ is just the set of all sequences with norm equal to 1. So clearly $$S_B^\infty \subset S_A^\infty$$ and $$D_B^\infty \subset D_A^\infty$$ (but they are not equal).

But wait, we're talking about sets, what's actually the topology on $$S_B^\infty$$? Following this question, let's use the weak topology, that we will call $$\tau_B$$: a set $$U\subseteq \mathbb{R}^\infty$$ is open iff $$U \cap \mathbb{R}^n$$ is open in $$\mathbb{R}^n$$ for every $$n$$.

Question 1. What is the relation between $$\tau_A$$ and $$\tau_B$$ on $$S^\infty_A$$? (or even better, on the larger set $$\ell^2$$?)

I'd say that $$\tau_A \subset \tau_B$$ because if we intersect an open disk $$\mathring{D}(0,r) = \{ x \in \ell^2 : ||x||_2 < r\}$$ with a finite-dimensional subspace, we get $$\mathring{D}(0,r) \cap \mathbb{R}^n = \mathring{D^n}(0,r)$$. However, $$\tau_A \neq \tau_B$$, because the convex hull (=finite linear convex combinations) of the points $$\{1 e_1, \frac{1}{2} e_2, \frac{1}{3} e_3, \ldots, \frac{1}{n} e_n, \ldots\}$$ is an open set in $$\tau_B$$ but not in $$\tau_A$$.

But sadly, I can't gain any further insight on how $$\tau_A$$ and $$\tau_B$$ behave on $$S_A^\infty$$.

Question 2. On the vector space $$V = \mathbb{R}^\infty$$, is $$\tau_B$$ (the weak topology in the sense of intersection with finite-dimensional things) the same as the weak topology (in the sense of duals, i.e. $$v_n \rightharpoonup v$$ iff for every $$f\in V'$$, $$f(v_n) \to f(v)$$)?

Since the 2-norm is continous on $$(\ell^2, \tau_A)$$, then $$S_A^\infty$$ and $$D^\infty_A$$ are closed in $$(\ell^2, \tau_A)$$. The closure of $$D^\infty_B$$ in $$(\ell^2, \tau_A)$$ is precisely $$\overline{D}^\infty_B = D^\infty_A$$, because:

1. It is possible to approximate every sequence $$x = (x_0, x_1, \ldots), ||x||_2 \leq 1$$ with the truncations $$(x_0, \ldots, x_n, 0,0, \ldots) \in D^\infty_B$$
2. Every sequence $$\{y^{(n)}\} \in D^\infty_B$$ that has limit in $$\ell^2$$, must have $$||y^{(n)}|| \leq 1$$.

With a similar argument, we can see that the closure of $$S^\infty_B$$ in $$(\ell, \tau_A)$$ is $$S^\infty_A$$, although the truncation of $$x \in S^\infty_A$$ have to be rescaled to have norm 1, so they are in $$S^\infty_B$$.

Question 3. Is it true that the closures of $$D^\infty_B, S^\infty_B$$ in $$(\ell^2, \tau_B)$$ are $$D^\infty_A, S^\infty_A$$?

In general, I would like to understand the differences between the spaces $$(S^\infty_B, \tau_B)$$ and $$(S^\infty_A, \tau_A)$$. For example, by the same first question, we know that both $$S^\infty_A$$ and $$S^\infty_B$$ are contractible. One way to do it, is to construct explicitly an homotopy between a generic $$x = (x_0, x_1, x_2, \ldots)$$ and $$(1,0,0,\ldots)$$.

1. We do $$f_t(x_0, x_1, \ldots) = (1-t)\cdot (x_0, x_1, x_2, \ldots) + t \cdot (0, x_0, x_1, \ldots)$$.
2. Then we use $$g_t(0, x_0, x_1, \ldots) = (1-t)(0,x_1,x_2,\ldots)+t(1,0,0,\ldots)$$. Combining $$f / |f|$$ and $$g/ |g|$$ we get the desired homotopy. Observe that if we start with $$x \in S^\infty_B$$, the homotopy has always image in $$S^\infty(B)$$.

For $$S^\infty_B$$, another interesting approach uses algebraic topology:

1. We inductively give a cell structure on $$S^n_B$$ as the union of $$S^{n-1}_B$$ and two $$n$$-cells (half-spheres).
2. The union of all $$S^n_B$$ gives a cell structure on $$S^\infty_B$$. Observe that the weak topology of cell complexes of $$S^\infty_B$$ is the same as $$\tau_B$$: one is the intersection with cells of size $$\leq n$$, the other one with $$\mathbb{R}^n$$, which are the same.
3. We have that $$H_n(S^\infty_B) = H_n (S^{n+1}_B)$$ for the theory of cell complexes. But $$H_n(S^{n+1}_B) = 0$$ (for example, by putting a friendlier cell structure on $$S^{n+1}_B$$, as one $$0$$-dimensional and one $$n+1$$-dimensional cell; the homology does not depend on cell structure). Therefore $$H_n(S^\infty_B)=0$$ for all $$n>0$$.
4. Using that $$S^\infty_B$$ is simply-connected, by Hurewicz theorem also $$\pi_n(S^\infty_B)=0$$ for all $$n$$. Then by Whitehead theorem, $$S^\infty_B$$ is contractible.

Wow, this is great! So we can deduce contractibleness from algebraic topology. It's cool, but it is also a bit overkill. Indeed, we could have deduced that $$H_n(S^\infty_B)$$ and $$\pi_n(S^\infty_B)$$ are all trivial from contractibility. And since also $$S_A^\infty$$ is contractible, it must have all trivial homology/homotopy groups.

However, maybe it is still interesting to study $$S^\infty_A$$ topologically? For example:
Question 4. It is possible to put a cell structure on $$(S^\infty_A, \tau_A)$$? It is possible to do in a way such that $$S^\infty_B$$ is a subcomplex of $$S^\infty_A$$?

## (TL;DR) What are the relations between the spaces $$D_A^\infty, D_B^\infty, S_A^\infty, S_B^\infty$$ with the topologies $$\tau_A, \tau_B$$?

• $\lVert x \rVert_2$ is divergent in general. Aug 6 at 0:42
• You misdefined $\ell^2$. Instead of $=1$ you need $< \infty$. Aug 6 at 8:47
• @KritikerderElche, corrected, thanks. Aug 6 at 11:01