# A non embeddable boolean algebra

I am starting to learn about boolean algebras and one excercise is to find two boolean algebras $$A$$ y $$B$$ such that $$\# A\leq \# B$$ but there is not an embedding from $$A$$ to $$B$$.

One example of a boolean algebra which is not a subalgebra is $$RO(\mathbb{R})\subseteq\mathcal{P}(\mathbb{R})$$ where $$RO(\mathbb{R})$$ are the regular open sets of $$\mathbb{R}$$ (with the usual topology). From this observation I was trying to prove that there is not an embedding $$e:RO(\mathbb{R})\to\mathcal{P}(\mathbb{R})$$ but I couldn't prove it. Is my claim true? If not, could you please give a hint to find such boolean algebras?

• @markvs Sorry for the lack of precision. The structure of $RO(\mathbb{R})$ is like the one indicated in this page planetmath.org/regularopenalgebra not using the usual set theoretic operations. Thanks for the hint, I wil try it.
– xyz
Nov 9, 2021 at 17:28
• @markvs That won't give an answer to the question: the Boolean algebra $\mathcal{P}(\mathbb{R})$ has cardinality $2^{2^{\aleph_0}}$, strictly larger than the cardinality $2^{\aleph_0}$ of the Boolean algebra of Borel sets (note that the OP wants $\# A\le \# B$). Nov 9, 2021 at 18:28

Here's one cute example: consider the Boolean algebras $$\mathcal{B}:=\mathcal{P}(\omega)$$ and $$\mathcal{A}:=\mathcal{P}(\omega)/\mathit{fin}$$, where $$\mathit{fin}$$ is the $$\mathcal{B}$$-ideal of finite sets. Clearly $$\vert\mathcal{A}\vert\le\vert\mathcal{B}\vert$$ (at least, assuming choice :P).

However, it turns out that there is no embedding from $$\mathcal{A}$$ into $$\mathcal{B}$$ for a simple structural reason: even though $$\mathcal{A}$$ is a quotient of $$\mathcal{B}$$, $$\mathcal{A}$$ is "wider" than $$\mathcal{B}$$!

## Wait, what?

Let the width of a Boolean algebra $$\mathcal{X}$$ be the supremum of the cardinalities of sets $$U\subseteq\mathcal{X}$$ such that for all $$u,v\in U$$ we have $$u\wedge v=0$$. (Incidentally this is not standard terminology - I don't know if there is a standard term for this.)

In $$\mathcal{B}$$, the equation "$$u\wedge v=0$$" just means that $$u$$ and $$v$$ are disjoint. Consequently we obviously have $$\mathit{width}(\mathcal{B})=\aleph_0$$. It's also easy to see that if $$\mathcal{X}$$ embeds into $$\mathcal{Y}$$ then $$\mathit{width}(\mathcal{X})\le\mathit{width}(\mathcal{Y})$$.

However, it turns out that $$\mathit{width}(\mathcal{A})=2^{\aleph_0}$$. This is quite surprising at first (at least it was for me), and amounts to the following combinatorial result:

There is a size-continuum set $$\mathcal{F}$$ of sets of natural numbers which are pairwise almost disjoint, that is, such that for $$u,v\in\mathcal{F}$$ we have $$u=v$$ or $$u\cap v$$ is finite.

This is a fun puzzle, so I'll hide a hint below:

Finally, note that if $$\mathcal{X}$$ is a quotient of $$\mathcal{Y}$$ then every antichain in $$\mathcal{X}$$ "comes from" an antichain in $$\mathcal{Y}$$. The definition of width may look odd at first, but it's exactly this stronger requirement "$$u\wedge v=0$$" that allows it to play interestingly with quotients: if we just looked at the supremum of the cardinalities of the antichains, things would be pretty boring.