Proof. Without loss of generality, we may assume that $\mathcal {I}$ is an ideal on $\omega $ .

Since $\mathcal {I}$ is not tall ( $\mathcal {I}\neq \mathrm {Fin}$ , respectively), there is $A\notin \mathcal {I}$ such that ${\mathcal {I}\restriction A=\mathrm {Fin}\restriction A}$ (an infinite $A\in \mathcal {I}$ , respectively). Let $g:((\{0\}\times A)\cup (\{1\}\times \omega ))\to A$ be any bijection. Then $f:2\times \omega \to \omega $ given by:

$$\begin{align*}f(i,n)=\begin{cases} g(i,n), & \text{if }(i,n)\in(\{0\}\times A)\cup(\{1\}\times\omega),\cr n, & \text{otherwise,} \end{cases} \end{align*}$$

for all $(i,n)\in 2\times \omega $ , witnesses $\mathcal {I}\cong \mathcal {I}\oplus \mathrm {Fin}$ ( $\mathcal {I}\cong \mathcal {I}\oplus \mathcal {P}(\omega )$ , respectively).

Proof. Since $\hat {\mathbb {u}}$ is unconditional, by [Reference Albiac and Kalton1, Proposition 3.1.3], there is $K\geq 1$ such that

$$\begin{align*}\left\| \sum_{i=0}^n c_i \mathbb{u}_i \right\|_X \leq K \left\| \sum_{i=0}^n d_i \mathbb{u}_i \right\|_X, \end{align*}$$

for all $n\in \omega $ and scalars $c_0,\ldots ,c_n,d_0,\ldots ,d_n$ such that $|c_i|\leq |d_i|$ , for $i\leq n$ .

Fix $n\in \omega $ . Then for any $m>n$

$$\begin{align*}\left\| \sum_{i=n}^m b_n \mathbb{u}_n \right\|_X \leq K \left\| \sum_{i=n}^m a_n \mathbb{u}_n \right\|_X \leq K^2 \left\| \sum_{i=n}^\infty a_n \mathbb{u}_n \right\|_X \xrightarrow{n \to \infty} 0. \end{align*}$$

Therefore $ \sum _{n=0}^\infty b_n \mathbb {u}_n $ is convergent.

Proof. Note that $\sum _{i=0}^k \alpha _i\mathbb {a}_i - \sum _{i=0}^{k'} \beta _i\mathbb {a}_i=(\alpha _l-\beta _l)\mathbb {a}_l-\mathbb {b}$ for some $\mathbb {b}\in \operatorname {span}\{\mathbb {a}_m:m \neq l\}$ . Thus

$$\begin{align*}\left\|\sum_{i=0}^k \alpha_i\mathbb{a}_i - \sum_{i=0}^{k'} \beta_i\mathbb{a}_i\right\|_X\geq \operatorname{dist}\left((\alpha_l-\beta_l)\mathbb{a}_l, \overline{\operatorname{span}\{\mathbb{a}_m:m \neq l\}}\right)>0. \end{align*}$$

Note that the distance above does not depend on k and $k'$ .

Proof. Note that if $n\neq m$ then the condition $\mathbb {a}^\star _n(\mathbb {a}_m)=0$ implies $\mathbb {a}_m\in \operatorname {Ker}(\mathbb {a}^\star _n)$ . By the continuity of $\mathbb {a}_n^\star $ , $\operatorname {Ker}(\mathbb {a}^\star _n)$ is closed, so $\overline {\operatorname {span}\{\mathbb {a}_m: m \neq n\}}\subseteq \operatorname {Ker}(\mathbb {a}^\star _n)$ . Since $\mathbb {a}_n^\star (\mathbb {a}_n)=1$ , we obtain $\mathbb {a}_n \notin \operatorname {Ker}\mathbb {a}^\star _n$ for every n. If $\beta _n\neq \mathbb {a}_n^\star (\mathbb {x})$ for some $\mathbb {x}\in X$ and $n\in \omega $ , then, by Lemma 2.2, there is some $\varepsilon>0$ such that

$$\begin{align*}\left\|\sum_{i=0}^k \mathbb{a}^\star_i(\mathbb{x}) \mathbb{a}_i- \sum_{i=0}^{k'} \beta_i \mathbb{a}_i \right\|>\varepsilon, \end{align*}$$

for every $k,k'>n$ . But since $\mathbb {x}=\sum _{n,\mathcal {I}} \mathbb {a}^\star _n(\mathbb {x}) \mathbb {a}_n$ and $\mathcal {I}\neq P(\omega )$ , there exists $k>n$ such that

$$\begin{align*}\left\|\mathbb{x}-\sum_{i=0}^k \mathbb{a}^\star_i(\mathbb{x}) \mathbb{a}_i\right\|<\frac{\varepsilon}{2}, \end{align*}$$

hence

$$\begin{align*}\left\|\mathbb{x}-\sum_{i=0}^{k'} \beta_i \mathbb{a}_i\right\|>\frac{\varepsilon}{2}, \end{align*}$$

for all $k'>n$ , contradicting $\mathbb {x}= \sum _{n,\mathcal {J}} \beta _i \mathbb {a}_n$ .

Proof. Follows from Proposition 2.3.

Proof. Follows from Proposition 2.3 and [Reference de Rancourt, Kania and Swaczyna6, Theorem A].

Proof. If Y is an arbitrary Banach space over $\mathbb {R}$ and $\hat {\mathbb {u}}$ is its $\mathrm {Fin}$ -Schauder basis, then consider:

$$\begin{align*}X=\left\{\mathbb{x}\in\mathbb{R}^\omega: \sum_{n\in\omega}x_n\mathbb{u}_n\in Y\right\}\!, \end{align*}$$

normed by:

$$\begin{align*}\|\mathbb{x}\|_X=\left\|\sum_{n\in\omega}x_n\mathbb{u}_n\right\|_Y, \end{align*}$$

for all $\mathbb {x}\in X$ . Then X is a Banach space and $\ell _1\subseteq X\subseteq c_0$ .

If $\hat {\mathbb {c}}$ is simple over $\hat {\mathbb {u}}$ , then $\mathcal {CR}(Y,\hat {\mathbb {c}})=\mathcal {CR}(X,\hat {\mathbb {a}})$ , where $\hat {\mathbb {a}}$ is simple over $\hat {\mathbb {e}}$ with the same witnessing D, $h,$ and $(\mathbb {b}_n)$ as for $\hat {\mathbb {c}}$ and $\hat {\mathbb {u}}$ .

Item (i) follows from the fact that X and Y are isometric (by the isometry that moves $\hat {\mathbb {e}}$ to $\hat {\mathbb {u}}$ ). To show item (ii), observe that for any $\mathbb {x}_n, \mathbb {x}\in X$ and any $k\in \mathbb {N}$ we have:

$$ \begin{align*} |\mathbb{u}^\star_{k+1}(\mathbb{x}_n-\mathbb{x})| & = \|\mathbb{u}^\star_{k+1}(\mathbb{x}_n-\mathbb{x})\mathbb{u}_{k+1}\|_X = \|S_{k+1}(\hat{\mathbb{u}}) (\mathbb{x}_{n}-\mathbb{x})- S_{k}(\hat{\mathbb{u}}) (\mathbb{x}_{n}-\mathbb{x})\|_X \cr & = \|(S_{k+1}-S_k)(\hat{\mathbb{u}}) (\mathbb{x}_{n}-\mathbb{x})\|_X \leq 2K \|\mathbb{x}_{n}-\mathbb{x}\|_X, \end{align*} $$

where K is the basis constant of $\hat {\mathbb {u}}$ (see [Reference Albiac and Kalton1, Proposition 1.1.4 and Definition 1.1.5]). The final claim is clear.

Proof. Observe that:

$$\begin{align*}S_n(\hat{\mathbb{a}})(\mathbb{x})=\sum_{i=0}^n (\mathbb{e}_i^\star-\mathbb{e}_{i+1}^\star) (\mathbb{x})\sum_{k=0}^i \mathbb{e}_k = \sum_{i=0}^n \mathbb{e}_i^\star (\mathbb{x}) \mathbb{e}_i - \mathbb{e}_{n+1}^\star(\mathbb{x}) \sum_{i=0}^n \mathbb{e}_i=S_n(\hat{\mathbb{e}})(\mathbb{x})-\mathbb{e}_{n+1}^\star(\mathbb{x}) \sum_{i=0}^n \mathbb{e}_i, \end{align*}$$

for any $\mathbb {x}\in \mathbb {R}^\omega $ . Thus:

$$\begin{align*}S_n(\hat{\mathbb{e}})(\mathbb{x})-S_n(\hat{\mathbb{a}})(\mathbb{x})=\mathbb{e}_{n+1}^\star(\mathbb{x}) \sum_{i=0}^n \mathbb{e}_i, \end{align*}$$

and $\hat {\mathbb {a}}$ is simple over $\hat {\mathbb {e}}$ (as witnessed by $D=\omega $ , $h(n)=n+1,$ and $\mathbb {b}_{n+1}=\sum _{i=0}^n \mathbb {e}_i$ ). Hence, applying Lemma 3.3, we have:

$$ \begin{align*} \mathcal{CR}(\ell_p,\hat{\mathbb{a}}) & =\left\{A\subseteq\omega: \sum_{n\in A}\frac{\mathbb{e}_{n+1}}{\|\sum_{i=0}^n \mathbb{e}_i\|_{\ell_p}}\in \ell_p\right\}=\left\{A\subseteq\omega: \sum_{n\in A}\frac{\mathbb{e}_{n+1}}{(n+1)^{1/p}}\in \ell_p\right\}\cr & =\left\{A\subseteq\omega: \sum_{n\in A}\frac{1}{n+1}<\infty\right\}=\mathcal{I}_{1/n}, \end{align*} $$

for every $1\leq p<\infty $ .

Proof. Let $(A_k)$ be any partition of $\omega $ into infinite sets such that ${A_0=\{2n:n\in \omega \}}$ .

For each $n\in \omega $ find $m_n\in \omega $ such that $n\in A_{m_n}$ and define:

$$\begin{align*}\mathbb{a}_n=\mathbb{e}_{n}+m_n\cdot \mathbb{e}_{n+1}\text{ and } \mathbb{a}_n^\star=\mathbb{e}_{n}^\star - m_{n-1}\cdot \mathbb{e}_{n-1}^\star, \end{align*}$$

(here we put $m_{-1}=0$ ). Then

$$ \begin{align*} \mathbb{a}_n^\star(\mathbb{a}_k) & = \mathbb{e}_{n}^\star(\mathbb{e}_{k})+m_k\mathbb{e}_{n}^\star(\mathbb{e}_{k+1}) - m_{n-1}\mathbb{e}_{n-1}^\star(\mathbb{e}_{k})- m_{n-1}m_k\mathbb{e}_{n-1}^\star(\mathbb{e}_{k+1})\cr & =\delta_{nk}+m_k\delta_{n(k+1)}-m_{n-1} \delta_{(n-1)k}-m_{n-1}m_k\delta_{(n-1)(k+1)}=\delta_{nk}, \end{align*} $$

since if $n-1=k+1$ , then one of $n-1,k\in A_0$ hence one of $m_{n-1},m_k$ equals $0$ .

Note that for any $\mathbb {x} \in \ell _p$ we have:

$$\begin{align*}S_n(\hat{\mathbb{e}})(\mathbb{x})-S_n(\hat{\mathbb{a}})(\mathbb{x})=\mathbb{e}_{n}^\star(\mathbb{x})m_n\mathbb{e}_{n+1}. \end{align*}$$

In particular, $\hat {\mathbb {a}}$ is simple over $\hat {\mathbb {e}}$ (as witnessed by $D=\{n\in \omega :m_n\neq 0\}=\omega \setminus A_0=\{2n+1:n\in \omega \}$ , $h(n)=n$ and $\mathbb {b}_n=m_n\mathbb {e}_{n+1}$ ). Hence, using Lemma 3.3, we have:

$$ \begin{align*} \mathcal{CR}(\ell_p,\hat{\mathbb{a}}) & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\sum_{n\in A\cap D}\frac{\mathbb{e}_{n}}{\left\|m_n\mathbb{e}_{n+1}\right\|_{\ell_p}}\in \ell_p\right\}\cr & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\sum_{n\in A\cap D}\frac{\mathbb{e}_{n}}{m_n}\in \ell_p\right\}\cr & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\sum_{n\in A\cap D}\frac{1}{m_n^p}<\infty\right\}\cr & =\left\{A\subseteq\omega:A\cap A_0\in\mathrm{Fin}\text{ and }\sum_{k\in\omega\setminus\{0\}}\frac{|A\cap A_k|}{k^p}<\infty\right\}\!. \end{align*} $$

The ideal $\mathcal {CR}(\ell _p,\hat {\mathbb {a}})$ is summable, since it is equal to $\mathcal {I}_{\mathbb {r}}$ for $\mathbb {r}=(r_n)$ given by:

$$\begin{align*}r_n = \left\{\begin{array}{@{}ll} 1 & \text{ if } n\in A_0,\\ \frac{1}{k^p} & \text{ if } n\in A_k \text{ for some } k>0. \end{array}\right. \end{align*}$$

To finish the proof, we will show that $1\leq p<q<\infty $ implies $\mathcal {CR}(\ell _q,\hat {\mathbb {a}})\not \subseteq \mathcal {CR}(\ell _p,\hat {\mathbb {a}})$ . Indeed, any set $A\subseteq \omega $ such that $A\cap A_0=\emptyset $ and $|A\cap A_k|=\lceil k^p\rceil $ , for $k\in \omega \setminus \{0\}$ , is in $\mathcal {CR}(\ell _q,\hat {\mathbb {a}})$ , but not in $\mathcal {CR}(\ell _p,\hat {\mathbb {a}})$ .

Proof. Define inductively the sequence $(c_k)\subseteq \mathbb {R}$ by $c_0=1$ and

$$\begin{align*}c_{k+1}=\frac{1}{s_k} \cdot \left\|\sum_{i=0}^k c_i\mathbb{u}_{t_i}\right\|_X \end{align*}$$

for all $k\in \omega $ . Set:

$$\begin{align*}\mathbb{a}_n = \left\{\begin{array}{@{}ll} \sum_{i=0}^{k} c_i\mathbb{u}_{t_i}, & \text{ if } n=t_k \text{ for some } k\in \omega,\\ \mathbb{u}_n, & \text{ otherwise}, \\ \end{array}\right. \end{align*}$$

and

$$\begin{align*}\mathbb{a}_n^\star = \left\{\begin{array}{@{}ll} \frac{\mathbb{u}_{t_k}^\star}{c_k} - \frac{u_{t_{k+1}}^\star}{c_{k+1}}, & \text{ if } n=t_k \text{ for some } k\in \omega,\\ \mathbb{u}_n^\star, & \text{ otherwise}. \end{array}\right. \end{align*}$$

Clearly, $\mathbb {a}_n^\star (\mathbb {a}_m)=\delta _{nm}$ for all $n,m\in \omega $ . Indeed, it is obvious whenever $n \notin \{t_k: k\in \omega \}$ or $m \notin \{t_k:k\in \omega \}$ , and the remaining follows from the case $t_k=k$ , which is easy.

Denote $T=\{t_k:k\in \omega \}$ and let $g:\omega \to \omega $ be given by:

$$\begin{align*}g(n)=k\ \Longleftrightarrow n\in[t_k,t_{k+1}). \end{align*}$$

Note that $h(n)=t_{g(n)+1}$ for all $n\in \omega $ (where the function h is defined in the formulation of this lemma). Observe that for any $\mathbb {x} \in X$ we have:

$$ \begin{align*} S_n(\hat{\mathbb{a}})(\mathbb{x}) & = \sum_{i \in (n+1)\setminus T} \mathbb{u}_i^\star (x) \mathbb{u}_i + \sum_{j=0}^{g(n)} \left(\frac{\mathbb{u}_{t_j}^\star}{c_j} - \frac{u_{t_{j+1}}^\star}{c_{j+1}}\right)(\mathbb{x})\sum_{i=0}^{j} c_i\mathbb{u}_{t_i} \cr & =S_n(\hat{\mathbb{u}})(\mathbb{x})-\mathbb{u}_{t_{g(n)+1}}^\star(\mathbb{x})\frac{\sum^{g(n)}_{i=0} c_i \mathbb{u}_{t_i}}{c_{g(n)+1}}. \end{align*} $$

Thus, $\hat {\mathbb {a}}$ is simple over $\hat {\mathbb {e}}$ as witnessed by $D=\omega $ , h and $(\mathbb {b}_{t_{k+1}})_{k\in \omega }\subseteq X\setminus \{0\}$ such that:

$$\begin{align*}\mathbb{b}_{t_{k+1}}=\frac{\sum^{k}_{i=0} c_i \mathbb{e}_{t_i}}{c_{k+1}}, \end{align*}$$

for all $k\in \omega $ . In particular, $\|\mathbb {b}_{t_{k+1}}\|_X=s_k$ , for all $k\in \omega $ .

Proof. Fix $1\leq p<\infty $ and apply Lemma 5.5 for $\mathbb {t}\in \omega ^\omega $ and $\mathbb {s}$ given by $s_k=r_k^{1/p}$ , for all $k\in \omega $ , to get $\hat {\mathbb {a}}$ simple over $\hat {\mathbb {e}}$ with the witnesses $D=\omega $ , $h:\omega \to \omega $ given by $h\restriction [t_k,t_{k+1})=t_{k+1}$ and some sequence $(\mathbb {b}_{t_{k+1}})_{k\in \omega }$ such that $\|\mathbb {b}_{t_{k+1}}\|_{\ell _p}=r_k^{1/p}$ .

Applying Lemma 3.3 and denoting:

$$\begin{align*}\tilde{A}=\left\{k\in\omega:A\cap [t_k,t_{k+1})\neq\emptyset\right\}\!, \end{align*}$$

for all $A\subseteq \omega $ , we have:

$$ \begin{align*} \mathcal{CR}(\ell_p,\hat{\mathbb{a}}) & =\left\{A\subseteq\omega: \sum_{k\in h[A]}\frac{\mathbb{e}_{k}}{\left\|\mathbb{b}_k\right\|_{\ell_p}}\in \ell_p\right\}\cr & =\left\{A\subseteq\omega: \sum_{k\in \tilde{A}}\mathbb{e}_{t_{k+1}}r_k^{1/p}\in \ell_p\right\}\cr & =\left\{A\subseteq\omega: \sum_{k\in \tilde{A}}r_k<\infty\right\}=\mathcal{I}^{\mathbb{t}}_{\mathbb {r}}.\\[-41pt] \end{align*} $$

Proof. It follows from Proposition 5.6 and [Reference Guzmán-González and Meza-Alcántara11, Theorem 1].

Proof. Let $t_0=0$ and $t_k=2^k$ for all $k\in \omega \setminus \{0\}$ . Denote $\mathbb {t}=(t_k)$ and $I_k=[t_k,t_{k+1})$ , for all $k\in \omega $ . In particular, $|I_0|=2$ and $|I_k|=2^k$ for all $k>0$ . By Proposition 5.6, for each $1\leq p<\infty $ there is $\hat {\mathbb {a}}$ such that $\mathcal {CR}(\ell _p,\hat {\mathbb {a}})=\mathcal {I}_{1/n}^{\mathbb {t}}$ . Therefore, we need to demonstrate that $\mathcal {I}_{1/n}^{\mathbb {t}}$ is non-summable and ${\mathbf {F}_\sigma }$ .

To verify that $\mathcal {I}_{1/n}^{\mathbb {t}}$ is ${\mathbf {F}_\sigma }$ , note that $\mathcal {I}_{1/n}^{\mathbb {t}}=\phi ^{-1}[\mathcal {I}_{1/n}]$ , where $\phi :\mathcal {P}(\omega )\to \mathcal {P}(\omega )$ given by $\phi (A)=\{k\in \omega : A\cap I_k\neq \emptyset \}$ , for all $A\subseteq \omega $ , is continuous.

Suppose towards contradiction that $\mathcal {I}_{1/n}^{\mathbb {t}}$ is a summable ideal, i.e., that there is $\mathbb {r}=(r_n)$ such that $r_n>0$ , $\sum _{n\in \omega }r_n=\infty $ and $\mathcal {I}_{1/n}^{\mathbb {t}}=\mathcal {I}_{\mathbb {r}}$ . Denote $d_k=\min \{r_n: n\in I_k\}$ and $n_k=\min \{n\in I_k: r_n=d_k\}$ . Observe that $Z=\{k\in \omega : 2^k\cdot d_k\geq 1\}$ is infinite, since otherwise we would get:

$$\begin{align*}\sum_{k\in\omega}r_{n_k}=\sum_{k\in Z}d_k+\sum_{k\in\omega\setminus Z}d_k<\sum_{k\in Z}d_k+\sum_{k\in\omega\setminus Z}\frac{1}{2^k}<\infty, \end{align*}$$

which shows that $\{n_k{\kern-1pt}:{\kern-1pt}k{\kern-1pt}\in{\kern-1pt} \omega \}{\kern-1pt}\in{\kern-1pt} \mathcal {I}_{\mathbb {r}}$ and contradicts $\mathcal {I}_{1/n}^{\mathbb {t}}{\kern-1pt}={\kern-1pt}\mathcal {I}_{\mathbb {r}}$ , as $\{n_k{\kern-1pt}:{\kern-1pt}k{\kern-1pt}\in{\kern-1pt} \omega \}{\kern-1pt}\notin{\kern-1pt} \mathcal {I}_{1/n}^{\mathbb {t}}$ . However, as Z is infinite, we can find an infinite $Z'\subset Z$ such that $Z'\in \mathcal {I}_{1/n}$ . Then $\bigcup _{k\in Z'}I_k\in \mathcal {I}_{1/n}^{\mathbb {t}}$ , but:

$$\begin{align*}\sum_{n\in \bigcup_{k\in Z'}I_k}r_n\geq \sum_{k\in Z'}|I_k|d_k\geq\sum_{k\in Z'}2^k\cdot d_k\geq \sum_{k\in Z'}1=\infty, \end{align*}$$

so $\bigcup _{k\in Z'}I_k\notin \mathcal {I}_{\mathbb {r}}$ , which again contradicts $\mathcal {I}_{1/n}^{\mathbb {t}}=\mathcal {I}_{\mathbb {r}}$ .

Proof. (b) $\implies $ (a): This is Proposition 5.6.

(a) $\implies $ (b): Fix some $\hat {\mathbb {a}} (D,h,(\mathbb {b}_n))$ -simple over $\hat {\mathbb {e}}$ . Find a partition $(I_n)$ of $\omega $ into consecutive intervals such that either $I_n=h^{-1}[\{k\}]$ for some $k\in \omega $ (note that in this case $I_n\subseteq D$ ) or $I_n=\{i\}$ for some $i\in \omega \setminus D$ . Denote $S=\{n\in \omega : I_n=h^{-1}[\{k\}]\text { for some }k\in \omega \}$ .

For each $n\in \omega $ put $t_n=\min I_n$ and

$$\begin{align*}r_n=\begin{cases} \left(\frac{1}{\|\mathbb{b}_n\|_{\ell_p}}\right)^p, & \text{if } n\in S,\\ 1, & \text{otherwise}. \end{cases} \end{align*}$$

Then, using Lemma 3.3, we get that $\mathcal {I}^{\mathbb {t}}_{\mathbb {r}}=\mathcal {CR}(\ell _p,\hat {\mathbb {a}})$ . Since $\mathcal {CR}(\ell _p,\hat {\mathbb {a}})=\mathcal {I}$ is an ideal, $\omega \notin \mathcal {CR}(\ell _p,\hat {\mathbb {a}})$ , which implies that $\sum _{n\in \omega }r_n=\infty $ .

Proof. Let $(A_k)$ and $\hat {\mathbb {a}}$ be as in the proof of Proposition 5.3, i.e., $(A_k)$ is a partition of $\omega $ into infinite sets such that $A_0=\{2n:n\in \omega \}$ , $\mathbb {a}_n=\mathbb {e}_{n}+m_n\cdot \mathbb {e}_{n+1}$ and $\mathbb {a}_n^\star =\mathbb {e}_{n}^\star - m_{n-1}\cdot \mathbb {e}_{n-1}^\star $ , where $m_n$ is given by $n\in A_{m_n}$ .

Define:

$$\begin{align*}\mathcal{J}=\left\{A\subset\omega:\forall_{k\in\omega}A\cap A_k\in\mathrm{Fin}\right\}\!. \end{align*}$$

Clearly, $\mathcal {J}$ is isomorphic with $\{\emptyset \}\otimes \mathrm {Fin}$ , as witnessed by any bijection $f:\omega ^2\to \omega $ such that $f[\{k\}\times \omega ]=A_k$ , for all $k\in \omega $ .

We will show that $\mathcal {CR}(c_0,\hat {\mathbb {a}})=\mathcal {J}$ . Similarly as in the proof of Proposition 5.3, for any $\mathbb {x} \in c_0$ we have:

$$\begin{align*}S_n(\hat{\mathbb{e}})(\mathbb{x})-S_n(\hat{\mathbb{a}})(\mathbb{x})=\mathbb{e}_{n}^\star(\mathbb{x})m_n\mathbb{e}_{n+1}, \end{align*}$$

i.e., $\hat {\mathbb {a}}$ is simple over $\hat {\mathbb {e}}$ as witnessed by $D=\{n\in \omega :m_n\neq 0\}=\omega \setminus A_0=\{2n+1:n\in \omega \}$ , $h(n)=n$ and $\mathbb {b}_n=m_n\mathbb {e}_{n+1}$ . Hence, applying Lemma 3.3, we have:

$$ \begin{align*} \mathcal{CR}(c_0,\hat{\mathbb{a}}) & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\sum_{n\in A\cap D}\frac{\mathbb{e}_{n}}{\left\|m_n\mathbb{e}_{n+1}\right\|_{c_0}}\in c_0\right\}\cr & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\sum_{n\in A\cap D}\frac{\mathbb{e}_{n}}{m_n}\in c_0\right\}\cr & =\left\{A\subseteq\omega: A\setminus D\in\mathrm{Fin}\text{ and }\lim_{n\in A\cap D}\frac{1}{m_n}=0\right\}=\mathcal{J}.\\[-39pt] \end{align*} $$

Proof. The claim is clear if $A\in \mathrm {Fin}$ , so assume that A is infinite. Denote:

$$\begin{align*}a=\max\left\{\frac{1}{\|\mathbb{b}_{h(n)}\|_{\mathfrak{J}}}:n\in A\right\}\!. \end{align*}$$

Put also:

$$\begin{align*}A_k=\left\{n\in A:\frac{1}{\|\mathbb{b}_{h(n)}\|_{\mathfrak{J}}}>\frac{1}{k+1}\right\}\!, \end{align*}$$

for all $k\in \omega $ . Observe that $A=\bigcup _{k\in \omega }A_k$ and $\lim _k \max h[A_{k}]=\infty $ (since A is infinite and h is interval-to-one).

Let $\mathbb {x}$ be such that $x_i=a$ for all $i\leq \max h[A_0]$ and $x_i=\frac {1}{k+1}$ for all $\max h[A_{k}]<i\leq \max h[A_{k+1}]$ and $k\in \omega $ . Observe that $\|\mathbb {x}\|_{\mathfrak {J}}=a^2$ , as for every $n\in \omega $ and $0\leq p_0<p_1<\ldots <p_n$ we have:

$$\begin{align*}\sum_{i=1}^n \left(x_{p_i}-x_{p_{i-1}}\right)^2\leq\left(\sum_{i=1}^n\left(x_{p_i}-x_{p_{i-1}}\right)\right)^2=\left(x_{p_0}-x_{p_{n}}\right)^2<a^2. \end{align*}$$

Hence, $\mathbb {x}\in \mathfrak {J}$ . We will show that $A\subseteq A_{\mathbb {x}}$ . Fix $n\in A$ . If $n\in A_0$ then $\frac {1}{\|\mathbb {b}_{h(n)}\|_{\mathfrak {J}}}\leq a$ and $|\mathbb {e}^\star _{h(n)}(\mathbb {x})|=|x_{h(n)}|=a$ (since $h(n)\leq \max h[A_0]$ ). On the other hand, if $n\in A_{k+1}\setminus A_k$ , for some $k\in \omega $ , then $\frac {1}{\|\mathbb {b}_{h(n)}\|_{\mathfrak {J}}}\leq \frac {1}{k+1}$ and $|\mathbb {e}^\star _{h(n)}(\mathbb {x})|=|x_{h(n)}|=\frac {1}{k+1}$ (since $h(n)\leq \max h[A_{k+1}]$ ). Hence, in both cases $n\in A_{\mathbb {x}}$ .

Proof. Let $g:\omega ^2\to \omega $ be the bijection witnessing that $\mathcal {I}$ is isomorphic with $\{\emptyset \}\otimes \mathrm {Fin}$ and denote $X_k=g[\{k\}\times \omega ]$ for all $k\in \omega $ . Put $t_k=k$ for all $k\in \omega $ and define the sequence $\mathbb {s}$ by:

$$\begin{align*}s_n=k+1\ \Longleftrightarrow\ n\in X_k \end{align*}$$

for all $n\in \omega $ . Apply Lemma 5.5 for $\mathbb {t}$ , $\mathbb {s}$ and $X=c_0$ ( $\mathbb {t}$ , $\mathbb {s}$ and $X=\mathfrak {J}$ , respectively) to get $\hat {\mathbb {a}}$ ( $\hat {\mathbb {a}}'$ , respectively) simple over $\hat {\mathbb {e}}$ with the witnesses $D=\omega $ , $h:\omega \to \omega $ given by $h(n)=n+1$ and some sequence $(\mathbb {b}_{n+1})_{n\in \omega }$ such that $\|\mathbb {b}_{n+1}\|_{c_0}=s_n$ for all $n\in \omega $ ( $(\mathbb {b}^{\prime }_{n+1})_{n\in \omega }$ such that $\|\mathbb {b}^{\prime }_{n+1}\|_{\mathfrak {J}}=s_n$ for all $n\in \omega $ , respectively).

In the case of the space $c_0$ , using Lemma 3.3 we get:

$$ \begin{align*} \mathcal{CR}(c_0,\hat{\mathbb{a}}) & =\left\{A\subseteq\omega: \sum_{n\in A}\frac{\mathbb{e}_{n+1}}{\left\|\mathbb{b}_{n+1}\right\|_{c_0}}\in c_0\right\}\cr & =\left\{A\subseteq\omega: \sum_{n\in A}\frac{\mathbb{e}_{n+1}}{s_n}\in c_0\right\}\cr & =\left\{A\subseteq\omega: \lim_{n\in A}\frac{1}{s_n}=0\right\}=\mathcal{I}. \end{align*} $$

In the case of the space $\mathfrak {J}$ , we cannot use Lemma 3.3; however, from Lemma 3.1 we know that $\mathcal {CR}(\mathfrak {J},\hat {\mathbb {a}}')$ is the ideal generated by $\mathrm {Fin}$ and all sets of the form

$$\begin{align*}A_{\mathbb{x}}=\left\{n\in \omega: |\mathbb{e}^\star_{n+1}(\mathbb{x})|\geq\frac{1}{\|\mathbb{b}^{\prime}_{n+1}\|_{\mathfrak{J}}}\right\}\!, \end{align*}$$

for $\mathbb {x}\in \mathfrak {J}$ . We need to show that $\mathcal {CR}(\mathfrak {J},\hat {\mathbb {a}}')=\mathcal {I}$ .

$\mathcal {CR}(\mathfrak {J},\hat {\mathbb {a}}')\subseteq \mathcal {I}$ : It suffices to show that $A_{\mathbb {x}}\in \mathcal {I}$ for all $\mathbb {x}\in \mathfrak {J}$ , so fix any such $\mathbb {x}$ . Since $\mathbb {x}\in \mathfrak {J}\subseteq c_0$ , we have:

$$\begin{align*}\lim_{n\in A_{\mathbb{x}}}\frac{1}{\|\mathbb{b}^{\prime}_{n+1}\|_{\mathfrak{J}}}=\lim_{n\in A_{\mathbb{x}}}\frac{1}{s_n}=0. \end{align*}$$

Hence, $A_{\mathbb {x}}\cap X_k\in \mathrm {Fin}$ for all $k\in \omega $ , which means that $A_{\mathbb {x}}\in \mathcal {I}$ .

$\mathcal {CR}(\mathfrak {J},\hat {\mathbb {a}}')\supseteq \mathcal {I}$ : Fix any $A\in \mathcal {I}$ . Then

$$\begin{align*}\lim_{n\in A}\frac{1}{\|\mathbb{b}^{\prime}_{n+1}\|_{\mathfrak{J}}}=\lim_{n\in A}\frac{1}{s_n}=0. \end{align*}$$

By Lemma 6.3, there is $\mathbb {x}\in \mathfrak {J}$ such that $A\subseteq A_{\mathbb {x}}$ , so $A\in \mathcal {CR}(\mathfrak {J},\hat {\mathbb {a}}')$ .

Proof. (c) $\implies $ (a) and (c) $\implies $ (b): If $\mathcal {I}\cong \mathrm {Fin}$ , then actually $\mathcal {I}=\mathrm {Fin}$ , so $\mathcal {CR}(c_0,\hat {\mathbb {e}})=\mathcal {I}$ and $\mathcal {CR}(\mathfrak {J},\hat {\mathbb {e}})=\mathcal {I}$ . On the other hand, if $\mathcal {I}\cong \mathrm {Fin}\oplus \mathcal {P}(\omega )$ or $\mathcal {I}\cong \{\emptyset \}\otimes \mathrm {Fin}$ , then it suffices to apply Theorem 4.1 or Proposition 6.4, respectively.

(a) $\implies $ (c) and (b) $\implies $ (c): Let $X=c_0$ or $X=\mathfrak {J}$ . Let $\hat {\mathbb {a}}$ be simple over $\hat {\mathbb {e}}$ . Then there are $D\subseteq \omega $ , an interval-to-one $h:D\to \omega $ and $(\mathbb {b}_{h(n)})\subseteq X\setminus \{0\}$ such that:

$$\begin{align*}S_n(\hat{\mathbb{u}})(\mathbb{x})-S_n(\hat{\mathbb{a}})(\mathbb{x})= \begin{cases} 0, & \text{if }n\in\omega\setminus D,\\ \mathbb{u}_{h(n)}^\star(\mathbb{x})\mathbb{b}_{h(n)}, & \text{if }n\in D, \end{cases} \end{align*}$$

for all $n\in \omega $ and $\mathbb {x}\in X$ .

Define $B_0=\{n\in D: \|\mathbb {b}_{h(n)}\|_{X}\geq 1\}$ and $B_i=\{n\in D: \frac {1}{k+1}\leq \|\mathbb {b}_{h(n)}\|_{X}<\frac {1}{k}\}$ for all $n\in \omega \setminus \{0\}$ . Then $(B_k)$ is a partition of D. Denote $T=\bigcup \{B_k: B_k\in \mathrm {Fin}\}$ and $S=\{k\in \omega : B_k\notin \mathrm {Fin}\}$ .

Observe that:

1. (i) $\mathcal {CR}(X,\hat {\mathbb {a}})\restriction \omega \setminus D=\mathrm {Fin}\restriction \omega \setminus D$ ;

2. (ii) $\mathcal {CR}(X,\hat {\mathbb {a}})\restriction B_k=\mathrm {Fin}\restriction B_k$ for every $k\in \omega $ ;

3. (iii) if $A\subseteq D$ and $A\cap B_k\in \mathrm {Fin}$ for all k, then $A\in \mathcal {CR}(X,\hat {\mathbb {a}})$ ;

4. (iv) $\mathcal {CR}(X,\hat {\mathbb {a}})\restriction T=\mathcal {P}(T)$ .

Indeed, in the case of $X=c_0$ all items follow from Lemma 3.3, while in the case of $X=\mathfrak {J}$ , item (ii) follows from the fact that $\mathfrak {J}\subseteq c_0$ and items (iii) and (iv) follow from Lemma 6.3.

There are four possibilities:

• • If $\omega \setminus T\in \mathrm {Fin}$ , then $\mathcal {CR}(X,\hat {\mathbb {a}})=\mathcal {P}(\omega )$ (by (iv)).

• • If $S\in \mathrm {Fin}$ and $T\in \mathrm {Fin}$ , then $\mathcal {CR}(X,\hat {\mathbb {a}})\cong \mathrm {Fin}$ (by (i) and (ii)).

• • If $S\in \mathrm {Fin}$ , $T\notin \mathrm {Fin}$ , but $\omega \setminus T\notin \mathrm {Fin}$ , then $\mathcal {CR}(X,\hat {\mathbb {a}})\cong \mathrm {Fin}\oplus \mathcal {P}(\omega )$ (by (i), (ii), and (iv)).

• • If $S\notin \mathrm {Fin}$ , then $\mathcal {CR}(X,\hat {\mathbb {a}})\cong \{\emptyset \}\otimes \mathrm {Fin}$ (by (ii), (iii), and Lemma 1.2).

Hence, if $\mathcal {CR}(X,\hat {\mathbb {a}})\cong \mathcal {I}$ and $\mathcal {I}$ is nontrivial, then $\mathcal {I}$ is isomorphic to $\mathrm {Fin}$ , $\mathrm {Fin}\oplus \mathcal {P}(\omega ),$ or $\{\emptyset \}\otimes \mathrm {Fin}$ .