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

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.

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 $|I| = |J|$.

• Nice one, thanks! – cats Mar 6 '14 at 1:22

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 $|I|=|J|$.

• unfortunately i don't know this stuff yet =/. this is an exercise in Conway's "A Course in Functional Analysis," chapter III, section 1, and there is very little material on banach spaces before this. – cats Feb 28 '14 at 7:04

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.

• Nice. Of course, extreme points are introduced only in Chapter V... – user127096 Mar 6 '14 at 0:11
• That's true, but after seeing this solution, I don't feel as if I just didn't know enough to do the problem, since the notion of extreme point is very intuitive and is just a sort of generalization of vertices of a (convex) polytope – cats Mar 6 '14 at 0:13
• In retrospect, I spent too much time thinking about how to control where characteristic functions went, and it makes sense that that didn't work, since those are not the extreme points on the unit ball. – cats Mar 6 '14 at 0:15
• Nice answer. I stole your idea of counting a collection of pairwise disjoint subsets for my answer, which I think is similar in spirit but avoids using extreme points. – Nate Eldredge Mar 6 '14 at 1:11

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.

• also don't know enough for this. thanks though – cats Feb 28 '14 at 7:04
• @lyj I know. IMHO, Conway is extremely poorly organized. The Banach–Stone theorem I am referring to (p. 172), for example, appears only way after the exercise you posted. Same goes for the Stone–Čech compactification (sec. V.6). – triple_sec Feb 28 '14 at 7:07
• yes, though most of the exercises so far i've been able to do without using anything in future sections, so i was hoping it would be possible for this one as well. in particular, the fact that the condition is the existence of an isometric isomorphism makes me think that there should be a simpler solution – cats Feb 28 '14 at 7:10
• There may be a solution that doesn't need future sections, but I doubt it's simple. What I meant by poor organization is that some exercises could be done way more simply if you already had had a solid background in later sections. I like chapter 5 of Folland (1999) on functional analysis so much better for this reason (even though that material is considerably thinner): It begins with the most general form (Banach spaces) and proceeds toward the more special (Hilbert spaces), which I have found much more efficient and fun. – triple_sec Feb 28 '14 at 7:15
• yes, a lot of exercises could be done quite easily using the uniform boundedness principle and such. i'll take a look at folland and see if i can make sense of one of the suggested solutions here. thanks! – cats Feb 28 '14 at 7:17