# Extremally disconnected space

Can one view $\ell_{\infty}^{4}$, $\mathbb{R}^{4}$ equipped with the $\ell_{\infty}$-norm, as a space of continuous functions on some extremally disconnected space? and what would the extremally disconnected space be?

Thanks!

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As far as I know the $\ell_\infty-$norm is $\sup\{\left|x_i\right|,\left|y_i\right| \} < \infty$, which induces the same topology as the usual distance on $\mathbb{R}^4$. Could you elaborate a little bit more on what you had in mind? – Andy Oct 1 '11 at 10:41
@Andy: I don't know what the motivation for the question is but I assume it's this: there is a theorem by Nachbin-Goodner-Kelley saying that a space of continuous functions $C(K)$ is a dual space if and only if $K$ is compact, Hausdorff and extremally disconnected. The finite-dimensional case is completely trivial but things get interesting in infinite dimensions. For example, $\ell^{\infty}(\mathbb{N}) = C(\beta\mathbb{N})$ is the space of continuous functions on the Stone-Čech compactification $\beta\mathbb{N}$ of $\mathbb{N}$ (essentially by definition of the Stone-Čech compactification). – t.b. Oct 1 '11 at 11:15
Oh ok, I get it now...I got here because of the topology tag, but I see now I'm not going to be able to contribute :) – Andy Oct 1 '11 at 11:53
@t.b. there exist compact, extremally disconnected spaces $K$ such that $C(K)$ is not a dual space. You need something more (i.e. to be hyperstonean). – Jorg Aug 12 '13 at 17:19

Take the four point space $X = \{0,1,2,3\}$ with the discrete topology. Then $C(X,\mathbb{R}) = \ell^{\infty}(X,\mathbb{R})$ is the space $\mathbb{R}^4$ equipped with the $\sup$-norm. Obviously, $\{0,1,2,3\}$ is extremally disconnected, as every subset is closed and open.
Added: Of course, this generalizes to all finite dimensions. If $n = \{0,1,2,\ldots,n-1\}$ is the $n$-point space with the discrete topology, then $C(n) = \ell^{\infty}(n)$ is $\mathbb{R}^n$ with the $\sup$-norm.
There is no wiggle room here: by a theorem of Banach (metric case) Stone (general case) two spaces $C(K)$ and $C(L)$ with $K$ and $L$ compact Hausdorff are isometrically isomorphic if and only if $K$ and $L$ are homeomorphic. The proof is not hard: if $T: C(K) \to C(L)$ is an isometric isomorphism then its adjoint is a homeomorphism from the unit ball of $C(L)^{\ast}$ to the one of $C(K)^{\ast}$ in the weak$^{\ast}$-topology. Then one finishes off by observing that $K$ and $L$ are stored as extremal points of the unit balls. Details can be found e.g. in Theorem 2 of §3 in Chapter V on page 115 of Day's Normed Linear Spaces, 3rd edition, Springer, 1973.