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 $ ?

  • 7
    $\begingroup$ Instead of putting negative marks better write an answer. Otherwise this will be a discouraging experience for MathsAlppila. The question seems to be reasonable and not entirely trivial. $\endgroup$ – Taras Banakh Jul 25 '17 at 19:32
  • $\begingroup$ I appreciate it Taras...in any case I would like to know a map if it exists... $\endgroup$ – MathsAlppila Jul 25 '17 at 19:34
  • $\begingroup$ @Qfwfq: If you're fixin' the title, might as well fix the body also... $\endgroup$ – Asaf Karagila Jul 25 '17 at 19:59
  • 4
    $\begingroup$ I don't have the leisure at the moment to develop this, but I'd be tempted to work on the other side of Stone duality. $\beta N$ is the Stone space of the Boolean ring $P N$ and $\beta N \times \beta N$ is the Stone space of the tensor product of commutative rings $P N \otimes P N$. My thinking is to show the latter can't possibly be complete as a Boolean algebra. $\endgroup$ – Todd Trimble Jul 25 '17 at 20:48
  • 5
    $\begingroup$ This seems related: Why is $\beta(\mathbb{N}\times\mathbb{N})\neq\beta(\mathbb{N})\times\beta(\mathbb{N})$? (math.SE) $\endgroup$ – Martin Sleziak Jul 26 '17 at 7:51

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.

  • $\begingroup$ Why is the unique continuous extension bijective? (I agree that the answer should be 'no'). $\endgroup$ – Todd Trimble Jul 25 '17 at 20:19
  • $\begingroup$ @ToddTrimble This a delicate place. I will add one more sentence in the proof. $\endgroup$ – Taras Banakh Jul 25 '17 at 20:23
  • $\begingroup$ I think here's what's going on: Let $h|_N : N \longrightarrow N \times N$ be the bijection. Then we could just define a bijection $f : N \times N \to N$ as $f(m,n):= h^{-1}(m,n)$. Then obviously $h\circ f$ is identity map from $N \times N$ to $N \times N$. Since $\beta(N) \cong \beta(N\times N)$, each of the maps $f$ and $h$ clearly extends to bijections. $\endgroup$ – MathsAlppila Jul 25 '17 at 20:29
  • 2
    $\begingroup$ No, $\omega^*$ is not homeomorphic to its square (in ZFC). The reason is that if $S$ is any countable discrete subset of $\omega^*$, then the closure of $S$ in $\omega^*$ is homeomorphic to $\beta S$. However, the square of $\omega^*$ contains the square $\beta S \times \beta S$ and, as noted above, this set is not homeomorphic to $\beta (S \times S)$. $\endgroup$ – Anonymous Jul 26 '17 at 1:19
  • $\begingroup$ @Anonymous Thank you for the answer to my question (formulated in the answer to the question of MathsAlppila). In fact, this answers also the question of MathsAlppila since the homeomorphness of $\beta\mathbb N$ to its square implies the homeomorphness of $\omega^*$ to its square. $\endgroup$ – Taras Banakh Jul 26 '17 at 5:19

(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.


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 $B$.

  • 1
    $\begingroup$ Another reason why $\beta\mathbb N\times\beta\mathbb N$ is not homeomorphic to $\beta\mathbb N$ is that the diagonal $\Delta$ of the square $\beta\mathbb N\times\beta\mathbb N$ coincides with the closure of the diagonal of $\mathbb N\times\mathbb N$ but is not open in $\beta\mathbb N\times\beta\mathbb N$. On the other hand, the closure of any subset $A\subset\mathbb N$ in $\beta\mathbb N$ is open in $\beta\mathbb N$. $\endgroup$ – Taras Banakh Jul 26 '17 at 5:24
  • 2
    $\begingroup$ @TarasBanakh It's not another reason, but just a restatement of the same reason. Indeed, when $X$ is a dense discrete open subset of a space $Y$, to say that $A$ has open closure in $Y$ is equivalent to say that $A$ and $X-A$ have disjoint closures. $\endgroup$ – YCor Jul 26 '17 at 11:57

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.]

  • $\begingroup$ Is it clear that the topology on $C(\beta\mathbb N)$ that makes that isomorphism work is the same as the Banach space topology on $\ell_\infty$? $\endgroup$ – მამუკა ჯიბლაძე Aug 2 '17 at 8:26
  • $\begingroup$ $K$ homeomorphic to $L$ implies $C(K)$ isomorphic to $C(L)$. $\endgroup$ – M.González Aug 2 '17 at 8:35
  • $\begingroup$ Sorry I don't understand why this helps. I mean this: the bijection $C(X\times Y,Z)\approx C(X,C(Y,Z))$ works for certain topology on $C(Y,Z)$. Is it clear that with $Y=\beta\mathbb N$ and $Z=\mathbb R$ one gets the same topology as in that result of Cembranos that you cite? $\endgroup$ – მამუკა ჯიბლაძე Aug 2 '17 at 9:23
  • $\begingroup$ For $K$ a compact space, $C(K)$ and $C(K,E)$ are endowed with their supremum norms, and the isomorphisms are considered for the topologies associated to this norm. $\endgroup$ – M.González Aug 2 '17 at 9:36
  • 1
    $\begingroup$ If $L$ is compact, then the supremum norm $\|f\|_\infty =\sup_{t\in L}|f(t)|$ induces the compact open topology in $C(L)$. $\endgroup$ – M.González Aug 2 '17 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.