TITLE: On Hamkins' answer to a problem by Michael Hardy
QUESTION [14 upvotes]: Based on a post by Michael Hardy and Hamkins' answer to it Andreas Blass, Will Brian, Joel Hamkins, Michael Hardy and Paul Larson introduced a new cardinal characteristic of the continuum $\mathfrak{rr}$ and as the main result they proved $\mathfrak{b}\leq\mathfrak{rr}\leq\mathop{\bf non}(\mathcal{M})$.(See the post by Hamkins on his homepage.)
The following question is motivated by a discussion I had with my friends Shahram Biglari and Mohammad Golshani.
Let $\mathfrak{i}\mathfrak{i}$ be the smallest cardinal such that there exists a family of injective functions $i:\omega\longrightarrow\omega$ of size $\mathfrak{i}\mathfrak{i}$ that guarantees the absolute convergence of a convergent real series $\sum\limits_{n=1}^{\infty} a_{n}$. Thus it's clear that $\mathfrak{r}\mathfrak{r}\geq\mathfrak{i}\mathfrak{i}$.

Question) What happens if one considers the set of all injective functions $i:
\omega\longrightarrow\omega$ instead of permutations. More precisely is it possible to have $\mathfrak{i}\mathfrak{i}<\mathfrak{r}\mathfrak{r}<\mathfrak{c}$?

Is it possible also to have $\mathfrak{i}\mathfrak{i}=\aleph_0$? Note that if $CH$ holds then $\mathfrak{r}\mathfrak{r}=\mathfrak{c}$.

REPLY [6 votes]: Let me show something about the subseries number, which I
suggested in the comments, and which I find very natural to
consider, in light of the following:
Fact. A series $\sum_n a_n$ is absolutely convergent just in
case every subseries $\sum_{n\in A} a_n$ is convergent, for every
$A\subset\mathbb{N}$.
So let us define the subseries number, denoted
$\newcommand\ss{\bf ß}\ss$ (German sharp s), to be the size of the
smallest family $\mathcal{A}$ of sets of natural numbers, such
that if a series $\sum_n a_n$ has the property that $\sum_{n\in
A}a_n$ converges for all $A\in\mathcal{A}$, then $\sum_n a_n$ is
absolutely convergent.
Clearly, $\frak{i\!i}\leq\ss$, since considering subseries amounts
to insisting on increasing injections in the context of the OP.
The splitting number $\newcommand\s{\frak{s}}\s$, in contrast,
is the smallest size of a splitting family $S$, which is a
family $\mathcal{S}$ of infinite sets of natural numbers, such
that if $B\subset\mathbb{N}$ is infinite, then there is some
$A\in\mathcal{S}$ such that $B\cap A$ and $B-A$ are both infinite,
and in this case we say that $A$ splits $B$ and that
$\mathcal{S}$ is a splitting family.
Meanwhile, ${\bf\text{non}}(\mathcal{M})$ is the size of the
smallest non-meager set in the space (it will be convenient to
consider) of all subsets of $\mathbb{N}$.
Theorem. $\s\leq\ss\leq{\bf\text{non}}(\mathcal{M})$.
Proof. For the first inequality, I shall show that every
subseries family $\mathcal{A}$, witnessing the defining property
of $\ss$, is also a splitting family. To see this (just as Will
had argued), consider any infinite set $B\subset\mathbb{N}$. Let
$\sum_n a_n$ be any conditionally convergent series whose nonzero
terms $a_n$ arise solely for $n\in B$. For example, $\sum_n a_n$
could be the alternating harmonic series, padded with $0$'s so
that the nonzero terms arise only on indices in $B$. Thus, $\sum_n
a_n$ converges, but only conditionally. So there must be some
$A\in\mathcal{A}$ such that $\sum_{n\in A} a_n$ diverges. Such a
set $A$ must have $B\cap A$ infinite, in order to have infinitely
many non-zero terms, and it must have $B-A$ infinite, in order to
differ infinitely from $\sum_n a_n$. So $A$ splits $B$, and
therefore $\mathcal{A}$ is a splitting family. So
$\s\leq|\mathcal{A}|$, as desired.
For the second inequality, I shall show that every nonmeager
family of subsets of $\mathbb{N}$ has the subseries property.
Suppose that $\mathcal{A}$ is such a nonmeager family, and we have
a conditionally convergent series $\sum_n a_n$. Since the positive
and negative terms each sum to $\infty$, it follows that any
finitely many terms of this series can be extended in such a way
to cause an additional huge oscillation; from this, it follows
that every conditionally convergent series has a comeager set of
subsets $A\subset\mathbb{N}$ for which $\sum_{n\in A}a_n$ does not
converge. Since $\mathcal{A}$ is non-meager, it must intersect
this set, and so there must be some $A\in\mathcal{A}$ for which
$\sum_{n\in A}a_n$ does not converge. So $\mathcal{A}$ has the
subseries property, as desired. QED