# Question on free Boolean algebras

Every Boolean algebra $A$ is isomorphic to a field of set. In particular, if $A$ is finite, then $A$ is isomorphic to the power set of its atoms.

Now, suppose that $A$ is free Boolean algebra with 2 free generators (or atoms). Because $A$ is a finite Boolean algebra, it is isomorphic to the power set of its atoms. However, $A$ has $16$ elements, whereas the power set of its atoms has only 4 elements.

Could anybody explain what I seem to miss.

• Actually two generators x and y give four atoms (x and y, x and not y, not x and y, not x and not y). – David C. Ullrich Sep 11 '15 at 14:20
• @DavidC.Ullrich I think your comment should be an answer. – Alex Kruckman Sep 14 '15 at 23:49

In particular, if $B$ is freely generated by $\{x,y\}$, it turns out that $B$ has four atoms, namely $x\wedge y$, $\neg x\wedge y$, $x\wedge \neg y$, and $\neg x\wedge\neg y$. More generally, the free Boolean algebra on a finite set with $n$ elements has $2^n$ atoms, namely all the ways to form meets of the generators or their complements.