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Let $A = \mathbb{R}^\mathbb{N}$ with the product topology. Let $B = \{ (x_n)_n \in A \mid \forall n : |x_n| \leq 1\} = [-1, 1]^\mathbb{N}$ be the $\infty$-dimensional cube. Let $C = \{ (x_n)_n \in A \mid \sum_{n=1}^\infty |x_n| \leq 1 \}$ be the $\infty$-dimensional cross-polytope. Let $\operatorname{d}B = \{(x_n)_n \in B \mid \exists n : |x_n| = 1\}$ be the pseudo-boundary of $B$. Let $\operatorname{d}C = \{(x_n)_n \in C \mid \sum_{n=1}^\infty |x_n| = 1 \}$ be the pseudo-boundary of $C$.

Are $B$ and $C$ homeomorphic? Are $\operatorname{d}B$ and $\operatorname{d}C$ homeomorphic? Does there exist a homeomorphism of $B$ and $C$ that restricts to a homeomorphism of $\operatorname{d}B$ and $\operatorname{d}C$?

A natural map would be $\varphi: \operatorname{d}C \to \operatorname{d}B, (x_n)_n \mapsto \frac{1}{\underset{m}{\operatorname{max}}|x_m|} \cdot (x_n)_n$. However, it is unclear to me whether this map is even continuous. Even more, the map is clearly not surjective. So this is not a homeomorphism of $\operatorname{d}B$ and $\operatorname{d}C$.

Smiley1000
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    Not sure if this is helpful, but let $c_0$, $\ell_1$, and $\ell_\infty$ denote the space of null, absolutely summable, and bounded sequences of real numbers, respectively. Each is a Banach space with its natural norm ($c_0$ and $\ell_\infty$ with the supremum norm $|\cdot|\infty$, $\ell_1$ with the norm $|(x_n)|_1=\sum_n |x_n|$). You're asking if the unit ball of $c_0^=\ell_1$ is weak$^$ homeomorphic to the unit ball of $\ell_1^*=\ell\infty$. The question on $dC$, $dB$ is the same question, but about the unit spheres. – user469053 Mar 26 '24 at 19:28
  • @user469053 This isn't quite correct as $\operatorname{d}B$ only contains elements that actually attain the supremum norm, so it is a little bit smaller than the unit sphere of $\ell^\infty$. – Smiley1000 Mar 26 '24 at 19:59
  • @user469053 Also, I don't think the weak* topology is the same as the topology of componentwise convergence. It is the same as the topology of pointwise convergence viewed as functions on $c_0$ and $\ell^1$, but not that is not quite the same as componentwise convergence in $\ell^1$ and $\ell^\infty$. – Smiley1000 Mar 26 '24 at 20:02
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    You are right about the attaining/not attaining the norm. But the weak$^$ topology on $B_{\ell_1}=B_{c_0^}$ is just coordinate-wise convergence, and the same is true for $B_{\ell_\infty}=B_{\ell_1^*}$. – user469053 Mar 26 '24 at 21:03
  • @user469053 You're right. On all of $(c_0)^* = \ell^1$ and $(\ell^1)^* = \ell^\infty$, the topologies of componentwise convergence and pointwise convergence do not agree, but they do agree on norm-bounded subsets such as $B_{(c_0)^} = B_{\ell^1}$ and $B_{(\ell^1)^} = B_{\ell^\infty}$. – Smiley1000 Mar 26 '24 at 21:32
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    To put this in a general context, if $(S_\lambda)$ is a net of bounded linear operators from a Banach space $X$ into a Banach space $Y$, then $$M:={x\in X:\lim_\lambda S_\lambda x \text{ exists}}$$ is closed. I've heard this referred to as a partial converse to the principle of uniform boundedness. By linearity, $M$ is a subspace of $X$. As a result, a bounded net $(x^_\lambda)$ in $X^$ is weak$^*$-convergent iff it converges pointwise on a subset with dense span, such as a Schauder basis. "Bounded" is necessary because we only assume the span is dense in $X$, rather than equal to $X$. – user469053 Mar 26 '24 at 21:46

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I don't know about $B$ and $C$ (I would guess they are homeomorphic but not by any simple explicit map) but $dB$ and $dC$ are not homeomorphic. To prove this, note first that the product topology on $dC$ is the same as the topology induced by the $\ell^1$ metric. The $\ell^1$ metric topology clearly contains the product topology; for the reverse direction, note that if $x\in dC$, then you can pick finitely many coordinates of $x$ whose absolute values add up to almost $1$, and then restricting those finitely many coordinates gives a product topology neighborhood of $x$ contained in a small $\ell^1$ ball around $x$.

In particular, $dC$ is completely metrizable, since it is just the unit sphere in the Banach space $\ell^1$. However, $dB$ is not completely metrizable because it does not satisfy the Baire category theorem: it is covered by the countable family of subsets $A_n=\{x\in dB: |x_n|=1\}$, which are closed and have empty interior in $dB$.

Eric Wofsey
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