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This is related to a couple recent MO/MSE questions of mine, namely 1,2. Belatedly, I've tweaked this post to remove an overly-ambitious secondary question; see the edit history if interested.

Let $\beta\mathbb{Z}$ be the set of all ultrafilters on $\mathbb{Z}$, and as usual conflate $n$ and $\{A\subseteq\mathbb{Z}:n\in A\}$ for $n\in\mathbb{Z}$. We can extend any binary operation on $\mathbb{Z}$ to a semicontinuous analogue on $\beta\mathbb{Z}$, at the cost of many (most?) algebraic properties. Ultrafilter addition is quite well studied (see e.g. Hindman/Strauss), but I've been able to find much less about ultrafilter subtraction: $$\mathcal{U}\widehat{-}\mathcal{W}=\{A\subseteq\mathbb{Z}: \{k:\{a: a-k\in A\}\in\mathcal{U}\}\in\mathcal{W}\}.$$

Say that an ultrafilter $\mathcal{W}$ is zeroid iff $\mathcal{W}=\mathcal{U}\widehat{-}\mathcal{U}$ for some $\mathcal{U}$. My question is:

Which ultrafilters are zeroid?

To make this actually answerable, I tentatively guess that $(i)$ $0$ is the only zeroid principle ultrafilter but not the only zeroid ultrafilter and $(ii)$ $p$-points are not zeroid; are these guesses true?

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    $\begingroup$ What about the “other” obvious subtraction, viꝫ. $\mathcal{U}\mathbin{\widetilde{-}}\mathcal{W} := \{A\subseteq\mathbb{Z}: \{k:\{a: k-a\in A\}\in\mathcal{W}\}\in\mathcal{U}\}$ (or, if you will, $(-\mathcal{W})\mathbin{\widehat{-}}(-\mathcal{U})$)? Doesn't it make sense to ask similar questions about it? $\endgroup$
    – Gro-Tsen
    Commented Jul 29, 2022 at 9:29
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    $\begingroup$ (My point being: since addition on $\mathbb{Z}$ is commutative, there is only one way to pass to ultrafilters, although we get a non-commutative operation. But since subtraction on $\mathbb{Z}$ is non-commutative, we need to choose the order in which we “ultrafilterize” the arguments, giving a priori two different operations on the ultrafilters. I have no idea whether they are the same, or similar, or very different.) $\endgroup$
    – Gro-Tsen
    Commented Jul 29, 2022 at 9:32
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    $\begingroup$ PS: Just an unimportant LaTeX/MathJAX point: when decorating a binary operation like \widehat{-}, you should wrap \mathbin{...} around it so as to give it proper spacing wrt its arguments: X\mathbin{\widehat{-}}Y produces $X\mathbin{\widehat{-}}Y$ which is nicer than X\widehat{-}Y giving $X\widehat{-}Y$. $\endgroup$
    – Gro-Tsen
    Commented Jul 29, 2022 at 9:35
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    $\begingroup$ @Gro-Tsen, actually commutativity doesn't help. The two ways of extending addition are different. $\endgroup$ Commented Jul 29, 2022 at 23:17
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    $\begingroup$ @Gro-Tsen, for one choice all left translations are continuous for the other all right. Only for principal ultrafilters are both translations continuous. This can be found in Hindman and Strauss Algebra in the Stone Cech compactification $\endgroup$ Commented Jul 29, 2022 at 23:37

1 Answer 1

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Both your guesses are correct. To see this, it's helpful to reformulate the way you're thinking about the subtraction operator on $\beta \mathbb Z$. Beginning with subtraction on $\mathbb Z$, you can first extend this to an operator $\beta \mathbb Z \times \mathbb Z \rightarrow \beta \mathbb Z$ by setting $\mathcal U - n = \{B-n :\, B \in \mathcal U\} = \{A \subseteq \mathbb Z :\, A+n \in \mathcal U\}$. Notice that this agrees with your definition of subtraction on $\beta \mathbb Z$ when we identify $n$ with the principle ultrafilter at $n$. So this is the "right" way to think of subtracting an integer from an ultrafilter. But then there is only one way to extend this to a semi-continuous operation $\beta \mathbb Z \times \beta \mathbb Z$: we must define $$\mathcal U - \mathcal W = \textstyle {\mathcal W}\text{-}\!\lim_n (\mathcal U-n).$$ This again agrees with your definition (it must!), but personally I find it much more intuitive to think of $\mathcal U - \mathcal W$ as a topological limit of the sequence $\mathcal U, \mathcal U - 1, \mathcal U - 2, \dots$.

This description of what's going on makes your two guesses plainly true. For $(i)$, note that if $\mathcal U$ is nonprincipal then so is $\mathcal U - n$ for all $n$, which means (because $\mathbb Z$ is open in $\beta \mathbb Z$) that the limit ${\mathcal W}\text{-}\!\lim_n (\mathcal U-n)$ is in $\beta Z \setminus \mathbb Z$ for any $\mathcal W$ (including $\mathcal W = \mathcal U$). For $(ii)$, note that ${\mathcal W}\text{-}\!\lim_n (\mathcal U-n)$ is a limit of a countable sequence of points in $\beta \mathbb Z \setminus \mathbb Z$, hence not a P-point. In fact, let me observe that $\mathcal U - \mathcal W$ is never a weak P-point (for the same reason), which shows that not every ultrafilter is representable as the difference of two others.

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  • $\begingroup$ I've cut Q2 so this fully answers my question. Belatedly, thanks very much! $\endgroup$ Commented Mar 5, 2023 at 19:32

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