Is $\beta \mathbb{N}$ homeomorphic to its own square?

Question

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 $ ?

Answer 1

The spaces $\beta \mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are not homeomorphic.

To derive a contradiction, assume that $\beta\mathbb N$ and $\beta\mathbb N\times \beta\mathbb N$ are homeomorphic. Since $\beta \mathbb N$ is homeomorphic to $\beta(\mathbb N\times\mathbb N)$, we conclude that there exists a homeomorphism $h:\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times \beta\mathbb N$. Taking into account that homeomorphisms preserve isolated points, we would conclude that $f=h|\mathbb N^2$ is a bijection of $\mathbb N\times \mathbb N$ and $h$ is a unique continuous extension of $f$. The bijective map $f^{-1}:\mathbb N^2\to\mathbb N^2$ extends to a homeomorphism $\beta(f^{-1}):\beta(\mathbb N^2)\to\beta(\mathbb N^2)$. Then the composition $H=h\circ\beta(f^{-1}):\beta(\mathbb N\times\mathbb N)\to\beta\mathbb N\times\beta\mathbb N$ is a homeomorphism extending the identity embedding $i:\mathbb N\times\mathbb N\to \mathbb N\times\mathbb N\subset\beta\mathbb N\times\beta\mathbb N$. Since $\mathbb N^2$ is dense in $\beta(\mathbb N\times\mathbb N)$, the map $H$ coincides with the unique continuous extension $\beta i$ of the identity embedding $i:\mathbb N\times\mathbb N\to\beta\mathbb N\times\beta\mathbb N$. To complete the proof, it remains to show that the map $\beta i$ is not injective.

Fix any free ultrafilter $\mathcal U_0$ on $\mathbb N$ and consider two distinct ultrafilters: $\mathcal U=\{\{(x,x):x\in U\}:U\in\mathcal U_0\}$ and $\mathcal V=\{\bigcup_{x\in U}\{x\}\times U_x:U\in\mathcal U_0,\;(U_x)_{x\in U}\in\mathcal U_0^U\}$ on $\mathbb N\times \mathbb N$. It can be shown that $H(\mathcal U)=\beta i(\mathcal U)=\beta i(\mathcal V)=H(\mathcal V)$. So, the map $H$ is not injective and can not be a homeomorphism.

Remark. By a result of Shelah and Velickovic, under PFA, each homeomorphism of $\omega^*=\beta\omega\setminus \omega$ is induced by some bijection between cofinite subsets of $\omega$. This implies that under PFA the space $\omega^*$ is not homeomorphic to its square.

Question. What happens under CH. Is $\omega^*$ homeomorphic to its square? The answer is affirmative if $(\omega^*)^2$ is an F-space, which means that two disjojnt open $F_\sigma$-sets in $(\omega^*)^2$ have disjoint closures.

Answer 2

(Just noticed - already done by Todd Trimble in a comment:)

A proof by Stone duality: the dual question is whether the Boolean algebras $\mathscr P\mathbb N$ and $\mathscr P\mathbb N\otimes\mathscr P\mathbb N$ are isomorphic. (Here "$\otimes$" is the coproduct in Boolean algebras.)

The answer is no since the former is complete and the latter is not: its completion is given by the canonical homomorphism $i:\mathscr P\mathbb N\otimes\mathscr P\mathbb N\to\mathscr P(\mathbb N\times\mathbb N)$, which can also serve to provide an explicit subset of $\mathscr P\mathbb N\otimes\mathscr P\mathbb N$ that does not posses a lub, by exhibiting an element of $\mathscr P(\mathbb N\times\mathbb N)$ not in the image of $i$.

Namely, the image of $i$ is the Boolean subalgebra of $\mathscr P(\mathbb N\times\mathbb N)$ generated by the rectangles $S\times T\subseteq\mathbb N\times \mathbb N$, for $S,T\subseteq\mathbb N$, while there clearly are subsets of $\mathbb N\times\mathbb N$ which are not finite Boolean combinations of such rectangles.

(Edit - found a simpler example: e. g. the diagonal $\mathbb N\subseteq \mathbb N\times\mathbb N$ is not a finite Boolean combination of rectangles; accordingly, the subset $$ \Delta:=\{\ \{n\}\otimes\{n\}\mid n=1,2,...\ \}\subseteq\mathscr P\mathbb N\otimes\mathscr P\mathbb N $$ does not have least upper bound. Indeed, an element $S_1\otimes T_1\lor\cdots\lor S_k\otimes T_k\in\mathscr P\mathbb N\otimes\mathscr P\mathbb N$ is an upper bound for $\Delta$ if and only if $\left(S_1\cap T_1\right)\cup\cdots\cup\left(S_k\cap T_k\right)=\mathbb N$. Then for some $i\in\{1,...,k\}$ there are $n_1,n_2\in S_i\cap T_i\setminus\bigcup_{j\ne i}S_j\cap T_j$ with $n_1\ne n_2$, and then in the above finite join we can replace $S_i\otimes T_i$ with $$\left(S_i\setminus\{n_1\}\otimes T_i\setminus\{n_1\}\right)\lor\left(S_i\setminus\{n_2\}\otimes T_i\setminus\{n_2\}\right),$$resulting in a strictly smaller upper bound.)

Turning back to "normal", this proof says that $\beta\mathbb N\times\beta\mathbb N$ is not extremally disconnected, while $\beta\mathbb N$ is.

Answer 3

The negative answer is equivalent to showing that there are two disjoint subsets $A,B$ of $\mathbf{N}^2$ with non-disjoint closures in $(\beta\mathbf{N})^2$. This be made explicit: take $A=\{(n,m):n=m\}$ and $B=\{(n,m):n>m\}$. Let $\omega$ be a non-principal ultrafilter on $\mathbf{N}$. Let $V$ be a neighborhood of $(\omega,\omega)$ in $(\beta\mathbf{N})^2$. Then $V$ contains $U\times U$ for some $U\in\omega$. The latter contains both points of $A$ and of $B$. Since this holds for every $V$, this shows that $(\omega,\omega)$ belongs to the closure of both $A$ and $B$.

Answer 4

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.]