First some definitions. By $\mathcal P(\mathbb N)$ we denote the family of all subsets of $\mathbb N$ endowed with the metrizable separable topology generated by the countable base consisting of the sets $[A;B]=\{C\subset \mathbb N:C\cap B=A\}$ where $A,B$ run over finite subsets of $\mathbb N$.

A subfamily $\mathcal F\subset\mathcal P(\mathbb N)$ is called hereditary if for any sets $F\in\mathcal F$ each subset of $F$ belongs to $\mathcal F$.

An ideal on $\mathbb N$ is a hereditary subfamily $\mathcal I\subset \mathcal P(\mathbb N)$, closed under unions.

An ideal $\mathcal I$ on $\mathbb N$ is called an $F_{\sigma\delta}$-ideal if $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$.

Problem 1. Can each $F_{\sigma\delta}$-ideal $\mathcal I\subset\mathcal P(\mathbb N)$ be written as $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some hereditary closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$?

We can ask also a more general

Problem 2. Can each hereditary $F_{\sigma\delta}$-set $\mathcal I\subset\mathcal P(\mathbb N)$ be written as $\mathcal I=\bigcap_{n=1}^\infty\bigcup_{m=1}^\infty\mathcal K_{n,m}$ for some hereditary closed subsets $\mathcal K_{n,m}$ of the compact topological space $\mathcal P(\mathbb N)$?

Problem 2 is a related to another open (?)

Problem 3. Does each Borel hereditary subset $\mathcal F\subset\mathcal P(\mathbb N)$ belong to the smallest subfamily of $\mathcal P(\mathbb N)$ containing all open hereditary sets and closed under taking countable unions and intersections?

  • $\begingroup$ The topology described in the post is the same as identifying $\mathcal P(\mathbb N)$ with the Cantor space $\mathcal C=\{0,1\}^{\mathbb N}$, right? (Here $\{0,1\}$ has the discrete topology and we take the product topology on $\mathcal C$.) $\endgroup$ Oct 13, 2017 at 13:51
  • $\begingroup$ @MartinSleziak Yes. This is the same topology (= pointwise convergence = Tychonoff product topology). $\endgroup$ Oct 13, 2017 at 14:00

2 Answers 2


In [1], Solecki proved (Theorem 2.1 combined with Lemma 2.4),

Theorem: $\mathcal{I} \subset \mathcal{P}(\mathbb{N})$ is an analytic P-ideal, if and only if, there is some $\varphi:\mathcal{P}(\mathbb{N})\rightarrow [0,\infty)$ satisfying the following,

  • $\varphi(\emptyset) = 0$,
  • $\varphi(A) \le \varphi(A \cup B) \le \varphi(A) + \varphi(B)$, and
  • $\varphi(A) = \lim_{n\rightarrow \infty} \varphi(A\cap n)$,

such that $\mathcal{I} = \{ A \subset \mathbb{N}: \limsup_{n} \varphi(A \backslash n)=0\}$. Furthermore, every analytic ideal on $\mathbb{N}$ is either a $P$-ideal, or finite-to-one reducible to the ideal generated by finite unions of the sets $\{ n\} \times \mathbb{N}.$

Remark: Any $\varphi$ which satisfies the hypothesis of the theorem is called a lower semi-continuous sub-measure on $\mathbb{N}$.

An easy corollary to the previous theorem provides a partial answer to Problem 1.

Every $F_{\sigma\delta}$ $P$-ideal on $\mathbb{N}$ can be written as $\bigcap_n \bigcup_m K_{n,m}$ for some closed and hereditary $K_{n,m}$.

Proof: Using the previous theorem, it follows that for each $F_{\sigma\delta}$ P-ideal $\mathcal{I}$ there is a lower semi-continuous $\varphi$ such that $\mathcal{I} = \{ A \subset \mathbb{N} : \limsup_{n} \varphi(A\backslash n) = 0\}$. Letting $K_{n,m} = \{ A \subset \mathbb{N}: \varphi(A\backslash m) \le \frac{1}{n} \}$, we have $\mathcal{I} = \bigcap_n \bigcup_m K_{n,m}$; so it only remains to show that each $K_{n,m}$ is closed and hereditary.

To this end note that, for each $A \in K_{n,m}$ and $B\subset A$ we have $B\backslash m \subset A\backslash m$, therefore using the properties of $\varphi$, it follows that, $\varphi(B\backslash m) \le \varphi(B\backslash m \cup A\backslash m)=\varphi(A\backslash m) \le \frac{1}{n}$ and so $B \in K_{n,m}.$

Next, assume $A \subset \mathbb{N}$ has the property that for every $k \in \mathbb{N}$ letting $a_k = A \cap k$ and $b_k =k$, we have $[a_k, b_k]\cap K_{n,m} \neq \emptyset$. Then, for every $k \ge 0$, $\varphi(a_k \backslash m) \le \frac{1}{n}$ (since the sequence is increasing.) It immediately follows that $\varphi(A \backslash m) = \lim_{k} \varphi(a_k \backslash m) \le \frac{1}{n}$ and $A \in K_{n,m}$.

[1] Solecki, Sławomir, Analytic ideals and their applications, Ann. Pure Appl. Logic 99, No.1-3, 51-72 (1999). ZBL0932.03060.

  • 1
    $\begingroup$ If we take any decomposition $\bigcup_{i\in\mathbb N} A_i=\mathbb N$ of $\mathbb N$ into countably many disjoint infinite subsets, then $$\mathcal I=\{A\subseteq \mathbb N; A\text{ intersects only finitely many $A_i$'s}\}$$ is not a P-ideal and it is $F_\sigma$.So the remark at the end of the current revision of the post ("it's possible that all $F_{\sigma\delta}$ ideals are P-ideals, but I'm not sure") is not true. $\endgroup$ Mar 16, 2018 at 19:51
  • 1
    $\begingroup$ @MartinSleziak Yes, of course you are right. I was implicitly restricting to ideals which are $F_{\sigma\delta}$ and not $F_{\sigma}$; since every $F_\sigma$ ideal has a closed hereditary presentation. $\endgroup$
    – Not Mike
    Mar 16, 2018 at 20:20
  • 1
    $\begingroup$ Thie ideal pointed by Martin Sleziak can be "mixed" with some other $F_{\sigma\delta}$ ideal which is not $F_\sigma$ and the resulting ideal will be $F_{\sigma\delta}$ but not $F_\sigma$ and not a $P$-ideal. $\endgroup$ Mar 17, 2018 at 9:13
  • $\begingroup$ @TarasBanakh Thank you for pointing that out. I guess I was confusing, not coverable by an $F_\sigma$ subideal with all countably generated sub-ideals being trivial; which now I that I say it, seems ridiculous. Anyway I've removed the remark at the end. $\endgroup$
    – Not Mike
    Mar 18, 2018 at 14:40

Let $\mathcal{I}$ be an $F_{\sigma \delta}$-set (not necessarily hereditary) in $\mathcal{P}(\mathbf{N})$, i.e., there exist a family of closed sets $\{F_{n,m}: n,m \in \mathbf{N}\}$ such that $$ \mathcal{I}=\bigcup_{n\ge 1}\bigcap_{m\ge 1}F_{n,m}. $$ Since $\mathcal{P}(\mathbf{N})$ is a zero-dimensional space with countable base $\mathcal{B}=\{B_k\}$ of clopen hereditary sets (made of cylinders), then it is easy to see that each $F_{n,m}$ can be written as finite union of countable intersection of elements in $\mathcal{B}$. Hence there exist $\{B_{n,m}(i,j): i\ge 1, j=1,\ldots,k\}\subseteq \mathcal{B}$ such that $$ F_{n,m}=\bigcup_{j=1}^k \bigcap_{i\ge 1} B_{n,m}(i,j). $$ for all $n,m$. Therefore $$ \begin{align} \mathcal{I}&=\bigcup_{n\ge 1}\bigcap_{m\ge 1}\bigcup_{j=1}^k \bigcap_{i\ge 1} B_{n,m}(i,j)\\ &=\bigcup_{n\ge 1}\bigcap_{m\ge 1}\bigcap_{i_1\ge 1} \cdots \bigcap_{i_k\ge 1}\left(B_{n,m}(i_1,1)\cup \cdots B_{n,m}(i_k,k)\right). \end{align} $$ Since finite unions of hereditary sets are hereditary, the answer to Problem 2 is affirmative.

  • $\begingroup$ The life is not so simple: the space $\mathcal P(\mathbb N)$ does not have a base consisting of hereditary sets: if it would be true, then we would get that each open set of $\mathcal P(\mathbb N)$ is hereditary and hence contains the empty set, which is not true. $\endgroup$ Oct 14, 2017 at 9:41

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