Existence of maximal analytic P-ideal

Question

An ideal $\mathcal{I}$ on the positive integers $\mathbf{N}$ is a P-ideal if for every sequence $(A_n)$ of sets in $\mathcal{I}$ there exists $A \in \mathcal{I}$ such that $A_n\setminus A$ is finite for all $n$.

Moreover, an ideal $\mathcal{I}$ is said to be analytic if (equipping $\mathcal{P}(\mathbf{N})$ with the Cantor-space topology and identifying set with their characteristic functions) it is a continuous image of a $G_\delta$ subset of the Cantor space.

By a result of Solecki (1999), an ideal $\mathcal{I}$ on the positive integers is an analytic P-ideal if and only if $$ \textstyle \mathcal{I}=\{X\subseteq \mathbf{N}: \lim_n \phi(X\setminus [1,n])=0\} $$ for some monotone subadditive function $\phi: \mathcal{P}(\mathbf{N}) \to [0,\infty]$ such that $\phi(\emptyset)=0$ and $\phi(X)=\lim_n \phi(X\cap [1,n])$ for all $X$.

Question 1. Does there exist a analytic P-ideal which is also maximal (i.e., its dual filter $\{X: X^c \in \mathcal{I}\}$ is an ultrafilter)?

And similarly:

Question 2. Let $\mathcal{I}$ be an analytic P-ideal and fix $X\notin \mathcal{I}$. Does there exist $Y\subseteq X$ such that $Y\notin \mathcal{I}$ and $X\setminus Y \notin \mathcal{I}$?

¹Sławomir Solecki: Analytic Ideals. Bull. Symbolic Logic Volume 2, Number 3 (1996), 339-348. doi: 10.2307/420994 JSTOR. Freely available here: https://www.math.ucla.edu/~asl/bsl/0203/0203-004.ps and http://www.math.ucla.edu/~asl/bsl/02-toc.htm

Answer 1

As already mentioned in the comments, a free ultrafilter considered as a subset of Cantor space (or Cantor set) cannot be analytic, so the answer to the Question 1 is No. (Even without the assumption that the given ideal is P-ideal; we get that no maximal ideal can be analytic.)

This follows immediately if we show that ultrafilter is not Lebesgue measurable or if we show that it does not have the Baire property. (Since both properties are true for any analytic set.)

Question 2 basically asks whether the restriction to the subset $X$ is a maximal ideal (an ultrafilter). Notice that $X\notin\mathcal I$ implies that $X$ is infinite, so this is rather similar to the original situation. Moreover, since $A\mapsto A\cap X$ is a continuous map $2^\omega\to 2^X$, the restriction will be analytic if the original ideal (filter) was. So the answer to Question 2 is No as well. This was pointed out in Jing Zhang's comment.

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Let me quote relevant parts from Blass, Andreas (2010), "Ultrafilters and set theory", Ultrafilters across mathematics, Contemporary Mathematics, 530, Providence, RI: American Mathematical Society, pp. 49–71, doi: 10.1090/conm/530/10440 (From the beginning of section 6. Ultrafilters, Descriptive Set Theory, and Determinacy, page 64.)

This paper is also mentioned as a reference in the current revision of the Wikipedia article Property of Baire, where the fact that ultrafilters do not have Baire property is also mentioned.

... we can ask about its behavior with respect to notions like Baire category and Lebesgue measure. The answer is that the behavior is bad.

Theorem 6.1 (Sierpinski [64]). A non-principal ultrafilter on $\omega$, regarded as a subset of $[0,1]$ is not Lebesgue measurable.

The proof uses the zero-one law for Lebesgue measure (see [56, Thm. 21.3]) to infer that, if a non-principal ultrafilter were measurable, its measure would be $0$ or $1$. But the measure-preserving reflection $x\mapsto 1-x \colon [0,1]\to[0,1]$ maps the ultrafilter to its complement (except for countably many points), so the measure of the ultrafilter would have to be $1/2$.

A similar argument, using the Baire category analog of the zero-one law [56, Thm. 21.4], shows that a non-principal ultrafilter cannot have the Baire property. (A set has the Baire property if it differs from some open set by a meager set.)

After this, the paper continuous with discussion of relation of ultrafilters to non-determined games.

The references mentioned above are:

[56] John Oxtoby, Measure and Category, Springer-Verlag, Graduate Texts in Mathematics 2 (1971).
[64] Waclaw Sierpinski, Fonctions additives non completement additives et fonctions non mesurables, Fund. Math. 30 (1938) 96–99; https://eudml.org/doc/212991

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A proof can also be found in Asaf Karagila: Zornian Functional Analysis or: How I Learned to Stop Worrying and Love the Axiom of Choice, http://karagila.org/2016/zornian-functional-analysis-or-how-i-learned-to-stop-worrying-and-love-the-axiom-of-choice/ (Wayback Machine) and http://karagila.org/wp-content/uploads/2016/10/axiom-of-choice-in-analysis.pdf (Wayback Machine)

Theorem 30 (Oxtoby’s Zero-One Law). Suppose that $T\subseteq 2^{\mathbb N}$ has the Baire property and is closed under finite modifications, then T is meager or $2^{\mathbb N}\setminus T$ is meager.

Theorem 31. Suppose that $\mu$ is a finitely additive probability measure on $\mathcal P(\mathbb N)$ which vanishes on finite sets. Then $T=\{a\in 2^{\mathbb N}; \mu(A)=0\}$ does not have the Baire property in the topological space $2^{\mathbb N}$

Theorem 31 is given here with proof. Schechter's Handbook of Analysis and Its Foundations, Th. 20.33, p. 544, is provided as a reference for Theorem 31. The proof given there is from the paper Miller and Živaljevic, Remarks on the zero-one law, Mathematica Slovaca, vol. 34 (1984), issue 4, pp. 375-384, https://eudml.org/doc/31650

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The proofs of the facts mentioned above and some related references can be also found online.

Ultrafilters are not measurable:

• Non-measurability of ultrafilter on $\omega$
• Relationships between AC, Ultrafilter Lemma/BPIT, Non-measurable sets
• Does the existence of a non-principal measure on ω imply that of a non Lebesgue measurable set?
• Lebesgue measurability of a set $A = \{\sigma_i \in B\ /\ {2^i}, B \in F\}$ for a a non-principal ultrafilter $F$

Ultrafilters do not have Baire property:

• If $\mathcal{U}$ is a non-principal ultrafilter on $\mathbb{N}$ then $\mathcal{U}$ does not have the Baire property in $2^\mathbb{N}$
• Ref. request: Additive probability measure on $\mathcal P({\bf N})$ supplies subset of $\mathbf R$ without Baire property