% LaTex2e
% A MAD Q-set A. Miller Dec 2002 last revised August 20, 2003
\documentclass[12pt]{article}
\usepackage{amssymb}
\textwidth=30cc
\baselineskip=16pt
\def\sq{\subseteq}
\def\om{\omega}
\def\aa{{\cal A}}
\def\bb{{\mathfrak b}}
\def\dd{{\mathfrak d}}
\def\cc{{\mathfrak c}}
\def\pp{{\mathfrak p}}
\def\reals{{\mathbb R}}
\def\dom{{\rm dom}}
\def\min{{\rm min}}
\def\swap{{\rm swap}}
\def\proof{\par\noindent Proof\par\noindent}
\def\poset{{\mathbb P}}
\def\qposet{{\widetilde{\mathbb Q}}}
\def\rposet{{\mathbb Q}}
\def\eposet{{\mathbb E}}
\def\forces{{| \kern -2pt \vdash}}
\def\force{\forces}
\def\res{\upharpoonright}
\def\qed{\par\noindent QED\par}
\def\rmor{\mbox{ or }}
\def\rmiff{\mbox{ iff }}
\def\rmand{\mbox{ and }}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{question}[theorem]{Question}
\newtheorem{conjecture}[theorem]{Conjecture}
\begin{document}
\begin{center}
{\large A MAD Q-set}
\end{center}
\begin{flushright}
Arnold W. Miller\footnote{
Thanks to the Fields Institute for Research in Mathematical Sciences
at the University of Toronto for their support during the time this paper
was written and to Juris Steprans who directed the special program in set
theory and analysis.
\par Mathematics Subject Classification 2000: 03E35
\par Fundamenta Mathematicae, 178(2003), 271-281.
}
\end{flushright}
\begin{quote}
\par\centerline{Abstract}\par
A MAD (maximal almost disjoint) family is an infinite
subset $\aa$ of the infinite subsets of
$\om=\{0,1,2,\ldots\}$ such that any two elements of
$\aa$ intersect in a finite set and every infinite subset
of $\om$ meets some element of $\aa$ in an infinite set.
A Q-set is an uncountable set of reals such that every
subset is a relative $G_\delta$ set.
It is shown that it is relatively consistent with
ZFC that there exists a MAD family which is also a Q-set in
the topology in inherits a subset of $P(\om)=2^{\om}$.
\end{quote}
In this paper we answer a question of Hru\v{s}\'{a}k by showing that it is
consistent that there exists a maximal almost disjoint family
$\aa\sq [\om]^\om$ which is also a Q-set. The reference
Hru\v{s}\'{a}k \cite{hrusak} contains some related problems.
A topological space is a Q-set iff
every subset is a $G_\delta$-set. His reason for asking this question
was because
in a certain argument involving a topological space $\Psi({\cal A})$ built
from a MAD family it would have been helpful to assume that a MAD family
cannot be a Q-set. Szeptycki \cite{szept} contains some results on van
Douwen's $\Psi$ and also on Q-sets.
Our construction is similar to that in Fleissner and Miller \cite{FM} where a
Q-set is obtained which is concentrated on the rationals. In Judah and Shelah
\cite{JS} it is shown consistent to have a Q-set while at the same time
$\bb=\dd=\om_1$. Their Q-set forcing has the Sack's property. Their forcing
is also used in Nowik and Weiss \cite{nowik} to construct a Q-set with certain
properties and also Gruenhage and Koszmider \cite{gruen} to construct a
topological space with certain properties. In our model as in \cite{FM} we have
that $\dd=\cc=\om_2$ and $\bb=\om_1$.
In Dow \cite{dow} and Brendle \cite{brendle} a type of Q-set forcing is used
which preserves towers (so $\pp=\om_1$)
which generalizes Hechler dominating real forcing, and
$\bb=\dd=\cc$.
\begin{theorem}
It is relative consistent with ZFC, that there exists a MAD family
$\aa\sq [\om]^\om$ which is also a Q-set.
\end{theorem}
\proof
We begin by forcing a generic MAD family and then we iterate
our Q-set forcing to make the generic MAD into a Q-set. The difficulty
is to ensure the family stays maximal.
\bigskip
Define: Let $\poset$ be the usual poset for forcing a MAD family:
\par\noindent $(p,q)\in \poset$ iff
\begin{enumerate}
\item $p:F\to 2^{N}$ for some finite $F\sq\om_1$ and $N<\om$
(write $F=\dom(p)$ and $N=N_p$)
\item $q$ a partial function from a subset of $[F]^2$ into $N$
\item if $q(\alpha,\beta)=n$, then for every $i$ with $n\leq ik.$$
\end{enumerate}
Condition (3) guarantees that for any $x\in 2^\om$ such
that $x\supseteq p(\alpha)$ that
$$\swap(x,s,k)\cap [t]=\emptyset.$$
\noindent The ordering is given by
$((p_1,q_1),r_1)\leq ((p_2,q_2),r_2)$ iff
$(p_1,q_1)\leq (p_2,q_2)$ and $r_1\supseteq r_2$.
\noindent Note that
$((p_1,q_1),r_1)$ and $((p_2,q_2),r_2)$ are compatible iff
there exists $p_3\leq p_1,p_2$ such that
$((p_3,q_1\cup q_2),r_1\cup r_2)$ is a condition.
\begin{center}
The $\om_2$ iteration.
\end{center}
Our iteration can be described as a suborder of the product
$$\poset\times \sum_{\alpha<\om_2}\eposet$$
where $\eposet$ is the set of all finite subsets of
$$\{(n,t):n<\om,t\in 2^{<\om}\}\cup
\{(n,(\alpha,s,k)): \alpha\in \om_1,
s\in 2^{<\om}, n, k <\om\}$$
and $\sum_{\alpha<\om_2}\eposet$ is the set of all
$r:\om_2\to \eposet$ such that $r(\delta)$ is trivial (i.e., the
empty set) for all but finitely many $\delta$.
By induction on $\beta\leq\om_2$ define
$$\poset_\beta\sq \poset\times\sum_{\alpha<\beta}\eposet$$
as follows:
\bigskip
Define. $\poset_0=\poset$, and suppose that we have
defined $\poset_\beta$ and we are also given a $\poset_\beta$
name $Y_\beta$
for a subset of $\om_1$, i.e.,
$$\forces_\beta Y_\beta\sq \om_1$$
Define. $((p,q),r)\in \poset_{\beta+1}$ iff
\begin{enumerate}
\item $((p,q),r\res\beta)\in \poset_{\beta}$,
\item $((p,q),r(\beta))\in \poset*\qposet$
\item $((p,q),r\res\beta)\forces_\beta \alpha\in Y_\beta$
\par whenever $(n,(\alpha,s,k))\in r(\beta)$ for some $n,s,k,\alpha$.
\end{enumerate}
For limit ordinals $\lambda\leq\om_2$ we define
$((p,q),r)\in \poset_\lambda$
iff for all $\beta<\lambda$ we have $((p,q),r\res\beta)\in \poset_\beta$
and for all but finitely many $\beta<\lambda$ we have that $r(\beta)$ is
the trivial condition (i.e. empty set).
Since the iteration of ccc forcing is ccc all of these forcings are ccc.
To see this directly we can argue as follows:
Standard arguments using $\Delta$ systems show that
$\poset_\beta$ has precalibre $\om_1$, i.e.,
any $\om_1$ sequence of conditions contain an $\om_1$
subsequence which is centered. Start with
$((p_\alpha,q_\alpha),r_\alpha)\in\poset_\beta$ for $\alpha<\om_1$.
We can find an uncountable $\Sigma\sq \om_1$
and finite sets $F$ and $H$ and $N<\om$ so that
\begin{enumerate}
\item $N_\alpha=N$ for all $\alpha\in\Sigma$,
\item $\dom(p_\alpha)\cap\dom(p_\beta)=F$ for $\alpha\not=\beta\in\Sigma$,
\item $\dom(r_\alpha)\cap\dom(r_\beta)=H$ for $\alpha\not=\beta\in\Sigma$,
\item $p_\alpha\res F$ are all the
same for $\alpha\in\Sigma$,
\item $q_\alpha\res [F]^2$ are all the
same for $\alpha\in\Sigma$, and
\item $r_\alpha\res H$ are all the
same with respect to $\{(n,s):n<\om,s\in 2^{<\om}\}$
for $\alpha\in\Sigma$.
\end{enumerate}
Then any two (or even finite subset) of them are compatible.
\bigskip
Assuming that the ground model satisfies the GCH
by the usual book keeping argument we can
arrange things
so that for any $Y\sq \om_1$ which appears in
$M[G_{\om_2}]$ there will be a name for it in the list
$Y_\alpha$ for some $\alpha<\om_2$. The simplest way
to do this is to take
$$\{\langle Z^\alpha_\beta:\beta<\om_1\rangle:\alpha<\om_2\}$$
which lists all $\om_1$ sequences of
countable subsets of $\poset\times \sum_{\alpha<\om_2}\eposet$ with
$\om_2$ repetitions and then define
$$Y_\alpha=\{\langle p, \check{\beta}\rangle: \beta<\om_1,
p\in Z^\alpha_\beta\cap
\poset_\alpha\}$$
If we define
$$x_\alpha=\bigcup\{s\in 2^{<\om}:\exists ((p,q),r)\in G,\;\;
s=p(\alpha)\}
\rmand X=\{x_\alpha:\alpha<\om_1\}$$
then $X$ will be the characteristic functions of an
almost disjoint family
$\aa=\{a_\alpha:\alpha<\om_1\}$. Furthermore if
we define the open sets
$$U_n^\beta=\cup\{[s]: \exists ((p,q),r)\in G\;\; (n,s)\in r(\beta)\}$$
then by the usual genericity argument
$$ \bigcap_{n<\om}U_n^\beta\cap X=\{x_\alpha:\alpha\notin Y_\beta^G\}$$
and so $X$ will be a $Q$-set.
\bigskip
The nontrivial part of our argument is to prove that $\aa$ remains
a maximal almost disjoint family. So let $\tau$ be a name for a
counterexample, i.e., suppose
$$((p_0,q_0),r_0)\forces \tau\in[\om]^\om
\rmand \forall \alpha<\om_1\; \tau\cap a_\alpha \mbox{ is finite }.$$
Let $\Sigma\sq\poset_{\om_2}$ be a countable set of
conditions extending $((p_0,q_0),r_0)$
such that for any $n\in\om\;\;$ $\Sigma$ contains a maximal antichain
beneath $((p_0,q_0),r_0)$ which decides $n\in\tau$.
Let $\alpha_0<\om_1$ be any ordinal not mentioned in any condition from
$\Sigma$. We show $a_{\alpha_0}\cap \tau$ is infinite.
Suppose for contradiction that we have $((p_1,q_1),r_1)\leq ((p_0,q_0),r_0)$,
and $N_1<\om$ such that
$$((p_1,q_1),r_1)\forces \tau\cap a_{\alpha_0}\sq N_1$$
Without loss of generality we may assume that $N_1=N_{p_1}$.
By tacking on strings of zeros to the conditions in $p_1$ we may assume
that every integer occurring in $r$ is bounded by $N_1-2$
(and not just as required by $N_1$). Let
$$F=\{\beta:\{\alpha_0,\beta\}\in\dom(q_1)\}$$
Define $r'\supseteq r_1$ as follows:
$$r'(\delta)=r_1(\delta)\cup \{(n,(\alpha_0,t',k+1)):
(n,(\alpha_0,t,k))\in r_1(\delta),t'\in A_{\delta,n}, t'\supseteq t\}$$
for each $\delta$ where
$$A_{\delta,n}=\{t'\in 2^{N_1-1}: t'\mbox{ is incomparable with all $s$ such
that }
(n,s)\in r_1(\delta)\}$$
Note that $((p_1,q_1),r')$ is a valid condition because $\alpha_0$ is
forced into $Y_\delta$ and $t'$ incomparable with all $s$ which might
be a problem. Let $G$ be a generic filter containing
$((p_1,q_1),r')$. Since $\tau^G$ is almost disjoint from each $a_\beta^G$
and infinite, there exists some $n_0\in\tau^G$ with $n_0>N_1$ and
$n_0\notin a_\beta^G$ for all $\beta\in F$. Let
$((p_2,q_2),r_2)\in \Sigma\cap G$ be so that
$$((p_2,q_2),r_2)\forces n_0\in \tau.$$
Since it is from $\Sigma$ it does not mention $\alpha_0$.
Let $((p^*,q^*),r^*)\in G$ be stronger than both $((p_1,q_1),r')$ and
$(p_2,q_2),r_2)$ and such that $N^*>n_0$. Note that
$((p^*,q_1\cup q_2),r'\cup r_2)$ is a valid condition. Any $\gamma$ that
needs to be forced into some $Y_\beta$ is already forced in by either
$((p_1,q_1),r'\res\beta)$ or $((p_2,q_2),r_2\res\beta)$.
If $p^*(\alpha_0)(n_0)=1$ then we already have a contradiction and there is
nothing to prove. So assume not, and define $p'$ to be exactly the same as
$p^*$ except $p'(\alpha_0)(n_0)=1$.
\bigskip
\noindent {\bf Claim}.
$((p',q_1\cup q_2),r\cup r_2)$ is a valid condition, extending
both $((p_1,q_1),r_1)$ and $((p_2,q_2),r_2)$.
Proof: Note that we have dropped the extra conditions from $r'$, these
were put there just to prove this Claim. The
fact that $p'$ extends both $p_1$ and $p_2$ uses that $n_0>N_{p_1}=N_1$ and
$\alpha_0$ is not in the domain of $p_2$. Similarly since
$q_2$ does not mention $\alpha_0$, so if
$\{\alpha_0,\beta\}\in\dom(q_1\cup q_2)$, then $\beta\in F$ and we
know that $p^*(\beta)(n)=0$ for each $\beta\in F$. So making
$p'(\alpha_0)(n)=1$ does not violate any promises of disjointedness
made in $q_1\cup q_2$.
So we have that $(p',q_1\cup q_2)\in\poset$.
Now fix
$\delta$ and we must check that
$$((p',q_1\cup q_2),r_1(\delta)\cup r_2(\delta))\in \poset*\qposet$$
We need to check condition (3)
(3) if $(n,(\alpha,s,k)),(n,t)\in r_1(\delta)\cup r_2(\delta)$
then either $s$ and $t$ are
incomparable or $s\sq t$ and
$$|\{i: |s|\leq i<|t|,\; t(i)\not=p'(\alpha)(i)\}|>k$$
Suppose it fails.
It can only fail if the $\alpha=\alpha_0$ and
since $r_2$ does not mention $\alpha_0$ it must be
that $(n,(\alpha_0,s,k))\in r_1(\delta)$ and $(n,t)\in r_2(\delta)$.
Also it must be that $s$ and $t$ are comparable with $s\sq t$ but
$$|\{i: |s|\leq i<|t|, \;t(i)\not=p'(\alpha_0)(i)\}|\leq k$$
Note also that $|t|>n_0>N_1$ because otherwise
$$\{i: |s|\leq i<|t|, \;t(i)\not=p'(\alpha_0)(i)\}=
\{i: |s|\leq i<|t|, \; t(i)\not=p^*(\alpha_0)(i)\}$$
But then
$$|\{i: |s|\leq i<|t|, \;\;t(i)\not=p^*(\alpha_0)(i)\}|> k$$
because
$((p^*,q_1\cup q_2),r_1(\delta)\cup r_2(\delta))\in \poset*\qposet$.
Now let $t'=t\res (N_1-1)$.
\medskip
\noindent Case 1. $t'$ is comparable with
some $s'$ such that $(n,s')\in r_1(\delta)$.
Recall that every integer occurring in $r_1(\delta)$ is bounded by
$N_1-1$. So it must be that $s'\sq t'$ but intuitively
this is easy because $r_1(\delta)$ is already asserting
$[s']\sq U_n^\delta$ and this implies $[t]\sq U_n^\delta$.
More formally,
$s'\sq t'$
and therefore $s'$ and $s$ are both initial strings of $t'$ and so
comparable, but then we know:
$$|\{i: |s|\leq i<|s'|k$$
But this is still true for $p'$ since we have not changed it
below $N_1$.
\medskip
\noindent Case 2. $t'\in A_{\delta,n}$ and
so we added $(\alpha_0,t',k+1)$ to
$r'(\delta)$.
But remember $((p^*,q_1\cup q_2),r'\cup r_2)$ is a valid
condition, which means that
$$|\{i: N_1\leq i<|t|, t(i)\not=p^*(\alpha_0)(i)\}| > k+1$$
but $kk$$
\begin{figure}
\unitlength=1.00mm
\begin{picture}(89.00,79.00)
\put(10.00,70.00){\line(1,-2){28.67}}
\put(38.67,13.00){\line(3,5){34.33}}
\put(13.00,42.00){\line(1,0){58.00}}
\put(10.00,42.00){\makebox(0,0)[cc]{$N_1$}}
\put(41.00,31.00){\line(-2,3){7.33}}
\put(33.67,42.00){\line(2,3){15.33}}
\put(49.00,65.00){\line(-3,4){10.67}}
\put(49.00,65.00){\line(1,1){12.00}}
\put(13.00,54.00){\line(1,0){58.00}}
\put(10.00,54.00){\makebox(0,0)[cc]{$n_0$}}
\put(43.00,34.00){\makebox(0,0)[cc]{$s$}}
\put(32.00,45.00){\makebox(0,0)[cc]{$t'$}}
\put(50.00,70.00){\makebox(0,0)[cc]{$t$}}
\put(10.00,79.00){\makebox(0,0)[cc]{$N^*$}}
\put(80.00,80.00){\makebox(0,0)[cc]{$[t]\sq U_n$ in $r_2(\delta)$}}
\put(88.00,32.00){\makebox(0,0)[cc]
{$\swap(x_{\alpha_0},s,k)\cap U_n=\emptyset$ in $r_1(\delta)$}}
\put(95.00,48.00){\makebox(0,0)[cc]
{$\swap(x_{\alpha_0},t',N_1-1)\cap U_n=\emptyset$ in $r'(\delta) $}}
\end{picture}
\caption{The swap}
\end{figure}
\noindent This proves that
$((p',q_1\cup q_2),r_1(\delta)\cup r_2(\delta))\in \poset*\qposet$
for every $\delta$.
Finally we must show that
$$((p',q_1 \cup q_2),(r_1\cup r_2)\res\beta)\forces_\beta
\gamma\in Y_\beta$$
whenever
$(n,(\gamma,s,k))\in (r_1\cup r_2)(\beta)$ for some $n,s,k$. But
by induction
$$((p',q_1\cup q_2),(r_1\cup r_2)\res\beta)$$ extends
both $((p_1,q_1),r_1\res\beta)$ and
$((p_2,q_2),r_2\res\beta)$, one of which does the required
forcing.
This proves the Claim. The theorem now follows from the contradiction
that
$$((p_1,q_1),r_1)\forces \tau\cap a_{\alpha_0}\sq N_1$$
$$((p_2,q_2),r_2)\forces n_0\in\tau$$
where $n_0>N_1$ and
$$((p',q_1\cup q_2),r_1\cup r_2)\forces n_0\in a_{\alpha_0}.$$
\qed
\bigskip
Remark. The usual Q-set forcing
kills the maximality of an almost disjoint family $X$. To see this
suppose $\{x_n:n<\om\}\sq X$ and conditions
are finite consistent sets of sentences of the form: ``$[s]\sq U_n$''
or ``$x\notin U_n$'' where $x\in X\setminus \{x_n:n<\om\}$. So
when we force we get a $G_\delta$ set so that
$$\cap_{n<\om}U_n\cap X=\{x_n:n<\om\}.$$
In the generic extension we can find $\{k_n:n<\om\}$ increasing
so that
$$k_{n+1}\notin\cup_{ik_n$ not in any $x_i$ for $i N_{p}$ and
$$(\hat{p},\hat{q})\forces [x_n\res N_{\hat{p}}]\sq U$$
Let $(p^*,q^*)$ extend both $(p',q')$ and $(\hat{p},\hat{q})$.
Change $p^*$ to $r$ with same domain but $r(\alpha)=p^*(n)$
and other coordinate all the same. But then
$(r,q'\cup \hat{q})$ is a common extension of both
$(p',q')$ and $(\hat{p},\hat{q})$. And this is a contradiction.
This proves the Claim and Theorem.
\qed
\begin{theorem}
CH implies exists a MAD family which is concentrated
on the finite subsets of $\om$ and is a $\lambda$-set
(i.e., every countable subset is a relative $G_\delta$.
\end{theorem}
\proof
It is easy to construct a MAD family $\{a_\alpha:\alpha<\om_1\}$ so
that if $f_\alpha:\om\to a_\alpha$ is the strictly increasing
enumeration of $\alpha$, then for every
$\alpha<\beta$ we have that $f_\alpha<^* f_\beta$ and
for every $g\in\om^\om$ there exists $\alpha<\om_1$
such that $g\leq^* f_\alpha$, i.e., they form a scale. Rothberger
(see Miller \cite{survey}) showed that any well-ordered subset
of $(\om^\om,\leq^*)$ is a $\lambda$-set and that any
$\om_1$ -ordered unbounded set is concentrated on the
rationals.
\qed
\bigskip
The same large cardinal results lead to the following
conjecture:
\begin{conjecture}
CH implies there exists a MAD family which is concentrated on
a countable subset of itself.
\end{conjecture}
Paul Szeptycki pointed out that the Q-set forcing using in Theorem 1
can be used to prove the following:
\begin{theorem} It is relatively consistent that there exists
a Q-set $X\sq [\om]^\om$ satisfying the property
that for every $a\in [\om]^\om$ for all but countably
many $x\in X$ we have that $|x\cap a|=|x\setminus a|=\om$,
ie $X$ is a strong splitting family.
\end{theorem}
\proof
We replace $\poset$ by the Cohen real partial order, i.e., just
drop the $q$'s from the $(p,q)$. We use the same $\poset*\qposet$.
Note that in the basic argument for $p'$ we could have flipped
$p'(\alpha_0)(n_0)=1-p^*(\alpha_0)(n_0)$ and $\alpha_0$ could be
any $\alpha<\om_1$ not mentioned in $\Sigma$.
\qed
Allan Dow asked if it is possible to have
$X=\{x_\alpha\in 2^\om:\alpha<\om_1\}$ and
$Y=\{y_\alpha\in 2^\om:\alpha<\om_1\}$ such that $x_\alpha=^*y_\alpha$
for every $\alpha<\om_1$, and $X$ a Q-set and $Y$ not a Q-set.
The answer is yes, in fact, we force something stronger:
\begin{theorem} \label{dowq} It is relatively to consistent to have
$X=\{x_\alpha\in 2^\om:\alpha<\om_1\}$ and $\{y_n\in 2^\om:n<\om\}$
such that $x_n=^*y_n$
for every $n<\om$, and $X$ a Q-set concentrated
on $\{y_n:n<\om\}$ (hence
$Y=\{y_n:n<\om\}\cup\{x_\alpha:\om\leq \alpha<\om_1\}$ is not a Q-set).
\end{theorem}
\proof
This is a variant of the forcing used in Fleissner-Miller \cite{FM}.
In that forcing we start by adding an $\om_1$ batch of Cohen
reals $X=\{x_\alpha\in 2^\om:\alpha<\om_1\}$. The Q-set forcing is
modified to always allow statements of the form $e\notin U_n^\alpha$
where $e$ is any eventually zero element of $2^\om$. With this
modification the Q-set $X$ is shown to be concentrated on the eventually
zero elements of $2^\om$.
We modify this slightly as follows.
Let $y_n$ be defined by $y_n(m)=x_n(m)$ except $y_n(n)=1-x_n(n)$, i.e.
we change only one coordinate. Now in the inductive construction of
$\poset_\alpha$ we use the $y_n$'s instead of the eventually zero
reals, i.e.
if $x\notin U^\alpha_m\in p(\alpha)$, then either
$x = y_n$ for some $n$ or $p\res\alpha\forces x\in Y_\alpha$.
The rest of the argument is the same as in \cite{FM}.
\qed
I am not sure how to do Theorem \ref{dowq} with a MAD family. This would
be interesting because it would show that the $\Psi$ space does
not determine whether or not you have a Q-set.
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\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 \\
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