% LaTeX2e  irr.tex
% Irredundant Generators
% Jonathan Cancino, Osvaldo Guzm\'an, Arnold W. Miller
% eprint June 2013  last revised May 2014

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\def\al{\alpha}
\def\base{{\mathcal B}}
\def\bb{{\mathfrak b}}
\def\be{\beta}
\def\cc{{\mathfrak c}}
\def\dd{{\mathfrak d}}
\def\de{\delta}
\def\es{\emptyset}
\def\forces{{\;\Vdash}}
\def\force{\forces}
\def\ga{\gamma}
\def\gg{\gamma}
\def\ii{{\mathcal I}}
\def\implies{{\to}}
\def\ka{\kappa}
\def\lam{\lambda}
\def\la{\langle}
\def\monk{{\mathfrak{s}_{mm}}}
\def\name#1{\stackrel{\circ}{#1}}
\def\om{\omega}
\def\pair(#1,#2){\la #1, #2 \ra}
\def\ph{\Phi}
\def\poset{{\mathbb P}}
\def\pow(#1){{\mathcal P}(#1)}
\def\pp{\mathfrak p}
\def\proof{\par\noindent Proof\par\noindent}
\def\pr{\prime}
\def\qed{\par\noindent QED\par\bigskip}
\def\qq{{\mathbb Q}}
\def\range{{\mbox{ range}}}
\def\rank{{\rm rank}}
\def\ra{\rangle}
\def\res{\upharpoonright}
\def\rmand{\mbox{ and }}
\def\rmforall{\mbox{ for all }}
\def\rmiff{\mbox{ iff }}
\def\rmif{\mbox{ if }}
\def\rmor{\mbox{ or }}
\def\rr{{\mathbb R}}
\def\si{\sigma}
\def\sm{\setminus}
\def\st{\;:\;}
\def\supp{{\mbox{supp}}}
\def\su{\subseteq}
\def\uu{{\mathcal U}}
\def\uu{{\mathfrak u}}
\def\vv{{\mathcal V}}
\def\zz{{\mathbb Z}}

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\def\jcaddress{\begin{flushleft}
Jonathan Cancino \\
jcancino@matmor.unam.mx\\
Centro de Ciencias Matem\'aticas, UNAM, \\
A.P. 61-3, Xangari,\\ 
Morelia, Michoac\'an, M\'exico.
\end{flushleft}}


\def\ogaddress{\begin{flushleft}
Osvaldo Guzm\'an \\
oguzman@matmor.unam.mx\\
Centro de Ciencias Matem\'aticas, UNAM, \\
A.P. 61-3, Xangari,\\ 
Morelia, Michoac\'an, M\'exico.
\end{flushleft}}


\def\awmaddress{\begin{flushleft}
Arnold W. Miller \\
miller@math.wisc.edu \\
http://www.math.wisc.edu/$\sim$miller\\
University of Wisconsin-Madison \\
Department of Mathematics, Van Vleck Hall \\
480 Lincoln Drive \\
Madison, Wisconsin 53706-1388 \\
\end{flushleft}}

\begin{document}

\title{Irredundant Generators}
\author{Jonathan Cancino, Osvaldo Guzm\'an, Arnold W. Miller}
\maketitle
\begin{abstract}
We say that $\mathcal{I}$ is an irredundant family if no element of
${\mathcal{I}}$ is a subset mod finite of a union of finitely many other
elements of ${\mathcal{I}}.$ We will show that the minimum size of a maximal
irredundant family is consistently bigger than both $\mathfrak{d}$ and
$\mathfrak{u},$ this answers a question of Donald Monk.
\end{abstract}


Let ${\mathfrak{s}_{mm}}$ be the minimal cardinality of a maximal irredundant
ideal generator, i.e., an infinite family
${\mathcal{B}}\su[\om]^{\om}$ such that no element of
${\mathcal{B}}$ is a subset mod
finite of a union of finitely many other elements of ${\mathcal{B}}$ and
${\mathcal{B}}$ is maximal with respect to this property, i.e., for any
$X\in\lbrack\om]^{\om}\sm{\mathcal{B}}$, it cannot be added to
${\mathcal{B}}$ and still be irredundant. This means that there is
$F\in\lbrack{\mathcal{B}}]^{\om}$ such that either $X\su^{\ast}\bigcup F$
or there is $B\in{\mathcal{B}}\sm F$ with $B\su^{\ast}X\cup\bigcup F$.
This concept also occurs in Monk \cite{monk2} where the
terminology ``ideal independent'' is used
instead of ``irredundant generator''. We will
compare $\mathfrak{s}_{mm}$ with the ultrafilter number and the dominating
number, for the definition and basic properties of the usual cardinal
invariants see Blass \cite{HandbookBlass}.

\bigskip 
In May 2013 at a conference at the Ben-Gurion University of the Negev
Donald Monk \cite{monk} asked if $\mathfrak{s}_{mm}$ was equal to
$\mathfrak{u}$ (this question was communicated to Arnold Miller by Juris
Steprans). The next proposition answers this question negatively.
In the rational perfect set model $\dd=\om_2$ and $\uu=\om_1$,
see Miller \cite{rat} and Blass-Shelah \cite{blass-shelah}.

\begin{prop}
$\max\left\{ \mathfrak{d,r}\right\} \leq \mathfrak{s}_{mm}$.
\end{prop}

\proof Given a maximal irredundant family $\mathcal{I}$, it is easy to see that
the following family of sets is a reaping family:
\begin{center}
$\{A\sm\bigcup F:F\in[\mathcal{I}]^{<\om}
\land A\in \mathcal{I}\sm F\}$
\end{center}

It remains to prove that  $\mathfrak{d\leq s}_{mm}.$
Assume otherwise that $\mathfrak{s}_{mm}<\mathfrak{d}$, and let $\mathcal{A}$
be a witness for this. Note that $\om =^{\ast }\bigcup \mathcal{A}$, so
we can assume that indeed the equality holds. Let
$\{A_{n}:n\in\om\}\su \mathcal{A}$ such that its union is $\om$.
Define $C_{0}=A_{0}$ and $C_{n+1}=A_{n+1}\sm \bigcup_{i\leq n}A_{i}$.
For each $F\in \lbrack \mathcal{A}]^{<\om}$ and
$B\in\mathcal{A}\sm (F\cup\{A_i:i<\om\})$, define a function as follows:
\[
\varphi _{F,B}(n)=\min \{k\in \om :(\exists j\geq n)(C_{j}\cap B\cap
k\sm \bigcup F\neq \emptyset )\}
\]
Since the family $\mathcal{A}$ is irredundant, the functions $\varphi_{F,B}$
are always well defined.
Let $h_0$ be an increasing function not dominated by
$$\{\varphi_{F,B}\;:\;F\in[\mathcal{A}]^{<\om},\; B\in\mathcal{A}\sm
(F\cup\{A_i:i<\om\})\}.$$
Define $D_n=C_n\sm h_0(n)$. Now for each
$F\in[\mathcal{A}]^{<\om}$,
whenever it is possible, define a function as follows:
\[
\tilde{\varphi}_F(n)=\min\{k\in\om:(\exists j\ge n)(D_j\cap k\sm
\bigcup F\neq\emptyset)\}
\]
This is  defined for $n$, otherwise
$$\bigcup_{j\geq n}D_j=\bigcup_{j\geq n}(C_j\sm h_0(j))\su \bigcup F$$
But then for some $j\geq n$ such that $A_j\notin F$ we would
have 
$$A_j\su^* \bigcup_{i<j}A_i \cup \bigcup F$$
which contradicts irreducibility.
%If this is not defined for $n$, then
%note that
%$$\bigcup F\su\bigcup_{i < n} C_i\cup\bigcup_{j\geq n}(C_j\cap h_0(j)).$$


Let $h_1>h_0$
be an increasing function not dominated by any totally defined
$\tilde{\varphi}_F$ for $F\in[\mathcal{A}]^{<\om}$ and such
that $C_n\cap[h_0(n),h_1(n))$ is nonempty for all $n$.

Let
\[
Y=\bigcup_{n\in\om} (C_n\cap[h_0(n),h_1(n)))=\bigcup_{n\in\om} D_n\cap
h_1(n)
\]

\noindent Let's see that $\mathcal{A}\cup \{Y\}$ is an irredundant family.

\bigskip\noindent
\textbf{Claim 1.} For all $F\in \lbrack \mathcal{A}]^{<\om }$,
$Y\nsubseteq ^{\ast }\bigcup F$.

If the function $\tilde{\varphi}_{F}$ is not defined, then
$Y\cap \bigcup F$ is finite. Otherwise, by the definition of the
function $\tilde{\varphi}_{F}$, if $\tilde{\varphi}_{F}(n)\leq h_{1}(n)$,
then for some $j\geq n$ we have $D_{j}\cap \tilde{\varphi}_{F}(n)\sm
\bigcup F\neq \emptyset $, which implies
$$\emptyset \neq D_{j}\cap
h_{1}(n)\sm \bigcup F\su D_{j}\cap h_{1}(j)\sm \bigcup
F\su Y.$$
Since this happens for infinitely $j$ and the family $
\{D_{j}:j\in \om \}$ is disjoint, we are done.

\bigskip\noindent
\textbf{Claim 2.} For any $F\in \lbrack \mathcal{A}]^{<\om }\sm
\{\emptyset \}$ and $B\in \mathcal{A}\sm F$, we have $B\nsubseteq
^{\ast }Y\cup \bigcup F$.

If $B=A_n$ for some $n$ this is clear. Otherwise,
by the definition of $\varphi _{F,B}$ and the
choice of $h_{0}$, we have that if $\varphi _{F,B}(n)\leq h_{0}(n)$, then
for some $j\geq n$,
$$\emptyset \;\;\neq\;\; C_{j}\cap B\cap \varphi
_{F,B}(n)\sm \bigcup F\;\;\su\;\; C_{j}\cap B\cap h_{0}(j)\sm
\bigcup F.$$
If $m\in C_{j}\cap B\cap h_{0}(j)\sm \bigcup F$, then $
m\notin Y\cup \bigcup F$. Since this happens infinitely many times, we are
done.
\qed


We will now show some results related to irredundant families.

\begin{prop}
If ${\mathcal{I}}$ is an ideal generated by a strictly ascending mod finite
sequence $A_{\al}\su\om$ for $\alpha<\om_{1}$, then ${\mathcal{I}
}$ is not generated by an irredundant family.
\end{prop}

\proof
So $\ii=\{B\;:\; \exists\al<\om_1 \;\; B\su^*A_\al\}$.
Suppose $\base\su\ii$ generates
$\ii$.
For each $\al$ choose $F_\al\su\base$ finite
with $A_\al\su^* \cup F_\al$.  Suppose $F_\al$ for $\al\in\gg$ is a
delta system for $\gg$ uncountable.  We may find $\al<\be$ in $\gg$ with
$\cup F_\al\su^* A_{\be}$.
Since $A_{\be}\su^*\cup F_\be$, for any $B\in (F_\al\sm F_\be)$,
$B\su^*\cup F_\be$ but this implies that $\base$ is redundant.
\qed


Although many ideals can be generated by an irredundant family, this is not the
case for the prime ideals.

\begin{prop}
A non-principal prime ideal $\ii$ on $\om$ cannot be irredundantly generated.
\end{prop}

\proof
Suppose $\base$ is an irredundant generator of $\ii$.
Let $\{A_n\;:\;n<\om\}\su\base$ be distinct.
By adding at most one
thing to each $A_n$ we may suppose $\cup_{n<\om}A_n$ is $\om$.
Let
$$B=\bigcup_n (A_{2n}\sm\cup_{i<2n}A_i) \rmand
C=\bigcup_n (A_{2n+1}\sm\cup_{i<2n+1}A_i) $$
and note these are complementary sets.  If $B\in\ii$ then
for some finite $F\su\base$ we have $B\su^*\cup F$.  But this
means $A_{2n}\su^*\bigcup F\cup \bigcup_{i<2n}A_i$
which contradicts irredundancy for $n$ large enough so
that $A_{2n}\notin F$.
Similarly if $C\in\ii$.
\qed

We will now show that $\mathfrak{s}_{mm}$ can be smaller than the
continuum, in fact this holds in the
side by side countable support Sacks model.

\begin{prop}
\label{sacks} In the side by side countable support Sacks model there is a
maximal irredundant generator of size $\om_{1}$. In this model the continuum
can be made arbitrarily large but $\monk=\uu=\dd=\om_{1}$.
\end{prop}

\proof
We are forcing with the countable support product of $\ka$-many
Sacks posets for any $\ka$ over a model of CH.

To get an irredundant generator which remains
maximal after forcing, we need only work with the
$\om$-product of Sacks forcing $\poset={\mathbb S}^\om$.

By Laver's combinatorial generalization of the Halpern-Lauchli
Theorem \cite{laver} for any $\poset$-name $\tau$ for a subset of $\om$
and $p\in\poset$ we may obtain $q\leq p$ and $Z\in [\om]^\om$
such that either
$$q\forces ``Z\su\tau \rmor q\forces Z\cap\tau=\es.\mbox{''}$$

As Laver points out one may use this to build a descending mod finite
sequence $Z_\al\in [\om]^\om$ for $\al<\om_1$ in the ground model
with the property that they generate a Ramsey ultrafilter in the extension.


\begin{lemma}
Given $(Y_{n}\in[\om]^{\om}\;:\;n<\om)$ pairwise disjoint in the ground model,
$\tau$ a $\poset$-name for a subset of $\om$, and $p\in\poset$, there are
$W_{n}\in[Y_{n}]^{\om}$ and $q\leq p$ such that
\[
q\forces ``\forall n \;\;(W_{n}\su\tau\rmor  W_{n}\cap\tau=\es).''.
\]
\end{lemma}

\proof
Let $(f_n:\om\to Y_n)_n$ be a sequence of bijections in the ground model
and define $\tau_n=f_n^{-1}(\tau)$.
Let $G$ be generic with $p\in G$.
By properness of $\poset$ in the generic extension for some $\al<\om_1$
$$\forall n \;\; ( Z_\al\su^*\tau_n^G\rmor Z_\al \cap\tau_n^G=^*\es)$$
Using the well-known fact that this forcing is $\om^\om$-bounding
(in Shelah's terminology) or weakly distributive (in Namba's terminology)
we may find $f\in\om^\om$ in the ground model
and take  $W_n=f_n(Z_\al)\sm f(n)$ so that
$$W_n\su\tau^G \rmor  W_n\cap\tau^G=\es \mbox{ for all }n.$$   Finally we
take $q\leq p$ to do the required forcing.
\qed

Working in the ground model
let $(p_\al,\tau_\al)$ for $\al<\om_1$
list with uncountable repetitions all pairs
$(p,\tau)$ for $p\in\poset$ and $\tau$ a canonical
$\poset$-name for a subset of $\om$.
Construct an increasing
family of irredundant countable families $\ii_\al$ for $\al<\om_1$.

At stage $\al$ given $\ii_\al=\{A_n:n<\om\}$.

Put $B_n=(A_n\sm\cup_{i<n}A_i)$ and
construct $Y_n\in [B_n]^\om$
so that $Y_n$ are infinite pairwise disjoint, $B_n\sm Y_n$ is infinite, and
$Y_n\cap A_k$ is finite for $k\neq n$.

By applying the Lemma we may obtain $(W_n\in [Y_n]^\om:n<\om)$
and $q\leq p_\al$ such that
$$q\forces ``\forall n\;\; (W_n\su\tau_\al
\rmor  W_n\cap\tau_\al=\es).\mbox{''}.$$

Take $W=\cup_{n<\om}(B_n\sm W_n)$ and
let $\ii_{\al+1}=\ii_\al\cup\{W\}$.
It is not hard to see that this family is indeed irredundant.
We claim that $\tau$ is forced by $q$ to never be added to
our irredundant family.  Let $G$ be generic with $q\in G$.

If for some $n$, $W_n\su\tau_\al^G$,
then $W\cup \tau_\al^G$ covers $B_n$
and hence $\tau^G$, $W$, $A_i$ for $i<n$ cover $A_n$.

If for all $n$ $W_n\cap\tau_\al^G=\es$, then
$\tau^G\su W$ since the $B_n$ partition $\om$,
and so the pair is redundant.

Hence $\ii=\bigcup_{\al<\om_1}\ii_\al$ will be a maximal irredundant
family in the ground model which remains a maximal irredundant
family in the generic extension.
\qed


Recall the following definition by Vojt\'{a}\v{s} \cite{Vojtas}

\begin{definition}
We say $\left( A,B,\rightarrow\right) $ is an \emph{invariant}
if,
\begin{enumerate}
 \item $\rightarrow$ $\su A\times B.$
 \item For every $a\in A$ there is $b\in B$ such that $a\rightarrow b$.
 \item There is no $b\in B$ such that $a\rightarrow b$ for all $a\in A.$
\end{enumerate}
\end{definition}


\qquad\

We say that $D\su B$ is dominating if for every $a\in A$ there is a
$d\in D$ such that $a\rightarrow d,$ so 3) means that $B$ is dominating
and 4) that no singleton is dominating. Given an invariant $\left(
A,B,\rightarrow\right)  $ we define it's \emph{evaluation} by
$\left\langle A,B,\rightarrow\right\rangle =min\left\{  \left\vert
D\right\vert : D\su B\text{ and }D\text{ is dominating}\right\}  .$
An invariant $\left(  A,B,\rightarrow\right)  $ is called Borel if $A,B$
and $\rightarrow$ are Borel subsets of a polish space. Most of the usual
(but not all) invariants are actually Borel invariants. In
\cite{ParametrizedDiamonds} for any Borel invariant $\left(
A,B,\rightarrow\right)  ,$ a guessing principle $\Diamond\left(
A,B,\rightarrow\right)  $ is defined and it is proved that it implies
$\left\langle A,B,\rightarrow\right\rangle \leq\om_{1}$ and it holds in
most of the natural models where this inequality holds. For our applications
in this note, we need to work in a slightly more general framework than the
one in \cite{ParametrizedDiamonds}.

\begin{definition}
We say an invariant $\left( A,B,\rightarrow\right)$ is an
$L\left(\mathbb{R}\right) $-\emph{invariant }if $A,B$ and $\rightarrow$
are subsets of Polish spaces and all three of them belong to
$L\left( \mathbb{R}\right).$
\end{definition}

Following \cite{ParametrizedDiamonds} we define the
following guessing principle for any
$L(\mathbb{R})$-invariant $(A,B,\to)$.

\begin{define}
$\Diamond_{L\left(\mathbb{R}\right)}\left(  A,B,\rightarrow\right)$
\par For every $C:2^{<\om_{1}}\rightarrow A$ such that
$C\upharpoonright2^{\alpha}\in L\left(  \mathbb{R}\right)  $ for all
$\alpha<\om_{1}$ there is a $g:\om_{1}\rightarrow B$ such that for
every $R\in$ $2^{\om_{1}}$ the set
$\left\{  \alpha\mid C\left(
R\upharpoonright\alpha\right)  \rightarrow g\left(  \alpha\right)
\right\}$
is stationary.
\end{define}

\qquad

Exactly as in the Borel case, $\Diamond_{L\left(  \mathbb{R}\right)  }\left(
A,B,\rightarrow\right)  $ implies $\left\langle A,B,\rightarrow
\right\rangle \leq\om_{1}.$ Given two $L\left(  \mathbb{R}\right)
$-invariants $\mathbb{A}=\left(  A_{-},A_{+},_{\mathbb{A}}\rightarrow
\right)  $ and $\mathbb{B=}\left(  B_{-},B_{+},_{\mathbb{B}}\rightarrow
\right)  $ we define the \emph{sequential composition }$\mathbb{A}
;\mathbb{B}=(  A_{-}\times Bor(  B_{-}^{A_{+}})  ,A_{+}\times
B_{+},\rightarrow)  $ where $Bor(  B_{-}^{A_{+}})  $
denotes the set of codes of all Borel functions from $A_{+}$ to $B_{-}$ and
$\left(  a_{-},f\right)  \rightarrow\left(  a_{+},b_{+}\right)  $ if
$a_{-\mathbb{A}}\rightarrow a_{+}$ and $f\left(  a_{-}\right)
_{\mathbb{B}}\rightarrow b_{+}.$ It is easy to see that $\mathbb{A}
;\mathbb{B}$ is an $L\left(  \mathbb{R}\right)  $-invariant and in
\cite{HandbookBlass} it is proved that $\left\langle \mathbb{A};\mathbb{B}
\right\rangle =max\left\{  \left\langle \mathbb{A}\right\rangle ,\left\langle
\mathbb{B}\right\rangle \right\}.$

As usual we will write $\mathfrak{d}$ instead of $\left(
\om^{\om},\om^{\om},\leq^{\ast}\right)  $ and $\mathfrak{r}
_{\sigma}$ instead of the invariant $\left(  \left(  \left[  \om\right]  ^{\om
}\right)  ^{\om},\left[  \om\right]  ^{\om},\text{is }\sigma
\text{-reaped}\right).$

\begin{proposition}
$\Diamond_{L\left( \mathbb{R}\right) }\left( \mathfrak{r}_{\sigma };
\mathfrak{d}\right) $ implies $\mathfrak{s}_{mm}=\om_{1}.$
\end{proposition}

\begin{proof}
We need to define a function $F$ into $[\om ]^{\om }\times (\om
^{\om })^{[\om ]^{\om }}$ such that for all $\alpha \in \om _{1}$
, $F\upharpoonright \alpha $ is in $L(\mathbb{R})$. For each $\alpha <\om
_{1}$, let $e_{\alpha }:\om \rightarrow \alpha $ be an enumeration of $
\alpha $ in $L(\mathbb{R})$. By a suitable coding, we can assume that the
domain of $F$ is the set
\[
\bigcup_{\alpha \in \om _{1}}[\om ]^{\om }\times ([\om ]^{\om
})^{\alpha }
\]
Given $(A,\vec{\mathcal{I}})\in [\om]^\om\times(\om^\om)^\alpha$
proceed as follows. If $\vec{\mathcal{I}}$ is not an irredundant family,
define $F(A,\vec{\mathcal{I}})=(\om,e)$, where $e(X)$ for $X\in[\om]
^\om$ is the enumeration of $X$. Otherwise, define $B^{\vec{\mathcal{I}}
}_n=I_{e_\alpha(n)}\sm\bigcup_{i<n}I_{e_{\alpha}(i)}$. For each $n$,
let $Z^{\vec{\mathcal{I}}}_n\su B^{\mathcal{I}}_n$ be an infinite
subset such that for all $\beta\neq e_\alpha(n)$, $Z^{\mathcal{I}}_n\cap
I_\beta$ is finite \footnote{$Z^{\vec{\mathcal{I}}}_n\su B^{\mathcal{I}
}_n$ should be found in a recursive way and should depend only on $\vec{
\mathcal{I}}$}, and let $\varphi_{\vec{\mathcal{I}},n}$ be a recursive
enumeration of $Z^{\vec{\mathcal{I}}}_n$. Then define $A_n=\varphi_{\vec{
\mathcal{I}},n}^{-1}[Z^{\vec{\mathcal{I}}}_n\cap A]$. Now define a function $
f_{A,\vec{\mathcal{I}}}:[\om]^\om\to\om^\om$ as follows: if $X\in
[\om]^\om$ reaps $A_n$ for all $n$, then define
\[
f_{A,\vec{\mathcal{I}}}(X)(n)=\min\{k\in\om: X\sm k\su
A_n\lor (X\sm k)\cap A_n=\emptyset\}
\]
Otherwise define $f_{A,\vec{\mathcal{I}}}(X)$ to be the identity function.
Finally, the value of $F$ in $(A,\vec{\mathcal{I}})$ is given by
$F(A,\vec{\mathcal{I}})=(\langle A_n:n\in\om\rangle, f_{A,\vec{\mathcal{I}}})$.
Let $g:\om_1\to[\om]^\om$ be a
$\Diamond_{L(\mathbb{R})}(\mathfrak{r}_\sigma;\mathfrak{d})$-guessing sequence
for $F$. We can assume that for all $
\alpha$ the set $A_\alpha$ in $g(\alpha)=(A_\alpha,h_\alpha)$ is coinfinite.
Recursively define an irredundant family as follows:
\begin{enumerate}
\item[1)] Start with a partition of $\om$ into infinitely many infinite sets
$\vec{\mathcal{I}}_\om= \langle I_n:n\in\om\rangle$.
\item[2)] Suppose we have defined $\vec{\mathcal{I}}_\alpha=\langle
I_\beta:\beta<\alpha\rangle$. Now define $I_\alpha$ as follows:
\[
I_\alpha=\bigcup_{n\in\om} B^{\vec{\mathcal{I}}_\alpha}_{n}\sm
\varphi^{\vec{\mathcal{I}}_\alpha}_n[A_\alpha\sm h_\alpha(n)]
\]
\end{enumerate}
Let $\vec{\mathcal{I}}_{\alpha+1}$ be the family
$\langle I_\beta:\beta\leq\alpha\rangle$. Finally, let
$\mathcal{I}=\langle I_\alpha:\alpha\in\om_1\rangle$
be the family obtained by the above
recursion. Let's see that $\mathcal{I}$ is a witness for $\mathfrak{s}_{mm}$.

\textbf{Claim 1.} $\mathcal{I}$ is an irredundant family. We proceed by
induction of $\alpha\in\om_1$. Clearly $\mathcal{I}_\om$ is
irredundant. Assume $\vec{\mathcal{I}}_\alpha$ is irredundant. Then $\vec{
\mathcal{I}}_{\alpha+1}$ is irredundant:
\begin{enumerate}
\item[a)] For all $H\in[\alpha]^{<\om}$, $I_\alpha\nsubseteq^*\bigcup H$.
Let $n\in\om$ be such that  $H$ is contained in
$\{e_\alpha(0),\ldots,e_\alpha(n)\}$,
so $\bigcup H\su\bigcup_{i\leq n} B^{{\mathcal{I}_\alpha}}_i$. By the
definition of  $I_\alpha$, $I_\alpha\sm\bigcup_{i\leq n} B^{\mathcal{I}
_\alpha}_i$ is infinite.
\item[b)] For all $\beta\in\alpha\sm H$, $I_\beta\nsubseteq^*
I_\alpha\cup\bigcup H$. Let $n$ be such that $\beta=e_\alpha(n)$. By the
choice of $Z^{\vec{\mathcal{I}}_\alpha}_n$,  we have that for any $
\ga\in\alpha\sm\{\beta\}$, $Z^{\vec{\mathcal{I}}_\alpha}_n\cap
I_\ga$ is finite, so in particular,  $Z^{\vec{\mathcal{I}}_\alpha}_n\cap
\bigcup H$ is finite. Also by the construction of $I_\alpha$,  $B^{\vec{
\mathcal{I}}_\alpha}_n\cap I_\alpha\cap\varphi^{\vec{\mathcal{I}}
_\alpha}_n[A_\alpha\sm h_\alpha(n)]$ is finite. This both  facts
together give $\varphi^{\vec{\mathcal{I}}_\alpha}_n[A_\alpha\sm
h_\alpha(n)]\sm I_\alpha\cup\bigcup H$ is infinite.  Since $\varphi^{
\vec{\mathcal{I}}_\alpha}_n[A_\alpha\sm h_\alpha(n)]\sm
I_\alpha\cup\bigcup H\su I_\beta\sm I_\alpha\cup\bigcup H$, we
are done.
\end{enumerate}

\textbf{Claim 2.} $\mathcal{I}$ is maximal. Pick any $X\in [\om]^\om$.
If $g$ guesses $(X,\langle I_\alpha:\alpha\in\om_1\rangle)$ in $\ga$,
then we have that $A_\ga$ $\sigma$-reaps $\langle X_n:n\in\om\rangle$
and $h_\ga$ almost dominates the function $l=f_{X,\mathcal{I}
_\ga}(A_\ga)$. There are two cases:
\begin{enumerate}
\item[i)] There are infinitely many $n\in\om$ such that $
A_\ga\su^* X_n$. Pick $n$ such that $l(n)\leq h_\ga(n)$. Then  $
A_\ga\sm h_\ga(n)\su X_n$, so $\varphi^{\vec{\mathcal{I}}
_\ga}_n[A_\ga\sm h_\ga(n)]\su X\cap B^{\vec{\mathcal{I}}
_\ga}_n$. Then by the definition  of $I_\ga$, $B^{\vec{\mathcal{I}}
_\ga}_n\su I_\ga\cup \varphi^{\vec{\mathcal{I}}
_\ga}_n[A_\ga\sm h_\ga(n)]\su I_\ga\cup X$, which
implies $I_{e_\ga(n)}\su X\cup
I_\ga\cup\bigcup_{i<n}I_{e_\ga(n)}$.
\item[ii)] For almost all $n\in \om $ $A_{\ga }\su ^{\ast
}\om \sm X_{n}$. Then for almost all $n$, $\varphi _{n}^{\vec{
\mathcal{I}}_{\ga }}[A_{\ga }\sm h_{\ga }(n)]\su Z_{n}^{
\vec{\mathcal{I}}_{\ga }}\sm X$, so for almost all $n$, $X\cap
Z_{n}^{\vec{\mathcal{I}}_{\ga }}\su I_{\ga }$, and for finitely
many $n$, $A_{\ga }\su ^{\ast }X_{n}$, so $\varphi _{n}^{\vec{
\mathcal{I}}_{\ga }}[A_{\ga }\sm h_{\ga }(n)]\su ^{\ast
}Z_{n}^{\vec{\mathcal{I}}}\cap X\su B_{n}\cap X$, which implies $
B_{n}\sm X\su ^{\ast }B_{n}\sm \varphi _{n}^{\vec{\mathcal{
I}}_{\ga }}[A_{\ga }\sm h_{\ga }(n)]\su I_{\ga }$.
Putting all this together we have that $X\su ^{\ast }I_{\ga }\cup
\bigcup_{i\leq k}B_{i}$, for some $k\in \om $.
\end{enumerate}\qed
\end{proof}


The following result was proved by Hiroaki Minami
\cite{DiamondPrinciplesinCichonsDiagram} for Borel invariants, however, the
proof for $L\left(\mathbb{R}\right)$-invariants is the same.

\begin{proposition}
Let $\left\langle \mathbb{P}_{\alpha},\mathbb{\dot{Q}}_{\alpha
}\mid\alpha\leq\om_{1}\right\rangle $ a finite support iteration of ccc
forcings and $\left( A,B,\rightarrow\right) $ be an $L\left( \mathbb{R}\right)$
-invariant with the following property:
For all $\alpha<\om_{1}$ there is  $b\in B\cap V\left[
G_{\alpha+1}\right] $ such that $a\rightarrow b$ for all $a\in A\cap V
\left[ G_{\alpha}\right]$. Then $\mathbb{P}_{\om_{1}}\Vdash
``\Diamond_{L\left( \mathbb{R}\right) }\left( A,B,\rightarrow\right)
\mbox{''}.$
\end{proposition}


With the previous proposition we can conclude the following:

\begin{corollary}
There is a finite support iteration of ccc forcings of length $\om_1$
such that
$\mathbb{P}_{\om_{1}}\Vdash``\mathfrak{s}_{mm}=\om_{1}$''.
\end{corollary}


\begin{proof}
Define
$\mathbb{P}_{\alpha},\mathbb{\dot{Q}}_{\alpha}$ for $\alpha<\om _{1}$
as follows. Let
$\mathbb{P}_{\alpha}\Vdash``\mathbb{\dot{Q}}
_{\alpha}=\mathbb{M}\left( \mathcal{\dot{U}}_{\alpha}\right) \ast \mathbb{
\dot{H}}$'' where $\mathcal{\dot{U}}_{\alpha}$ is the name
of any ultrafilter, $\mathbb{M}\left( \mathcal{\dot{U}}_{\alpha}\right) $ is
its Mathias forcing and $\mathbb{H}$ is Hechler forcing. Using the previous
proposition, it is easy to show that $\Diamond_{L\left( \mathbb{R}\right)
}\left( \mathfrak{r}_{\sigma};\mathfrak{d}\right) $ holds in $\mathbb{P}
_{\om_{1}}$ and then $\mathfrak{s}_{mm}$ is equal to $\om_{1}$ in the
extension.\qed
\end{proof}

In \cite{ParametrizedDiamonds} it is shown that for any Borel invariant
$\left(  A,B,\rightarrow\right)  ,$ most countable support iterations of
proper forcings that force $\left\langle A,B,\rightarrow\right\rangle
\leq\om_{1}$ will also force $\Diamond\left(  A,B,\rightarrow\right)
.$ This is also true for $L\left(  \mathbb{R}\right)  $-invariants and will be
proved in \cite{ParametrizedDiamondsII}.

\begin{proposition}
Let $\ \langle\mathbb{Q}_{\alpha}\mid\alpha\in\om_{2}\rangle$ be a
sequence of Borel proper partial orders where each $\mathbb{Q}_{\alpha}$ is
forcing equivalent to $\wp\left( 2\right) ^{+}\times\mathbb{Q}_{\alpha}$ and
let $\mathbb{P}_{\om_{2}}$ be the countable support iteration of this
sequence. If $\left( A,B,\rightarrow\right) $ is an $L\left( \mathbb{R}
\right) $-invariant and $\mathbb{P}_{\om_{2}}\mathbb{\Vdash}
``\left\langle A,B,\rightarrow\right\rangle
\leq\om_{1}''$ then $\mathbb{P}_{\om_{2}}\Vdash``
\Diamond_{L\left( \mathbb{R}\right) }\left( A,B,\rightarrow\right)
''$.
\end{proposition}

With the previous propositions we can conclude the following,

\begin{corollary}
There is a model where $\mathfrak{s}_{mm}<non\left( \mathcal{M}\right)$, hence
$\mathfrak{s}_{mm}<\mathfrak{i}.$
\end{corollary}


\proof Do a countable support iteration of fat tree forcings
(see \cite{idealizedForcing}, section 4.4.3)
over a model of $\mathsf{CH}$.
It can be proved that this forcings preserve Ramsey ultrafilters, so
in the final model we have $\mathfrak{r}_\sigma=\om_1$. These forcings are
also $\om^\om$-bounding,
so in the generic extension we have $\mathfrak{d}=\om_1$. This implies that
$\diamondsuit_{L(\mathbb{R})}(\mathfrak{r}_\sigma;\mathfrak{d})$ holds in the
generic extension, and then
$\mathfrak{s}_{mm}=\om_1$ in this model. Since this forcings add
eventually different reals then
$non(\mathcal{M})=\om_2$ holds.
Since $\mathfrak{i}\ge cof(\mathcal{M})$ (\cite{CombinatoricsRationals}),
we are done.
\qed


By the previous results it might be conjecture that $\mathfrak{s}
_{mm}=max\left\{  \mathfrak{d},\mathfrak{r}_{\sigma}\right\}  $ (note this
equality holds in all the Cohen, random, Hechler, Sacks, Laver, Mathias and
Miller models) but we will now show this is not the case. Let $\kappa$ be a
measurable cardinal and $\mathcal{U}$ be a $\kappa$ complete ultrafilter.
Given a partial order $\mathbb{P}$ we denote by $\mathbb{P}^{\kappa
}/\mathcal{U}$ its ultrapower and for every $f:\kappa\rightarrow
\mathbb{P}$ we denote by $\left[  f\right]  $ its equivalence class mod
$\mathcal{U}.$ For some\ background about forcing with ultrapowers see
 \cite{MADFamiliesandUltrafilters}. It
follows by the \L o\v{s} theorem that if $\mathbb{P}$ is ccc then
$\mathbb{P}^{\mu}/\mathcal{U}$ is also ccc and $\mathbb{P}$ regularly embeds
in its ultrapower. The next lemma shows that big irredundant families are
destroyed when taking ultrapowers and the proof is very similar to the lemma 4
of \cite{MADFamiliesandUltrafilters}, so we skip the proof.

\begin{lemma}
Let $\mathbb{P}$ be ccc and $\mathcal{\dot{A}}$ be a $\mathbb{P}$-name for an
irredundant family of size at least $\kappa.$ Then $\mathbb{P}^{\kappa
}/\mathcal{U}$ forces $\mathcal{\dot{A}}$ is not maximal.
\end{lemma}

\qquad\

With the previous lemma it is possible to show the following,

\begin{proposition}
Assume $V\models\mathsf{GCH}$, $\kappa$ is a measurable cardinal and $
\mathcal{U}$ is a $\kappa$-complete ultrafilter. Then there is $\mathbb{P}
_{\kappa^{+}\kappa^{++}}$ a ccc partial order such that $\mathbb{P}
_{\kappa^{+}\kappa^{++}}\Vdash``\kappa^{+}=\mathfrak{d=r}_{\sigma }=
\mathfrak{u<s}_{mm}=\mathfrak{c=}$ $\kappa^{++}\mbox{''}.$
\end{proposition}

\proof
The partial order $\mathbb{P}_{\kappa ^{+}\kappa ^{++}}$ is the forcing
constructed in \cite{MADFamiliesandUltrafilters} theorem 1 that forces $
\mathfrak{u<a.}$ We construct a matrix iteration $\left\langle \mathbb{P}
_{\alpha \beta }\mid \alpha \leq \kappa ^{+},\beta \leq \kappa
^{++}\right\rangle $ where $\left\langle \mathbb{P}_{\alpha 0}\mid \alpha
\leq \kappa ^{+}\right\rangle $ is a finite support iteration of Laver
forcings based on some ultrafilters, $\mathbb{P}_{\alpha \beta +1}=\mathbb{P}
_{\alpha \beta }^{\kappa }/\mathcal{U}$ and some ``
amalgamated limit''\ is taken at limit stages. We refer to
\cite{MADFamiliesandUltrafilters} for details. In that paper it is shown
that
$$\mathbb{P}_{\kappa ^{+}\kappa ^{++}}\Vdash ``\kappa ^{+}=\mathfrak{d=r}
_{\sigma }=\mathfrak{u<a}=\mathfrak{c=}\kappa ^{++}\mbox{''}$$
and following the proof of $\mathbb{P}_{\kappa ^{+}\kappa ^{++}}\Vdash ``
\mathfrak{a}=\kappa ^{++}\mbox{''} $ and using the previous lemma
it is possible to show that $\mathbb{P}_{\kappa ^{+}\kappa ^{++}}\Vdash ``
\mathfrak{s}_{mm}$ $\mathfrak{=}$ $\kappa ^{++}\mbox{''}$.
\qed


However we do not know the answer to the following questions,

\begin{question}
Is $\mathfrak{u\leq s}_{mm}?$
\end{question}

\begin{question}
Is $\mathfrak{s}_{mm}\leq\mathfrak{i?}$
\end{question}


We would like to remark that in \cite{Shelah} Shelah built a model of
$\mathfrak{i<u}$ so in that model one of the questions has a
negative answer, but we do not know which one.

\bigskip
\noindent\textbf{Acknowledgments. }
The first and the second authors would like to thank
Michael Hru\v{s}\'{a}k
for communicating Monk's question to them, as well as for
several important observations and hours of stimulating conversations.

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\jcaddress

\ogaddress

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