$l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$

Question

Is it possible for $l^\infty (I)$ and $l^{\infty} (J)$ to be isometrically isomorphic with the cardinality of $I$ not equal to the cardinality of $J$? I'm able to show that if $1\le p < \infty,$ then $l^{p} (I)$ and $l^p (J)$ are isometrically isomorphic iff $|I| = |J|,$ but this relies on constructing a dense subset of $l^p (I)$ with cardinality equal to that of $I.$ Obviously this fails in $l^{\infty}$ since $l^\infty$ isn't separable, but I'm having trouble coming up with an example.

Edit: This is an exercise in Conway's "A Course in Functional Analysis," Chapter III, Section 1, just after the introduction to Banach spaces, so it seems that it can be done from first principles.

Answer 1

Here's a solution from first principles. As in lyj's answer, we use $\ell^\infty(I)$ to detect the number of pairwise disjoint subsets of $I$.

Let $S$ be the unit sphere of $\ell^\infty(I)$. For a set $E \subset S$, let me say that $E$ is dispersed if for every distinct $f,g \in E$, $\|f + g\| \le 1$ and $\|f - g\| \le 1$.

Lemma. If $E \subset S$ is dispersed, then $|E| \le |I|$.

Proof. For $f \in S$, let $\psi(f) = \{ x \in I : |f(x)| > 1/2\}$. Clearly $\psi(f)$ is nonempty. Now suppose $f,g \in S$ and $x \in \psi(f) \cap \psi(g)$. If $f(x)$ and $g(x)$ have the same sign, then clearly $|f(x) + g(x)| > 1$, and if $f(x)$ and $g(x)$ have opposite signs, then $|f(x) - g(x)| > 1$. So if $\|f + g\| \le 1$ and $\|f - g\| \le 1$, we see that $\psi(f) \cap \psi(g) = \emptyset$. Thus since $E$ is dispersed, the sets $\{ \psi(f) : f \in E\}$ are nonempty and pairwise disjoint. Choosing for each $f$ a point of $\psi(f)$, we have an injection from $E$ into $I$.

Now the solution follows quickly. Let $I,J$ be sets and suppose $T : \ell^\infty(I) \to \ell^\infty(J)$ is an isometry. Let $E \subset \ell^\infty(I)$ be the set of indicators of singletons: $E = \{ 1_{\{x\}} : x \in I\}$. Clearly $E$ is dispersed. Since $T$ is a linear isometry, $T(E)$ is also dispersed. Then by our lemma, we have $|T(E)| \le |J|$. But since $T$ is an injection, $|T(E)| = |E| = |I|$. So $|I| \le |J|$. If $T$ is an isometric isomorphism, then by symmetry $|I| \ge |J|$ as well, and the Cantor-Bernstein-Schroeder theorem gives us $|I| = |J|$.

Answer 2

As any von Neumann algebra $\ell_\infty(X)$ has unique predual, and in our case this is $\ell_1(X)$. Therefore $\ell_\infty(I)\cong\ell_\infty(J)$ implies $\ell_1(I)\cong\ell_1(J)$. As you already proved, the last isomorphism is possible iff $|I|=|J|$.

Answer 3

Note that $\ell^{\infty}(I)=BC(I)$ (set of bounded continuous functions) when $I$ is endowed with the discrete topology. In turn, $BC(I)\cong BC(\beta(I))$, where $\beta(\cdot)$ denotes Stone–Čech compactification. Therefore, if $\ell^{\infty}(I)\cong\ell^{\infty}(J)$, then $BC(\beta(I))\cong BC(\beta(J))$. The Banach–Stone theorem, in turn, implies that $\beta(I)$ and $\beta(J)$ are homeomorphic, from which it follows that $I$ and $J$ are homeomorphic, too.

Answer 4

A solution, by my professor:

Consider the closed unit ball in $\ell^{\infty}(I).$ The extreme points of this are functions $f: I \to \{-1, 1\}.$ Notice that if $\phi: \ell^{\infty}(I) \to \ell^{\infty}(J)$ is an isometric isomorphism, then $\phi$ takes extreme points to extreme points. We may assume without loss of generality that $\phi(\chi_I) = \chi_J,$ since if not, we can multiply each coordinate in $\phi(\chi_I)$ appropriately to get another isometry. Here, $\chi_I$ is the characteristic function on $I$ (so all $1$s).

Now, for any other extreme point $f \in \ell^{\infty}(I),$ we have $\frac{\chi_I + f}{2} = \chi_A$ for some $A\subset I.$ Moreover, we can get any characteristic function this way. Then $\phi(\chi_A) = \frac{\chi_J + \phi(f)}{2} = \chi_B$ for some $B \subset J$ since $\phi(f)$ is an extreme point in $\ell^{\infty}(J).$ We'll denote $B$ by $\varphi(A).$

Note then that for disjoint $A, A^{\prime}\subset I,$ if $\varphi(A) = B$ and $\varphi(A^{\prime}) = B^{\prime},$ then $B\cap B^{\prime} = \emptyset$ since $\phi(\chi_A + \chi_{A^{\prime}}) = \chi_B + \chi_{B^{\prime}}$ has norm $1.$ We now consider the set $\mathcal{B} = \{\varphi(\{i\}): i\in I\}.$ Since $\varphi({i})\cap \varphi({i^{\prime}}) = \emptyset$ for $i\neq i^{\prime},\, |\mathcal{B}| \le |J|$ (any collection of pairwise disjoint subsets has cardinality less than that of the set since we can construct an injection by arbitrarily choosing an element from each such subset). But $\varphi$ is a bijection from $I$ to $\mathcal{B},$ so $|I| \le |J|.$ Similarly, $|J| \le |I|,$ so we are done.