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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?

Thanks in advance!

xyz
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    @markvs Sorry for the lack of precision. The structure of $RO(\mathbb{R})$ is like the one indicated in this page https://planetmath.org/regularopenalgebra not using the usual set theoretic operations. Thanks for the hint, I wil try it. – xyz Nov 09 '21 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 $# A\le # B$). – Noah Schweber Nov 09 '21 at 18:28

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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:

Think about Dedekind cuts.

Complete solutions have appeared in various places across this site (search for "almost disjoint family" or similar); here is one of mine, where I break the problem into separate (hidden) hints.

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.

Noah Schweber
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