Automorphisms of a boolean algebra

Question

Let $A = P(\mathbb{N})$ be the powerset of the natural numbers. We can look at $A$ as the Boolean aglebra - having in mind the obvious operations on elements of $A$.

What I am interested in knowing is if perhaps the following holds:

If $f:A\mapsto A$ is an automorphism of $A$ then $f(\{x\}) = \{y\}$ where $x,y \in \mathbb{N}$

In other words, an automorphism of $A$ preserves singletons (see this for a definition of homomorphism ).

I was not able to find a proof of this fact but neither a counterexample.

Is anyone able to settle this question for me?

Answer 1

Atoms of the Boolean algebra $\mathcal P(\mathbb N)$ are precisely the singletons.

The property of being an atom (atomicity, if there is such a word) is preserved by isomorphisms. (In particular, by automorphisms).

---

If you only want homomorphism from $\mathcal P(\mathbb N)$ to itself, then you can take any ultrafilter $\mathcal F$ on $\mathbb N$ and put $$\varphi(A)= \begin{cases} \mathbb N & A\in\mathcal F, \\ \emptyset & \text{otherwise}. \end{cases} $$

This is a homomorphism, which does not map singletons to singletons.

Answer 2

Suppose that $f(\{x\})$ is not a singleton, then there are two distinct $u,v\in f(\{x\})$ therefore $\{u\},\{v\}\subseteq f(\{x\})$. This means that $f^{-1}(\{u\})$ and $f^{-1}(\{v\})\subseteq\{x\}$ and that both are non-empty.

This is a contradiction that this implies $u=v$.