# Representation of free Boolean sigma-algebras

By a theorem of Loomis and Sikorski, for every Boolean $\sigma$-algebra $\mathfrak{A}$ there exists a $\sigma$-field of sets $\mathcal{F}$ and a $\sigma$-ideal $\Delta$ such that $\mathfrak{A}$ is isomorphic to $\mathcal{F}/\Delta$.

More precisely, let $X$ be the Stone space of $\mathfrak{A}$, $\mathcal{F}$ be the least $\sigma$-field (of subsets of $X$) containing all open-closed subsets of $X$, and $\Delta$ be the $\sigma$-ideal of all subsets of $\mathcal{F}$ of first category in $X$. Then $\mathfrak{A}$ is isomorphic to $\mathcal{F}/\Delta$.

Does the above theorem hold when $\mathfrak{A}$ is a free Boolean $\sigma$-algebra? In other words, if $X$ denotes the Cantor set, $\mathcal{F}$ the least $\sigma$-field (of subsets of $X$) containing all open-closed subsets of $X$, and $\Delta$ the $\sigma$-ideal of all subsets of $\mathcal{F}$ of first category in $X$, is $\mathfrak{A}$ is isomorphic to $\mathcal{F}/\Delta$.

There appears to be some confusion in the question, which I will try to dispel.

As a free Boolean $\sigma$-algebra $A$ is a Boolean $\sigma$-algebra, the Loomis-Sikorski theorem certainly applies. However, what you say after "in other words" is not equivalent to your second paragraph, and is not true.

The clopen sets of the Cantor space $2^\omega$ do form the free Boolean algebra $A$ on countably many generators, and the $\sigma$-field $B$ generated by these clopens (equivalently, because $2^\omega$ is compact, metrizable and zero dimensional, the Borel sets) is the free Boolean $\sigma$-algebra on countably many generators.

But note that $2^\omega$ is the Stone space of $A$, not $B$. We can see that it is not the Stone space of $B$, because every point of $2^\omega$ defines a $\sigma$-ultrafilter on $B$, i.e. one closed under countable intersections, but $B$ has many ultrafilters that are only closed under finite intersections (to see this, just take a partition of $B$ into countably many sets and define an ultrafilter using a non-principal ultrafilter on $\omega$). If $2^\omega$ were the Stone space of $B$, every ultrafilter would be determined by a point.

You have to take the Stone space of $B$, not $B$'s embedding in the Stone space of $A$, to get the representation described in your second paragraph. Now, you might wonder what happens if you take the quotient of $B$ by the $\sigma$-ideal of meagre sets in $2^\omega$. The resulting Boolean $\sigma$-algebra is actually a complete Boolean algebra, isomorphic to the algebra of regular open sets of $2^\omega$. Since $B$ is not a complete Boolean algebra, this is another way of seeing that the procedure described in your third paragraph does not produce $B$.

• Thank you for the clarification. It makes more sense now. So, if I understand correctly, every free Boolean algebra on a countable set of free generators is isomorphic the field of all open-closed sets of the Cantor space $2^\omega$ (which is countable, atomless, but also incomplete). The completion this field of open-closed sets, i.e. the regular open algebra of $2^\omega$, is in turn isomorphic to the Borel algebra of $2^\omega$ mod meagre sets in $2^\omega$ or, equivalently, the $\sigma$-algebra generated by the open-closed sets of $2^\omega$ mod meagre sets. – user111723 Sep 4 '17 at 16:47
• Yes. You can find a proof of this in Halmos's Lectures on Boolean Algebras, section 13, theorem 4. But note that it is only because $2^\omega$ is metrizable that the $\sigma$-algebra generated by the clopen sets is the Borel $\sigma$-algebra. – Robert Furber Sep 5 '17 at 22:24
• In fact, for the spectrum of $B$, the two algebras differ - if we take the $\sigma$-algebra generated by the clopens (or equivalently the $\sigma$-algebra of Baire sets) modulo meagre sets, we get $B$ back again, and if we take Borel sets modulo meagre sets (or we could also use sets with the Baire property (a very different thing from Baire sets) modulo meagre sets), we get the completion of $B$. – Robert Furber Sep 5 '17 at 22:25
• Sorry, Robert, by spectrum, do you mean the Cantor space without any reference to its metrizability? Also, I am not sure to fully understand what you mean when you say: " if we take the σ-algebra generated by the clopens (or equivalently the σ-algebra of Baire sets) modulo meagre sets, we get B back again". Finally, what is the difference between Borel sets mod meagre sets and sets with BP mod meagre sets? Thank you in advance. – user111723 Sep 7 '17 at 10:49
• Spectrum is another name for Stone space, and the name I more usually use (I must have lapsed into it without realizing that you might not know what it meant). So the spectrum of $B$ is a complicated non-metrizable space not homeomorphic to the Cantor space. By "we get $B$ back again", I mean that the map taking an $B$ to its corresponding equivalence class of Baire sets modulo meagre sets is an isomorphism (exactly the statement of Loomis-Sikorski). – Robert Furber Sep 9 '17 at 8:57