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

Solution 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 $|I| = |J|$.

Solution 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 $|I|=|J|$.

Solution 3:

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.

Solution 4:

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.