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Theorem: $\mathfrak{b} \leq \kappa \leq non(M)$.

Theorem: $\mathfrak{b} \leq \kappa \leq \operatorname{non}(M)$.

To each $b \in \mathcal B$, we associate a (recursively defined) function $f_b: \mathbb N \rightarrow \mathbb N$ as follows: $$f_b(n) = \max\{b(0),\dots,b(n),b^{-1}(0),\dots,b^{-1}(n)\}+1.$$ Because $|\mathcal B| = \lambda < \mathfrak{b}$, there is some $h: \mathbb N \rightarrow \mathbb N$ such that, for every $b \in \mathcal B$, $h(n) > f_b(n)$ for all but finitely many $n$.

To each $b \in \mathcal B$, we associate a (recursively defined) function $f_b: \mathbb N \rightarrow \mathbb N$ as follows: $$f_b^0(n) = \max\{b(m) : m \leq f_b(n-1)\}+1,$$ $$f_b(n) = \max\{b^{-1}(m) : m \leq f_b^0(n)\}+1.$$ Because $|\mathcal B| = \lambda < \mathfrak{b}$, there is some $h: \mathbb N \rightarrow \mathbb N$ such that, for every $b \in \mathcal B$, $h(n) > f_b(n)$ for all but finitely many $n$.

Let $f \in \mathcal B$. Since no member of $\mathcal B$ jumbles $A$, the nonzero terms of $\langle b_n \rangle$, $\langle a_n \rangle$, and $\langle f(a_n) \rangle$ are all in the same order, except perhaps for finitely many terms. Thus $$\sum_{n \in \mathbb N}b_n = \sum_{n \in \mathbb N}a_n = \sum_{n \in \mathbb N}f(a_n).$$ Therefore no $f \in \mathcal B$ can rearrange $\langle a_n \rangle$ sufficiently to change the value of its sum (even though such a rearrangement is clear possible, because $\langle b_n \rangle$ is only conditionally convergent). Since $\mathcal B$ was an arbitrary family of bijections with $|\mathcal B| < \mathfrak{j}$, this finishes the proof. QED

For the record, I can see no real reason to think that $\mathfrak{j} = \kappa$. In fact, I'll leave it as an exercise to show that the family $\mathcal B$ defined in Part II of my proof has the following property: for every $f \in \mathcal B$, if $\sum_{n \in \mathbb N}a_n$ converges, then $\sum_{n \in \mathbb N}f(a_n)$ converges also, and to the same value. In other words, we have a jumbling family that doesn't change the value of any sums!

Let $f \in \mathcal B$. Since no member of $\mathcal B$ jumbles $A$, the nonzero terms of $\langle b_n \rangle$, $\langle a_n \rangle$, and $\langle f(a_n) \rangle$ are all in the same order, except perhaps for finitely many terms. Thus $$\sum_{n \in \mathbb N}b_n = \sum_{n \in \mathbb N}a_n = \sum_{n \in \mathbb N}a_{f(n)}.$$ Therefore no $f \in \mathcal B$ can rearrange $\langle a_n \rangle$ sufficiently to change the value of its sum (even though such a rearrangement is clear possible, because $\langle b_n \rangle$ is only conditionally convergent). Since $\mathcal B$ was an arbitrary family of bijections with $|\mathcal B| < \mathfrak{j}$, this finishes the proof. QED

For the record, I can see no real reason to think that $\mathfrak{j} = \kappa$. In fact, I'll leave it as an exercise to show that the family $\mathcal B$ defined in Part II of my proof has the following property: for every $f \in \mathcal B$, if $\sum_{n \in \mathbb N}a_n$ converges, then $\sum_{n \in \mathbb N}a_{f(n)}$ converges also, and to the same value. In other words, we have a jumbling family that doesn't change the value of any sums!