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 $\tau_A, \tau_B$?

Related questions: 1) Unit sphere in $\mathbb{R}^\infty$ is contractible?, 2) Infinite-Dimensional Unit Ball, Sphere, and Disc Question, 3) Prove: The weak closure of the unit sphere is the unit ball.

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


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