# A free boolean algebra

Consider the following definition:

The boolean algebra $A$ is generated freely with the subset $G \subseteq A$ if for every boolean algebra $B$ and map $f:G \mapsto B$ there is precisely one homomorphism $\overline{f}:A \mapsto B$ extending $f.$ That is $\overline{f}(x) = f(x)$ for all $x \in G.$

I would like to check if $A = P(\mathbb{N})$ is generated freely by some subset $G.$

Clearly if $G$ generates $A$ then $G$ is not the empty set and it does not contain 0 or 1. Define a mapping $f$ such that a fixed $x \in G$ is mapped to an atom if $x$ is not an atom, and to a non atom element otherwise.

Since any automorphism preserves atoms there is no way to extedn $f$ to a homomorphism from $A$ to $A$.

The above reasoning (if correct) would therefore imply that $A$ is not generated freely.

Am I mistaken?

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What if $G$ itself has no atoms at all? –  Asaf Karagila Jun 5 '12 at 21:24
I've edited the answer. G clearly has some element. If $G$ has no atoms at all then take any element and construct a map taking the element to an atom. –  Jernej Jun 5 '12 at 21:30
I'm quite sure $\mathscr{P}(X)$ is a free boolean algebra for any finite set $X$. Where have you used the infiniteness of $\mathbb{N}$? –  Zhen Lin Jun 5 '12 at 21:32
Suppose $\{x\},\{x,y\}\in G$. Then you send $\{x,y\}$ to some atom and $\{x\}$ to a whole other set, this would not preserve the fact that $\{x\}\cap\{x,y\}=\{x\}\implies f(\{x\})\cap f(\{x,y\})=f(\{x\})$. Your new map cannot be extended to a homomorphism, at all. –  Asaf Karagila Jun 5 '12 at 21:33
@Zhen Lin According to wikipedia every free bolean algbebra has $2^{2^n}$ elements which would imply that $P(X)$ is not always free even for finite $X$? Am I missing something? –  Jernej Jun 5 '12 at 21:37

Indeed $P(\mathbb{N})$ is not free. All infinite freely generated Boolean algebras are atomless, while $A$ is atomic. Furthermore, only the Boolean algebras that are generated by a finite set $G$ are finite (with $2^{2^{|G|}}$ many elements -if I'm not mistaken).

To see that if $G$ is infinite then $A$ is, is trivial. To see that if $G$ is finite then $A$ is finite you just need to check that $A$ is the Lindenbaum algebra of propositional logic with $|G|$ many atoms (check that this algebra satisfies the requirements).

To see that if $G$ is infinite then $A$ is atomless you do the following: Take an atomless Boolean algebra $B$ with $|G|$ many elements and define a mapping $f:G\to B$ such that $f[G]$ is dense, i.e. for every element $b\in B$ there is some element in $c\in f[G]$ such that $c\leq b$. Then take the homomorphism $\bar{f}$ that extends $f$. If $A$ has an atom $a$, take $f(a)$, since $B$ is atomless, there is some $d<f(a)$ and there is some $c\in G$ such that $f(c)\leq d$. Then $0<f(a)\land f(c)<f(a)$ while $f(a\land c)$ is either $0$ or $f(a)$.

Actually you can completely describe what the freely generated Boolean algebra looks like, though it's a bit troublesome. You can find this is, for example, in Jonhstone's "Stone Spaces" (at the very end of the first chapter). The basic idea behind the construction lies in the fact that all propositions can be written in disjunctive normal form.

P.S.: I don't think your argument is correct, because you didn't seem to use the infiniteness of $\omega$ and as I have argued there are atomic freely generated Boolean algebras.

EDIT: The problem with your argument is that you assume that $\bar{f}$ will be an automorphism but this may not be the case. All that is required is that $\bar{f}$ is a homomorphism, which doesn't need to preserve atoms.

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Just a small question. Why is there always an atomless boolean algebra of given cardinality |G| –  Jernej Jun 6 '12 at 10:55
@Azoo: That's a simple Lowenheim-Skolem argument. We know there is one atomless Boolean algebra. By the upwards and downwards Lowenheim-Skolem theorems we can find such an algebra of any size. Actually the unique (up to isomorphism) atomless Boolean algebra of countable size is the free algebra generated by a countable set. –  Apostolos Jun 6 '12 at 11:15