TITLE: Is $\beta \mathbb{N}$ homeomorphic to its own square?
QUESTION [23 upvotes]: Let $\mathbb{N}$ be the set of natural numbers and $\beta \mathbb N$ denotes the Stone-Cech compactification of $\mathbb N$. 
Is it then true that $\beta \mathbb N\cong \beta \mathbb N \times \beta \mathbb N $ ?

REPLY [5 votes]: An indirect argument: 
Since the Banach space of continuous functions $C(\beta\mathbb{N})$ is isomorphic to $\ell_\infty$, it contains no complemented copies of $c_0$. 
Since $C(\beta\mathbb{N}\times\beta\mathbb{N})$ is isomorphic to  $C\big(\beta \mathbb{N},C(\beta\mathbb{N})\big)$, 
it contains a complemented copy of $c_0$. See [P. Cembranos. 
$C(K,E)$ contains a complemented copy of $c_0$. Proc. Amer. Math. Soc. 
91 (1984), 556-558.]