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It is well known that there cannot be a homeomorphism between $\mathbb{R}$ and $\mathbb{R^2}$ because if we remove the origin from $\mathbb{R}$ the space becomes disconnected, but if we remove the origin from $\mathbb{R^2}$ the resulting space is connected.

My question is, does there exist an analogous result showing that there cannot be a homeomorphism between $C([0,1])$ and $C([0,1]^2)$ where the sets of continuous functions have the topology induced by the sup-norm $\|f\| = \max_{x} |f(x)|$?

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    $\begingroup$ Kadec proved that any two infinite dimensional Banach spaces are homeomorphic. Long before that Millutin proved that $C(K)$ is linearly homeomorphic to $C(K)$ if $K$ is an uncountable compact metric space. See the book by Albiac and Kalton for a proof. $\endgroup$ Sep 6, 2016 at 20:07
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    $\begingroup$ Any two infinite-dimensional separable Banach spaces, of course. Kadec, M. I. A proof of the topological equivalence of all separable infinite-dimensional Banach spaces. (Russian) Funkcional. Anal. i Priložen. 1 1967 61–70. $\endgroup$ Sep 6, 2016 at 20:09
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    $\begingroup$ @BillJohnson, should one of those $C(K)$'s be, say, $C([0, 1])$? $\endgroup$
    – LSpice
    Sep 6, 2016 at 20:40
  • $\begingroup$ Incidentally, Torunczyk later generalized Kadets's result by showing that any two infinite dimensional locally convex Frechet spaces with the same density character are homeomorphic. $\endgroup$
    – Nik Weaver
    Sep 7, 2016 at 2:53
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    $\begingroup$ By the way, for the topology of pointwise convergence the homeomorphness of the spaces $C_p([0,1])$ and $C_p([0,1]^2)$ is an old open problem of Arkhangelski. In 1999 Robert Cauty by a true tour de force proved that for any $n\in\mathbb N$ the function space $C_p([0,1]^n)$ is not homeomorphic to $C_p([0,1]^\omega)$ but his method cannot be adapted to prove that $C_p([0,1])$ and $C_p([0,1]^2)$ are not homeomorphic. $\endgroup$ Feb 28, 2018 at 16:23

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