1 Introduction

Let \(\mathscr {X}\) be the set of positive real sequences \(x=(x_n)\) with divergent series \(\sum _n x_n\). For each \(x \in \mathscr {X}\), let \(\mathcal {I}_x\) the collection of sets of positive integers A such that the (possibly finite) subseries indexed by A is convergent, that is,

$$\begin{aligned} \mathcal {I}_x:=\left\{ A\subseteq \mathbf {N}: \sum _{n \in A}x_n < \infty \right\} . \end{aligned}$$
(1)

Note that each \(\mathcal {I}_x\) is closed under finite unions and subsets, i.e., it is an ideal. Moreover, it contains the collection \(\mathrm {Fin}\) of finite sets \(A\subseteq \mathbf {N}\), and it is different from the power set \(\mathcal {P}(\mathbf {N})\). Following [4], a collection of sets of the type (1) is called summable ideal. It is not difficult to see that every infinite set of positive integers contains an infinite subset in \(\mathcal {I}_x\) if and only if \(\lim _n x_n=0\). Accordingly, define

$$\begin{aligned} \mathscr {Z}:=\mathscr {X} \cap c_0=\{x \in \mathscr {X}: \lim _{n\rightarrow \infty } x_n=0\}. \end{aligned}$$

It is known that the families \(\mathcal {I}_x\) defined in (1) are “small”, both in the measure-theoretic sense and the categorical sense, meaning that “almost all” subseries diverge, see [3, 6, 13, 16]. Related results in the context of filter convergence have been given in [1, 2, 10]. The set of limits of convergent subseries of a given series \(\sum _n x_n\), which is usually called “achievement set”, has been studied in [7, 9, 12]. Of special interest have been specific subseries of the harmonic series \(\sum _n \frac{1}{n}\); see, e.g., [11, 14, 15, 17].

Roughly, the question that we are going to answer is the following: Is it true that for each \(x \in \mathscr {Z}\) there exists \(y \in \mathscr {X}\) such that \(\mathcal {I}_x=\mathcal {I}_y\) and \((y_n)\) “does not oscillates too much”?

Hoping for a characterization of the class of summable ideals \(\mathcal {I}_x\) with \(x \in \mathscr {Z}\), M. Filipczak and G. Horbaczewska asked recently in [5] the following:

Question 1.1

Is it true that for each \(x \in \mathscr {Z}\) there exists \(y \in \mathscr {X}\) such that \(\mathcal {I}_x=\mathcal {I}_y\) and

$$\begin{aligned} \forall n\in \mathbf {N},\quad \frac{y_{n+1}}{y_n} \ge \frac{n}{n+2}\,\, ? \end{aligned}$$

We show in Theorem 1.3 below that the answer is negative in a strong sense. To this aim, define

$$\begin{aligned} \mathscr {Y}:=\left\{ y \in \mathscr {X}: \liminf _{n\rightarrow \infty }\, \frac{y_{n+1}}{y_n} >0\right\} , \end{aligned}$$

and note that \(A-1:=\{a-1: a \in A\}\) belongs to \(\mathcal {I}_y\) whenever \(y \in \mathscr {Y}\) and \(A \in \mathcal {I}_y\), cf. (2) below. Let also \(\sim\) be the equivalence relation on \(\mathscr {X}\) so that two sequences are identified if they generate the same ideal, hence

$$\begin{aligned} \forall x,y \in \mathscr {X}, \quad x\sim y \Longleftrightarrow \left( \forall A\subseteq \mathbf {N}, \,\,\sum _{n \in A}x_n<\infty \Longleftrightarrow \sum _{n \in A}y_n<\infty \right) . \end{aligned}$$

First, we show that the set of pairs \((x,y) \in \mathscr {X}^2\) such that x is \(\sim\)-equivalent to y is topologically well behaved:

Proposition 1.2

\(\sim\) is a coanalytic relation on \(\mathscr {X}\).

Then, we answer Question 1.1 by showing that:

Theorem 1.3

There exists \(x \in \mathscr {Z}\) such that \(x\not \sim y\) for all \(y \in \mathscr {Y}\).

In light of the explicit example which will be given in the proof of Theorem 1.3, one may ask about the topological largeness of the set of such sequences. To be precise, is it true that

$$\begin{aligned} \mathscr {A}:=\left\{ x \in \mathscr {Z}: \forall y \in \mathscr {Y}, x\not \sim y\right\} \end{aligned}$$

is a set of second Baire category, i.e., not topologically small? Note that the question is really meaningful since \(\mathscr {Z}\) is completely metrizable (hence by Baire’s category theorem \(\mathscr {Z}\) is not meager in itself): this follows by Alexandrov’s theorem [8, Theorem 3.11] and the fact that

$$\begin{aligned} \mathscr {Z}=\bigcap _{n\ge 1}\left\{ x\in c_0: x_n>0\right\} \cap \bigcap _{m\ge 1}\bigcup _{k\ge 1}\left\{ x \in c_0: x_1+\cdots +x_k>m\right\} \end{aligned}$$

is a \(G_\delta\)-subset of the Polish space \(c_0\). With the premises, we show that \(\mathscr {A}\) is comeager, that is, \(\mathscr {Z}\setminus \mathscr {A}\) is a set of first Baire category:

Theorem 1.4

\(\mathscr {A}\) is comeager in \(\mathscr {Z}\). In particular, \(\mathscr {A}\) is uncountable.

We remark that Theorem 1.4 gives an additional information on relation \(\sim\). Since it is a coanalytic equivalence relation by Proposition 1.2, we can appeal to the deep theorem by Silver [8, Theorem 35.20] which states that every coanalytic equivalence relation on a Polish space either has countably many equivalence classes or there is a perfect set consisting of non-equivalent pairs. Thanks to Theorem 1.4, the latter holds for the relation \(\sim\) in a strong form. Indeed, every pair in \(\mathscr {A}\times \mathscr {Y}\) does not belong to \(\sim\), where \(\mathscr {A}\) is comeager (hence it contains a \(G_\delta\)-comeager subset) and \(\mathscr {Y}\) is an uncountable \(F_\sigma\)-set. Therefore \(\mathscr {A}\times \mathscr {Y}\) contains a product of two perfect sets by [8, Theorem 13.6].

Lastly, on a similar direction, we strenghten the fact that \(\mathscr {A}\) is uncountable by proving that exist uncountably many sequences in \(\mathscr {A}\) which generate pairwise nonisomorphic ideals (here, recall that two ideals \(\mathcal {I}, \mathcal {J}\) are isomorphic if there exists a bijection \(f: \mathbf {N}\rightarrow \mathbf {N}\) such that \(f[A] \in \mathcal {I}\) if and only if \(A \in \mathcal {J}\) for all \(A\subseteq \mathbf {N}\)).

Theorem 1.5

There are \(\mathfrak {c}\) sequences in \(\mathscr {A}\) which generate pairwise nonisomorphic ideals.

Hereafter, we use the convention that \(\sum _{n\ge 1}a_n \ll \sum _{n\ge 1}b_n\), with each \(a_n,b_n>0\), is a shorthand for the existence of \(C>0\) such that \(\sum _{n\le k}a_n \le C\sum _{n\le k}b_n\) for all \(k \in \mathbf {N}\).

2 Proof of Proposition 1.2

Equivalently, we have to show that the set \(E:=\left\{ (x,y) \in \mathscr {X}^2: x\not \sim y\right\}\) is analytic in \(\mathscr {X}^2\). For, note that E is the projection on \(\mathscr {X}^2\) of \(E_1 \cup E_2\), where

$$\begin{aligned} E_1:=\left\{ (A,x,y) \in \mathcal {P}(\mathbf {N}) \times \mathscr {X}^2: A \in \mathcal {I}_x\setminus \mathcal {I}_y\right\} \end{aligned}$$

and, similarly,

$$\begin{aligned} E_2:=\left\{ (A,x,y) \in \mathcal {P}(\mathbf {N}) \times \mathscr {X}^2: A \in \mathcal {I}_y\setminus \mathcal {I}_x\right\} . \end{aligned}$$

Now, for each \(n \in \mathbf {N}\), define the functions \(\alpha _n, \beta _n: \mathcal {P}(\mathbf {N}) \times \mathscr {X}^2\rightarrow \mathbf {R}\) by \(\alpha _n(A,x,y)=\sum x_t\) and \(\beta _n(A,x,y)=\sum y_t\), where each sum is extended over all \(t \in A\) such that \(t\le n\). Since they are continuous, the set \((\alpha _n \le k):=\{(A,x,y)\in \mathcal {P}(\mathbf {N}) \times \mathscr {X}^2: \alpha _n(A,x,y) \le k\}\) is closed and \((\beta _n>k)\) is open for all \(n,k \in \mathbf {N}\). Therefore

$$\begin{aligned} E_1=\left( \bigcup _{k\ge 1}\bigcap _{n\ge 1}(\alpha _n \le k)\right) \cap \left( \bigcap _{k\ge 1}\bigcup _{n\ge 1}(\beta _n > k)\right) \end{aligned}$$

is the intersection of an \(F_\sigma\)-set and a \(G_\delta\)-set, hence it is Borel. Analogously, \(E_2\) is Borel. This proves that E is analytic subset of \(\mathscr {X}^2\).

3 Proof of Theorem 1.3

Define the sequence \(x=(x_n)\) so that \(x_n=\frac{1}{n}\) if n is even and \(x_n=\frac{1}{n\log (n+1)}\) if n is odd. Note that \(\lim _{n}x_n=0\) and that \(\sum _n x_n=\infty\), hence \(x \in \mathscr {Z}\). At this point, fix \(y \in \mathscr {Y}\) such that \(\kappa :=\liminf _n y_{n+1}/y_n>0\) and let us show that \(\mathcal {I}_x \ne \mathcal {I}_y\).

Let \(\mathbf {P}\) be the set of prime numbers, with increasing enumeration \((p_n)\). By the prime number theorem we have \(p_n\) is asymptotically equal to \(n\log (n)\) as \(n\rightarrow \infty\), hence

$$\begin{aligned} \sum _{n \in \mathbf {P}}x_n=\sum _{n\ge 1} x_{p_n} \ll \sum _{n\ge 1} \frac{1}{p_n\log (p_n)}\ll \sum _{n\ge 2} \frac{1}{n\log ^2(n)}<\infty , \end{aligned}$$

with the consequence that \(\mathbf {P} \in \mathcal {I}_x\). In addition, \(\mathbf {P}-1 \notin \mathcal {I}_x\) because

$$\begin{aligned} \sum _{n \in \mathbf {P}-1}x_n=\sum _{n\ge 1} x_{p_n-1}\gg \sum _{n\ge 1} \frac{1}{p_n} \gg \sum _{n\ge 2}\frac{1}{n\log (n)}=\infty . \end{aligned}$$

Lastly, suppose for the sake of contradiction that \(\mathcal {I}_x=\mathcal {I}_y\). Then we should have that \(\mathbf {P} \in \mathcal {I}_y\) and, at the same time, \(\mathbf {P}-1\notin \mathcal {I}_y\). The latter means that

$$\begin{aligned} \sum _{n\ge 1}y_{p_n-1}=\infty . \end{aligned}$$

However, this implies that

$$\begin{aligned} \sum _{n\ge 1}y_{p_n} \gg \sum _{n\ge 1} \kappa \, y_{p_n-1}=\infty , \end{aligned}$$
(2)

contradicting that \(\mathbf {P} \in \mathcal {I}_y\).

4 Proof of Theorem 1.4

Consider the Banach–Mazur game defined as follows: Players I and II choose alternatively nonempty open subsets of \(\mathscr {Z}\) as a nonincreasing chain \(U_1 \supseteq V_1 \supseteq U_2 \supseteq V_2 \supseteq \cdots\), where Player I chooses the sets \(U_m\). Player II has a winning strategy if \(\bigcap _m V_m\subseteq \mathscr {A}\). Thanks to [8, Theorem 8.33], Player II has a winning strategy if and only if \(\mathscr {A}\) is comeager. Hence, the rest of the proof consists in showing that Player II has a winning strategy.

Note that the open neighborhood of a sequence \(x \in \mathscr {Z}\) with radius \(\varepsilon >0\) satisfies

$$\begin{aligned} B_\varepsilon (x):=\{y \in \mathscr {Z}: \Vert x-y\Vert<\varepsilon \} \supseteq \left\{ y \in \mathscr {Z}: \forall n \in \mathbf {N}, \, |x_n-y_n| < \frac{\varepsilon }{2}\right\} . \end{aligned}$$

Since \(x \in \mathscr {Z}\subseteq c_0\), there exists \(k_0=k_0(x,\varepsilon ) \in \mathbf {N}\) such that \(x_n<\frac{\varepsilon }{2}\) for all \(n\ge k_0\). Hence

$$\begin{aligned} B_\varepsilon (x) \supseteq W_\varepsilon (x):=\left\{ y \in \mathscr {Z}: \forall n \ge k_0(x,\varepsilon ), \, y_n< \frac{\varepsilon }{2} \text { and }\forall n<k_0(x,\varepsilon ),|x_n-y_n|<\frac{\varepsilon }{2}\right\} . \end{aligned}$$
(3)

For each \(m \in \mathbf {N}\), suppose that the nonempty open set \(U_m\) has been fixed by Player I. Hence, \(U_m\) contains an open ball \(B_{\varepsilon _m}(x^{(m)})\), for some \(x^{(m)} \in \mathscr {Z}\) and \(\varepsilon _m>0\). In particular, thanks to (3), there exists a sufficiently large integer \(k_0=k_0(x^{(m)}, \varepsilon _m) \in \mathbf {N}\) such that \(y_n<\frac{\varepsilon _m}{2}\) for all \(y \in W_{\varepsilon _m}(x^{(m)})\) and \(n \ge k_0\). Without loss of generality, let us suppose that \(k_0\) is even.

At this point, let \(x^\star\) be the sequence in \(\mathscr {A}\) defined in the proof of Theorem 1.3. Then, for each \(m \in \mathbf {N}\), let \(t_m\ge k_0(x^{(m)},\frac{\varepsilon _m}{2})\) be an integer such that \(\max \{x^\star _{p_m},x^\star _{p_m-1}\}<t_m\cdot {\frac{\varepsilon _m}{4}}\) (we recall that \(p_m\) stands for the mth prime number), and define the positive real

$$\begin{aligned} \delta _m:=\min \left\{ \frac{1}{m^2t_m}, \frac{\varepsilon _m}{4}-\frac{\max \{x^\star _{p_m},x^\star _{p_m-1}\}}{t_m} \right\} . \end{aligned}$$

Note that \(\lim _m \delta _m=0\). Without loss of generality, we can assume also that \(k_0(x^{(m+1)}, \varepsilon _{m+1})>k_0(x^{(m)}, \varepsilon _m)+2t_m\) for all \(m \in \mathbf {N}\). Now, define the set \(I_m=\mathbf {N}\cap [k_0(x^{(m)}, \varepsilon _m), k_0(x^{(m)}, \varepsilon _m)+2t_m)\) (hence the sets \(I_m\) are pairwise disjoint) and let \(z^{(m)}\) be the sequence such that

$$\begin{aligned} \forall n \in \mathbf {N} \quad z^{(m)}_n= {\left\{ \begin{array}{ll} x^\star _{p_m-1}/t_m \,\,\,\,\,\, &{} \text { if }n \in I_m \text { and } n \text { even},\\ x^\star _{p_m}/t_m \,\,\,\,\,\, &{} \text { if }n \in I_m \text { and } n \text { odd},\\ x^{(m)}_n \,\,\,\,\,\, &{} \text { if }n \notin I_m.\\ \end{array}\right. } \end{aligned}$$

Lastly, set \(V_m:=B_{\delta _m}(z^{(m)})\) and note that, for each sequence \(y \in V_m\), we have by construction:

  1. (i)

    \(|y_n-x^{(m)}_n|=|y_n-z^{(m)}_n|<\delta _m<\frac{\varepsilon _m}{2}\) if \(n<\min I_m\);

  2. (ii)

    \(y_n< z_n^{(m)}+\delta _m \le \max \{x^\star _{p_m-1},x^\star _{p_m}\}/t_m+\delta _m < \frac{\varepsilon _m}{2}\) if \(n\in I_m\);

  3. (iii)

    \(y_n< z_n^{(m)}+\delta _m=x_n^{(m)}+\delta _m < \frac{\varepsilon _m}{4}+\frac{\varepsilon _m}{4}=\frac{\varepsilon _m}{2}\) if \(n>\max I_m\).

(In the last point, we used that if \(n>\max I_m\ge t_m \ge k_0(x^{(m)},\frac{\varepsilon _m}{2})\) then \(x_n^{(m)} \le \frac{1}{2}\cdot \frac{\varepsilon _m}{2}\).) This implies that \(y \in W_{\varepsilon _m}(x^{(m)})\). To sum up, we obtain

$$\begin{aligned} \forall m \in \mathbf {N}, \quad V_m \subseteq W_{\varepsilon _m}(x^{(m)}) \subseteq B_{\varepsilon _m}(x^{(m)}) \subseteq U_m, \end{aligned}$$

hence \(V_m\) is a nonempty open set contained in \(U_m\). In addition, the sequence of centers \((z^{(m)})\) is a Cauchy sequence in the complete metric space \(\mathscr {Z}\). Indeed, since the sequence \((V_n)\) is decreasing, we have \(\Vert z^{(m)}-z^{(m^\prime )}\Vert \le \delta _m \le \frac{1}{m^2}\) for all integers \(m^\prime \ge m \ge 1\). Hence \((z^{(m)})\) is convergent to some \(z \in \mathscr {Z}\) and it is straighforward to see that \(\{z\}=\bigcap _m V_m\).

To complete the proof, we need to show that \(z \in \mathscr {A}\). Set \(A:=\left( \bigcup _m I_m\right) \setminus 2\mathbf {N}\). Proceeding as in the proof of Theorem 1.3, we see that

$$\begin{aligned} \sum _{n \in A}z_n=\sum _{m\ge 1}\sum _{n \in I_m \setminus 2\mathbf {N}}z_n \le \sum _{m\ge 1}\sum _{n \in I_m \setminus 2\mathbf {N}}(z_n^{(m)}+\delta _m)\le \sum _{m\ge 1}|I_m|\left( \frac{x^\star _{p_m}}{t_m}+\frac{1}{m^2t_m}\right) < \infty , \end{aligned}$$

hence \(A \in \mathcal {I}_z\). Similarly, \(A-1 \notin \mathcal {I}_z\) since

$$\begin{aligned} \sum _{n \in A-1}z_n \ge \sum _{m\ge 1}\sum _{n \in I_m \cap 2\mathbf {N}}(z_n^{(m)}-\delta _m) \gg \sum _{m\ge 1}|I_m|\left( \frac{x^\star _{p_m-1}}{t_m}-\frac{1}{m^2t_m}\right) = \infty . \end{aligned}$$

Now, fix \(y \in \mathscr {Y}\) such that \(\kappa :=\liminf _n y_{n+1}/y_n>0\) and suppose that \(\mathcal {I}_z=\mathcal {I}_y\). Then we would have that \(A \in \mathcal {I}_y\) and \(A-1 \notin \mathcal {I}_y\), which is impossible reasoning as in (2).

5 Proof of Theorem 1.5

Let \(x^\star\) be the sequence defined in the proof of Theorem 1.3. For each \(r \in (0,1]\), let \(x^{(r)}\) be the sequence defined by \(x^{(r)}_n=(x^\star _n)^r\) for all \(n \in \mathbf {N}\). Replacing the set of primes \(\mathbf {P}\) with

$$\begin{aligned} \textstyle \mathbf {P}_r:=\left\{ 2\left\lfloor \frac{p_n^{1/r}}{2}\right\rfloor +1: n \in \mathbf {N}\right\} , \end{aligned}$$

we obtain that

$$\begin{aligned} \sum _{n \in \mathbf {P}_r}x_n^{(r)} \ll \sum _{n\ge 1}\left( \frac{1}{p_n^{1/r}\log (p_n^{1/r})}\right) ^r \ll \sum _{n\ge 1}\frac{1}{n \,(\log (n))^{1+r}}<\infty , \end{aligned}$$

and \(\sum _{n \in \mathbf {P}_r-1}x_n^{(r)}\gg \sum _{n\ge 1}\frac{1}{p_n}=\infty\). Reasoning as in the proof of Theorem 1.3, we conclude that \(x^{(r)} \in \mathscr {A}\) for each \(r \in (0,1]\).

To complete the proof, fix reals r, s such that \(0<r<s\le 1\). Then, it is sufficient to show that the ideals generated by \(x^{(r)}\) and \(x^{(s)}\) are not isomorphic. To this aim, let \(f: \mathbf {N} \rightarrow \mathbf {N}\) be a bijection and assume for the sake contradiction that

$$\begin{aligned} \forall A\subseteq \mathbf {N}, \quad \sum _{n \in A}x^{(r)}_{f(n)}<\infty \,\, \text { if and only if }\,\,\sum _{n \in A}x^{(s)}_{n}<\infty . \end{aligned}$$
(4)

Fix \(t \in (1,\frac{s}{r})\) and define \(S:=\{n \in \mathbf {N}: f(n)>n^t\}\) and \(T:=\mathbf {N}\setminus S\). We have \(\sum _{n \in S}\frac{1}{f(n)}\le \sum _{n \in S}\frac{1}{n^t}<\infty\). Considering that f is a bijection and the harmonic series is divergent, we obtain that \(\sum _{n \in T}\frac{1}{f(n)}=\infty\) (in particular, T is infinite). In addition, since \(r<1\) and

$$\begin{aligned} \frac{1}{f(n)} \ll \frac{1}{(f(n) \log (f(n)+1))^r} \le x^{(r)}_{f(n)} \le \frac{1}{f^r(n)}, \end{aligned}$$

we get that \(\sum _{n \in T}x^{(r)}_{f(n)}=\infty\) and, thanks to (4), also that \(\sum _{n \in T}x^{(s)}_{n}=\infty\). Note that, if \(n \in T\), then \(f(n) \le n^t\), which implies that

$$\begin{aligned} \forall n \in T, \quad \frac{x^{(s)}_n}{x^{(r)}_{f(n)}} \le \frac{1/n^s}{1/(f(n) \log (f(n)+1))^r} \ll \frac{(\log n)^r}{n^{s-tr}}, \end{aligned}$$

which has limit 0 if \(n\rightarrow \infty\) (and belongs to T). In particular, for each \(k \in \mathbf {N}\), there exists \(n_k \in \mathbf {N}\) such that \(x^{(s)}_n/x^{(r)}_{f(n)}\le \frac{1}{k^2}\) for all \(n\ge n_k\). Let \((A_k)\) be a sequence of finite subsets of T defined recursively as follows: for each \(k \in \mathbf {N}\), let \(A_k\) be a finite subset of T such that \(\min A_k\ge n_k+ \max A_{k-1}\) and \(\sum _{n \in A_k}x^{(r)}_{f(n)} \in (\frac{1}{2},1)\) where, by convention, we assume \(\max A_0:=0\) (note that it is really possible to define such sequence). Finally, define \(A:=\bigcup _k A_k\) so that we obtain

$$\begin{aligned} \sum _{n \in A}x^{(r)}_{f(n)}=\sum _{k\ge 1}\sum _{n \in A_k}x^{(r)}_{f(n)}=\infty \,\,\,\text { and }\,\,\, \sum _{n \in A}x^{(s)}_{n} \le \sum _{k\ge 1}\sum _{n \in A_k}\frac{x^{(r)}_{f(n)}}{k^2}\le \sum _{k\ge 1}\frac{1}{k^2} <\infty . \end{aligned}$$

This contradicts (4), concluding the proof.

6 Concluding remarks

We remark that the ideal \(\mathcal {I}_x\) defined in the proof above is just (an isomorphic copy of) the Fubini sum \(\mathcal {I}_{s}\oplus \mathcal {I}_{t}\), where \(s,t \in \mathscr {Z}\) are sequences defined by \(s_n=\frac{1}{n}\) and \(t_n=\frac{1}{n\log (n+1)}\) for all \(n \in \mathbf {N}\). Here, we recall that the Fubini sum of two ideals \(\mathcal {I}\) and \(\mathcal {J}\) on \(\mathbf {N}\) is the ideal \(\mathcal {I}\oplus \mathcal {J}\) on \(\{0,1\}\times \mathbf {N}\) of all sets A such that \(\{n \in \mathbf {N}: (0,n) \in A\} \in \mathcal {I}\) and \(\{n \in \mathbf {N}: (1,n) \in A\} \in \mathcal {J}\), cf. e.g. [4, p. 8]

Also, some comments are in order about the simplifications. The assumption that the sequence x has positive elements (instead of nonnegative elements) is rather innocuous. Indeed, in the opposite, if \(x_n=0\) for infinitely many n, then the summable ideal \(\mathcal {I}_x\) would be (isomorphic to) the Fubini sum \(\mathcal {P}(\mathbf {N}) \oplus \mathcal {I}_y\), for some \(y \in \mathscr {X}\). Lastly, also the hypothesis \(x \in \mathscr {Z}\) in Question 1.1 (instead of \(x \in \mathscr {X}\)) has a similar justification. Indeed, if \(x \in \mathscr {X} \setminus \mathscr {Z}\), then \(\mathcal {I}_x\) would be the Fubini sum \(\mathrm {Fin} \oplus \mathcal {I}_y\), for some \(y \in \mathscr {Z}\).

We conclude with two open questions:

Question 6.1

Is it true that for each \(x \in \mathscr {Z}\) there are (possibly infinite) sequences \(y^1,y^2,\ldots \in \mathscr {Y}\) such that \(\mathcal {I}_x=\oplus _i \mathcal {I}_{y^i}\)?

Question 6.2

Is it true that \(\mathscr {A}\) is an analytic subset of \(\mathscr {Z}\)?