Does a homeomorphism between Sierpinski space powers imply equality of exponents?

Let $$2 = \{0, 1\}$$, and let $$[2] = \{\emptyset, \{1\}, \{0, 1\}\}$$ denote the Sierpinski space. Let $$m, n$$ be arbitrary cardinal numbers, and $$[2]^m \approx [2]^n$$; i.e. that these product topologies are homeomorphic.

Is $$m = n$$?

Notes

• The Weak Generalized Continuum Hypothesis (WGCH) states that $$|2^m| = |2^n|$$ implies $$m = n$$. In general this does not hold.
• Suppose WGCH holds. Since $$[2]^m \approx [2]^n$$, then $$|2^m| = |2^n|$$ because a homeomorphism bijects the underlying sets. By WGCH, $$m = n$$. Hence the result is true under WGCH.
• The question is whether the result can be proved without WGCH.
• If one could show that $$|2^m| = |2^n|$$ implies $$[2]^m \approx [2]^n$$, then that would show that the stated problem is equivalent to WGCH.
• Pretty sure for infinite $m$, you can retrieve $m$ from the topology on $[2]^m$ as the smallest cardinal s.t. the intersection of that many open sets can be a singleton. Commented Jul 24 at 6:50
• Yes, that sounds like it works. Good thinking.
– kaba
Commented Jul 24 at 7:08
• $m$ is also the weight of $[2]^m$ for infinite $m$ (the smallest cardinality of a basis). Commented Jul 24 at 7:40
• @Adayah Nice! Something along the lines: create a subbasis from those $m$ sets where exactly one coordinate is $1$ (+ the whole set), and then the set of all finite intersections of that infinite subbasis has the same same cardinality.
– kaba
Commented Jul 24 at 7:50
• That suffices for upper bound for weight, but then there is still the lower bound left.
– kaba
Commented Jul 24 at 7:56

Let me formalize my comment:

Proposition: Let $$m$$ be an infinite cardinal, then in $$[2]^m$$, the only singleton that is an intersection of open sets is $$\{1\}^m$$. Furthermore, the least cardinality of a collection of open sets whose intersection is $$\{1\}^m$$ is $$m$$.

Proof: First, assume to the contrary that some other singleton $$\{f\}$$ is an intersection of open sets $$O_i$$. Then $$f(\kappa) = 0$$ for some $$\kappa < m$$. But then any $$O_i$$, which is an open neighborhood of $$f$$, must contain a basic open neighborhood of the form $$\prod_{\lambda < m} A^i_\lambda$$ where $$A^i_\kappa = \{0, 1\}$$ (since the only open neighborhood of $$0$$ in $$[2]$$ is $$\{0, 1\}$$) and $$f(\lambda) \in A^i_\lambda$$ for all $$\lambda \neq \kappa$$. Thus, if we let $$f’ \in [2]^m$$ be defined by $$f’(\lambda) = f(\lambda)$$ whenever $$\lambda \neq \kappa$$ and $$f’(\kappa) = 1$$, we must have $$f’ \in \cap_i O_i$$, contradicting the assumption that $$\cap_i O_i$$ is a singleton.

Now, we observe that $$\{1\}^m$$ is indeed an intersection of $$m$$ open sets, namely the intersection of $$O_\kappa = \prod_{\lambda < \kappa} 2 \times \{1\} \times \prod_{\kappa < \lambda < m} 2$$ for $$\kappa < m$$. $$m$$ is also the least cardinal with this property. Indeed, assume $$\{1\}^m = \cap_{i \in I} O_i$$. Then each $$O_i$$ necessarily contains a basic open neighborhood $$\prod_{\lambda < m} B^i_\lambda$$, where for each $$i$$, for all but finitely many $$\lambda$$ we have $$B^i_\lambda = 2$$ and the remaining $$B^i_\lambda$$ equal $$\{1\}$$. Thus, we have a one-to-finite map $$I \ni i \mapsto \{\lambda < m: B^i_\lambda \neq 2\}$$. We must have $$\cup_{i \in I} \{\lambda < m: B^i_\lambda \neq 2\} = m$$, as, otherwise, assume $$\kappa < m$$ is not in the set on the LHS. Then $$B^i_\kappa = 2$$ for all $$i$$. If we let $$f \in [2]^m$$ be defined by $$f(\lambda) = 1$$ for all $$\lambda \neq \kappa$$ and $$f(\kappa) = 0$$, then we easily see that $$f \in \cap_i O_i$$, contradicting the assumption that $$\cap_i O_i = \{1\}^m$$. Hence, $$m \leq |I| \times \aleph_0$$. As $$m$$ is infinite, this implies $$m \leq |I|$$. $$\square$$

The desired result is now an easy consequence of the proposition above.

• Nice! In the latter paragraph, the uses of $[2]$ (topology) should be replaced with $2$ (underlying set).
– kaba
Commented Jul 24 at 7:44
• @kaba True - fixed. Commented Jul 24 at 8:23

It suffices to show that $$m$$ is the weight of $$[2]^m$$ for infinite $$m$$. Given that, if $$[2]^m \approx [2]^n$$, then these spaces must have equal weights and so $$m = n$$.

For $$i < m$$ let $$\pi : [2]^m \to [2]$$ denote the projection onto the $$i$$-th coordinate. The family

$$\mathcal{C} = \{ \pi_i^{-1}[ \{ 1 \} ] : i < m \}$$

is a subbasis of $$[2]^m$$ of size $$m$$, so its finite intersections form a basis of size $$m$$.

Now assume for contradiction that $$\mathcal{B}$$ is a basis of $$[2]^m$$ of size less than $$m$$. For any $$i < m$$ there is $$B_i \in \mathcal{B}$$ such that $$\varnothing \neq B_i \subseteq \pi_i^{-1}[ \{ 1 \} ]$$. Then there is $$B \in \mathcal{B}$$ such that $$B_i = B$$ for infinitely many $$i < m$$. It follows that

$$\varnothing \neq B \subseteq \bigcap_{i \in I} \pi_i^{-1}[ \{ 1 \} ]$$

for some infinite set $$I \subseteq m$$. But the intersection on the right has empty interior, which is a contradiction.

• I feel like I'm going to get corrected too in regard to $2$ vs $[2]$, so let me preemptively ask: since $[2]^m \approx [2]^n$ is a correct statement (because it's from the OP), doesn't this mean that $[2]$ denotes the whole topological space rather than just the topology? Conventionally it's not the topologies, but the topological spaces that can be homeomorphic. Commented Jul 24 at 11:45
• Exactly, $[2]$ is the whole topological space (i.e. the set of open subsets). Often a space is given as a pair $(X, \mathcal{T})$ where $\mathcal{T}$ is a topology on $X$ (e.g. $(2, [2])$). But $X$ is always the union of $\mathcal{T}$, so it is redundant. I tend to state things in terms of the topology (e.g. that $[2]$ is compact, rather than $(2, [2])$ is compact) simply because it makes things shorter.
– kaba
Commented Jul 24 at 19:36
• (Also, I don't say the underlying set is compact (e.g. "$2$ is compact"), because then the topology is not clear. Being explicit on the topologies has been necessary on my current work, which routinely assigns different topologies to the same set in the same proof. Some other work can get away with being implicit though.)
– kaba
Commented Jul 24 at 20:10
• The set $\mathcal{C}$ is missing the whole set $2^m$. Otherwise, looks good!
– kaba
Commented Jul 24 at 20:15
• I believe a basis for $[2]^m$ in the arbitrary case is given by $\mathcal{U} = \{\{x \in 2^m : \forall i \in I: x_i = 1\} : I \in \hat{\mathcal{P}}(m)\}$, where $\hat{\mathcal{P}}(m)$ denotes the set of finite subsets of $m$. These are all distinct sets, so $|\mathcal{U}| = |\hat{\mathcal{P}}(m)|$, which for finite $m$ is $2^m$, and for infinite $m$ just $m$. This basis has also minimum cardinality, so in general $w([2]^m) = |\hat{\mathcal{P}}(m)|$.
– kaba
Commented Jul 24 at 22:14