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](https://doi.org/10.2307/420994)
[JSTOR](http://www.jstor.org/stable/420994). Freely available here: [https://www.math.ucla.edu/~asl/bsl/0203/0203-004.ps](https://www.math.ucla.edu/%7Easl/bsl/0203/0203-004.ps) and
[http://www.math.ucla.edu/~asl/bsl/02-toc.htm](http://www.math.ucla.edu/%7Easl/bsl/02-toc.htm)



Existence of maximal analytic P-ideal
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](https://en.wikipedia.org/wiki/Property_of_Baire).* (Since both properties are true for any [analytic set](https://en.wikipedia.org/wiki/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](https://mathoverflow.net/questions/270756/existence-of-maximal-analytic-p-ideal/270822#comment669986_270822).




---


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](https://doi.org/10.1090%2Fconm%2F530%2F10440)
(From the beginning of section 6. Ultrafilters, Descriptive Set Theory, and Determinacy, [page 64](https://books.google.com/books?id=_2ZHnqMwGwYC&pg=PA64).)


This paper is also mentioned as a reference in the [current revision](https://en.wikipedia.org/w/index.php?title=Property_of_Baire&oldid=772272069) of the Wikipedia article [Property of Baire](https://en.wikipedia.org/wiki/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>
> 
> 
> 




---


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](http://web.archive.org/web/20190311171701/http://karagila.org/2016/zornian-functional-analysis-or-how-i-learned-to-stop-worrying-and-love-the-axiom-of-choice/)) and <http://karagila.org/wp-content/uploads/2016/10/axiom-of-choice-in-analysis.pdf> ([Wayback Machine](http://web.archive.org/web/20190311171642/http://karagila.org/wp-content/uploads/2016/10/axiom-of-choice-in-analysis.pdf))



> 
> **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](https://books.google.com/books?id=eqUv3Bcd56EC&pg=PA544), 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>




---


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 $\omega$](https://math.stackexchange.com/q/1130615)
* [Relationships between AC, Ultrafilter Lemma/BPIT, Non-measurable sets](https://math.stackexchange.com/q/259247)
* [Does the existence of a non-principal measure on ω imply that of a non Lebesgue measurable set?](https://mathoverflow.net/questions/256133/does-the-existence-of-a-non-principal-measure-on-%CF%89-imply-that-of-a-non-lebesgue)
* [Lebesgue measurability of a set $A = \{\sigma\_i \in B\ /\ {2^i}, B \in F\}$ for a a non-principal ultrafilter $F$](https://math.stackexchange.com/q/1018302)


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 $2^\mathbb{N}$](https://math.stackexchange.com/q/1180042)
* [Ref. request: Additive probability measure on $\mathcal P({\bf N})$ supplies subset of $\mathbf R$ without Baire property](https://mathoverflow.net/questions/209067/ref-request-additive-probability-measure-on-mathcal-p-bf-n-supplies-sub)


