# Addition and Rudin-Keisler ordering in $\beta \omega$

$$\DeclareMathOperator{\RK}{\mathrm{RK}}$$Let $$\beta\omega$$ be the Stone-Cech compactification of $$\omega$$ with the discrete topology. We can endow $$\beta\omega$$ with an addition operation that extends the addition on $$\omega$$ in a natural way. (The same statement holds for multiplication.)

The Rudin-Keisler preorder on $$\beta\omega$$ is defined by $${\bf p} \leq_{\RK} {\bf q} :\;\Leftrightarrow\; (\exists f:\omega\to\omega)\,(\forall U\in{\bf p})\, f^{-1}(U)\in {\bf q}$$ for $${\bf p},{\bf q}\in\beta\omega$$.

Question. Given $${\bf a},{\bf b}, {\bf c}\in\beta\omega$$, do we have

1. $${\bf a} \leq_{\RK} {\bf a+b}$$, and
2. if $${\bf a}\leq_{\RK}{\bf b}$$, then $$({\bf a}+{\bf c}) \leq_{\RK} ({\bf b}+{\bf c})$$?
• Sanity check: is ${\bf a+b}=\{N\subseteq\omega:\{x:\{y: x+y\in N\}\in{\bf b}\}\in{\bf a}\}$? "The set of subsets of $\omega$ such that for $\bf a$ many $x\in\omega$ there are $\bf b$ many $y\in\omega$ with $x+y\in N$" – Alessandro Codenotti Nov 11 '20 at 11:00
• That's correct @AlessandroCodenotti – Dominic van der Zypen Nov 11 '20 at 11:21
• In other words $x+\mathbf{b}=\lim_{y\in\mathbf{b}}(x+y)$, and $\mathbf{a}+\mathbf{b}=\lim_{x\in\mathbf{a}}(x+\mathbf{b})$. – YCor Nov 11 '20 at 12:03
• Does the Rudin-Keisler preorder extend the order on $\mathbb N$? Does your question have a positive answer if $a$ or $b$ is in $\mathbb N$? – Tri Nov 12 '20 at 6:52
• @Tri all principal ultrafilters are RK-isomorphic (look at the bijection of $\omega$ swapping the two points whose singletons are in the ultrafilters) – Alessandro Codenotti Nov 12 '20 at 9:04

## 1 Answer

For Question 1: First, there are idempotent ultrafilters, so some ultrafilters satisfy 1 in a very strong form. But 1 does not hold in general. The reason is that the semigroup $$\beta\omega-\omega$$ has subsemigroups $$G$$ that are groups of cardinality $$2^{\mathfrak c}$$ where $$\mathfrak c$$ is the cardinal of the continuum. (See Hindman and Strauss, "Algebra in the Stone-Cech Compactification", Theorems 2.7(d) and 2,25, and remember that all infinite closed subsets of $$\beta\omega-\omega$$ have cardinality $$2^{\mathfrak c}$$.) If the answer to Question 1 were always affirmative, then all elements of $$G$$ would be RK-below each other (and therefore isomorphic, but I don't need that). That's impossible as any ultrafilter is RK-above only $$\mathfrak c$$ others.

I expect the answer to Question 2 to be negative also, but I'll need to think some more to prove it. I also expect that there's an easier proof for Question 1. Finally, the answer to Question 1 becomes positive if either a or b is assumed to be a P-point.

Edit: A negative answer to Question 2 follows from the continuum hypothesis. Under CH (or certain weaker hypotheses, but not in ZFC alone), there exist P-point ultrafilters. Let a be one of these, and let b be an idempotent ultrafilter Rudin-Keisler above a. (It's provable in ZFC that any non-principal ultrafilter on $$\omega$$ is below an idempotent one.) Because a is a P-point, any sum $$\mathbf a+\mathbf c$$ (with c noprincipal) is isomorphic to the tensor product (also called Fubini product) $$\mathbf a\otimes\mathbf c=\{S\subseteq\omega\times\omega:\{x:\{y:(x,y)\in S\}\in\mathbf c\}\in\mathbf a\},$$ and is therefore strictly above both a and c in the Rudin-Keisler order. In particular, $$\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b$$. But since b is idempotent, this gives $$\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b+\mathbf b$$ even though $$\mathbf a\leq_{\text{RK}}\mathbf b$$.

• Thanks for these results! While your answer is worth accepting, I'll wait for little while before doing so -- maybe an answer to question 2 can be found. – Dominic van der Zypen Nov 11 '20 at 20:06