Whether an isotone bijection from a power set lattice to another sends singletons to singletons

Question

By the work of Paul Cohen (on the continuum hypothesis), one can neither prove nor disprove from the axioms of ZFC that a bijection $f$ from the power set $\mathcal{P}(S)$ of a set $S$ to the power set $\mathcal{P}(T)$ of a set $T$ implies a bijection from $S$ to $T$.

Question. What if $f$ is also isotone, meaning that if $X \subseteq Y \subseteq S$ then $f(X) \subseteq f(Y)$? More precisely, does an isotone $f$ send singletons to singletons? Or is this also independent from ZFC?

In fact, I don't know of a single example of an isotone bijection from a power set lattice to another (or even into itself) that is not an order isomorphism (namely, whose functional inverse is not isotone too). This may be relevant, considering that if $f$ is isotone, then $f^{-1}(\{y\})$ is a singleton for each $y \in T$.

It is not difficult to show that an isotone bijection from one complete lattice to another need not be an order isomorphism. In a way, I'm asking whether power set lattices are special in this regard.

Answer 1

Here is a more general fact:

Proposition: Any monotone bijection $f: A \to B$ between two Boolean algebras is an isomorphism.

Claim: $f(0) = 0$ and $f(1) = 1$.

Proof of Claim: For all $a \in A$, $0 \leq a \leq 1$ implies $f(0) \leq f(a) \leq f(1)$. Since $f$ is surjective, the result follows. $\square$

Claim: $f(a^c) = f(a)^c$ for all $a \in A$.

Proof of Claim: Since $f$ is surjective, there exists $b \in A$ s.t. $f(b) = f(a)^c$. But then by monotonicity, $f(a \wedge b) \leq f(a) \wedge f(b) = f(a) \wedge f(a)^c = 0$, so $f(a \wedge b) = 0$. As $f(0) = 0$, by injectivity, $a \wedge b = 0$. Similarly, $a \vee b = 1$. Thus, $b = a^c$, i.e., $f(a^c) = f(a)^c$. $\square$

Proof of Proposition: It suffices to show that $f(a) \leq f(b)$ implies $a \leq b$ for all $a, b \in A$. As $f(a) \leq f(b)$, we have $f(a \wedge b^c) \leq f(a) \wedge f(b^c) = f(a) \wedge f(b)^c = 0$, so $f(a \wedge b^c) = 0$ and by injectivity, $a \wedge b^c = 0$, i.e., $a \leq b$. $\square$

Since the lattice of all subsets of a given set is a Boolean algebra, and any (order-)isomorphism between lattices of all subsets clearly has to preserve singletons, the result follows.

Answer 2

Yes, such a mapping necessarily sends singletons to singletons.

Let $f\colon\mathcal P(S)\to\mathcal P(T)$ be a monotone bijection (or more generally, a surjective strictly monotone function). By monotony, $f(\varnothing)\subseteq f(X)\subseteq f(S)$ for all $X\subseteq S$, hence using the surjectivity of $f$, $f(\varnothing)=\varnothing$ and $f(S)=T$. Also, as noted in the question, $f$-inverses of singletons must be singletons: if $f(X)=\{t\}$, we must have $X\ne\varnothing$; if we further assume for contradiction that there exists $\varnothing\ne X'\subsetneq X$, then $\varnothing\ne f(X')\subsetneq\{t\}$, which is impossible.

Thus, there is $S'\subseteq S$ and a bijection $g\colon S'\to T$ such that $f(\{s\})=\{g(s)\}$ for all $s\in S'$. But then by monotony, $\{t\}\subseteq f(S')$ for all $t\in\operatorname{im}(g)=T$, i.e., $f(S')=T=f(S)$, which implies $S'=S$.

Furthermore, we have $f(X)=g[X]$ for all $X\subseteq S$, hence $f$ is an order isomorphism. To see this, note that for any $s\in S$, $f(S\smallsetminus\{s\})$ includes $f(\{s'\})=\{g(s')\}$ for all $s'\ne s$, but it is strictly contained in $T=f(S)$, hence $f(S\smallsetminus\{s\})=T\smallsetminus\{g(s)\}$. Then for any $X\subseteq S$, we have $\{g(s)\}\subseteq f(X)\subseteq T\smallsetminus\{g(s')\}$ for all $s\in X$ and $s'\notin X$, whence $f(X)=g[X]$.