Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

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 $u=v$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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