text:
\section{Introduction}
One of the central concepts in the field of models of Peano
Arithmetic is the \emph{standard system} of a model of {\rm PA}. The
\emph{standard system} of a model of {\rm PA} is the collection of
subsets of the natural numbers that arise as intersections of the
definable sets of the model with its \emph{standard part} $\mathbb {N}$. The
notion of a \emph{Scott set} captures three key properties of
standard systems.
\begin{definition}
$\mathfrak{X}\subseteq \mathcal{P}(\mathbb {N})$ is a \emph{Scott set} if
\begin{itemize}
\item [(1)] $\mathfrak{X}$ is a Boolean algebra of sets.
\item [(2)] If $A\in\mathfrak{X}$ and $B$ is Turing computable from
$A$, then $B\in\mathfrak{X}$.
\item [(3)] If $T$ is an infinite binary tree coded by a set in
$\mathfrak{X}$, then $\mathfrak{X}$ has a set coding some path through $T$.
\end{itemize}
\end{definition}
In 1962, Scott showed that every standard system is a Scott set and
the partial converse that every \emph{countable} Scott set is the
standard system of a model of {\rm PA} \cite{scott:ssy}. The
question of whether \emph{every} Scott is the standard system of a
model of {\rm PA} became known in the folklore as Scott's Problem.
In 1982, Knight and Nadel extended Scott's result to Scott sets of
size $\omega_1$ \cite{knight:scott}. They showed that every Scott
set of size $\omega_1$ is the standard system of a model of {\rm
PA}. It has proved very difficult to make further progress on
Scott's Problem. My approach, following Engstr\"om
\cite{engstrom:thesis} and suggested several years earlier by
Hamkins, Marker, etc., has been to use the set theoretic techniques
of forcing and the forcing axioms.

We can associate with every family of reals $\mathfrak{X}$, the poset
$\mathfrak{X}/\mathrm{Fin}$ which consists of the infinite sets of $\mathfrak{X}$ under the
ordering of \emph{almost inclusion}. Engstr\"om in
\cite{engstrom:thesis} introduced the use of this poset in
connection with Scott's Problem. A family of reals is
\emph{arithmetically closed} if whenever $A$ is in it and $B$ is
arithmetically definable from $A$, then $B$ is also in it. The
arithmetic closure of $\mathfrak{X}$ is an essential ingredient in the
constructions that make the posets $\mathfrak{X}/\mathrm{Fin}$ useful in investigating
properties of uncountable models of {\rm PA} (for details of the
constructions, see \cite{gitman:scott}). For this reason, whenever
we view a family of reals as a poset we will always assume
arithmetic closure. A family $\mathfrak{X}$ is \emph{proper} if it is
arithmetically closed and the poset $\mathfrak{X}/\mathrm{Fin}$ is proper. A family
$\mathfrak{X}$ is piecewise proper if it is the union of a chain of proper
families each of which has size $\leq\omega_1$. I showed in
\cite{gitman:scott} that under the Proper Forcing Axiom ({\rm PFA}),
every proper or piecewise proper family of reals is the standard
system of a model of {\rm PA}. I will give an extended discussion of
properness and the {\rm PFA} in Section \ref{sec:proper}.

Throughout the paper, I equate \emph{reals} with subsets of $\mathbb {N}$. It
is easy to see that every countable arithmetically closed family of
reals is proper and $\mathcal{P}(\mathbb {N})$ is proper as well (see Section
\ref{sec:forcing}). Every arithmetically closed family of size
$\leq\omega_1$ is trivially piecewise proper since it is the union
of a chain of countable arithmetically closed families. It becomes
much more difficult to find instances of uncountable proper families
of reals other than $\mathcal{P}(\mathbb {N})$. Also, it was not clear for a while
whether there are piecewise proper families of of size larger than
$\omega_1$. My main results are:

\begin{theorem}
If {\rm CH} holds, then $\mathcal{P}^V(\mathbb {N})/\mathrm{Fin}$ remains proper in any
generic extension by a c.c.c.\ poset.
\end{theorem}

\begin{theorem}
There is a generic extension of $V$ by a c.c.c.\ poset, which
contains continuum many proper families of reals of size $\omega_1$.
\end{theorem}

\begin{theorem}
There is a generic extension of $V$ by a c.c.c.\ poset, which
contains continuum many piecewise proper families of reals of size
$\omega_2$.
\end{theorem}

\section{Proper Posets and the PFA}\label{sec:proper}
Proper posets were invented by Shelah, who sought a class of
$\omega_1$ preserving posets that would extend the c.c.c.\ and
countably closed classes of posets and be preserved under iterations
with countable support. The Proper Forcing Axiom ({\rm PFA}) was
introduced by Baumgartner who showed that it is consistent by
assuming the existence of a supercompact cardinal
\cite{baumgartner:pfa}. This remains the best known upper bound on
the consistency of {\rm PFA}.

Recall that for a cardinal $\lambda$, the set $H_\lambda$ is the
collection of all sets whose transitive closure has size less than
$\lambda$. Let $\mathbb{P}$ be a poset and $\lambda$ be a cardinal greater
than $2^{|\mathbb{P}|}$. Since we can always take an isomorphic copy of $\mathbb{P}$
on the cardinal $|\mathbb{P}|$, we can assume without loss of generality
that $\mathbb{P}$ and $\mathcal{P}(\mathbb{P})$ are elements of $H_\lambda$. In
particular, we want to ensure that all dense subsets of $\mathbb{P}$ are in
$H_\lambda$. Let $M$ be a countable elementary submodel of
$H_\lambda$ containing $\mathbb{P}$ as an element. If $G$ is a filter on
$\mathbb{P}$, we say that $G$ is $M$-\emph{generic} if for every maximal
antichain $A\in M$ of $\mathbb{P}$, the intersection $G\cap A\cap
M\neq\emptyset$. It must be explicitly specified what $M$-generic
means in this context since the usual notion of generic filters
makes sense only for transitive structures and $M$ is not
necessarily transitive. This definition of $M$-generic is closely
related to the definition for transitive structures. To see this,
let $M^*$ be the Mostowski collapse of $M$ and $\mathbb{P}^*$ be the image
of $\mathbb{P}$ under the collapse. Let $G^*\subseteq \mathbb{P}^*$ be the pointwise
image of $G\cap M$ under the collapse. Then $G$ is $M$-generic if
and only if $G^*$ is $M^*$-generic for $\mathbb{P}^*$ in the usual sense.

Later we will need the following important characterization of
$M$-generic filters.

\begin{theorem} If\/ $\mathbb{P}$ is a poset in $M\prec H_\lambda$, then a
$V$-generic filter $G\subseteq\mathbb{P}$ is $M$-generic if and only if
$M\cap \text{\emph{Ord}}=M[G]\cap\text{\emph{Ord}}$. \emph{(See, for
example, \cite{shelah:proper}, p. 105)}\label{th:equproper}
\end{theorem}

\begin{definition} Let $\mathbb{P}\in H_\lambda$ be a poset and $M$ be an
elementary submodel of $H_\lambda$ containing $\mathbb{P}$. Then a condition
$q\in\mathbb{P}$ is $M$-\emph{generic} if and only if every $V$-generic
filter $G\subseteq\mathbb{P}$ containing $q$ is $M$-generic.
\end{definition}
\begin{definition} A poset $\mathbb{P}$ is \emph{proper} if for every
$\lambda> 2^{|\mathbb{P}|}$ and every countable $M\prec H_\lambda$
containing $\mathbb{P}$, for every $p\in\mathbb{P}\cap M$, there is an $M$-generic
condition below $p$.
\end{definition}

When proving that a poset is proper it is often easier to use the
following equivalent characterization which appears in
\cite{shelah:proper} (p.\ 102).
\begin{theorem}\label{th:equivproper}
A poset $\mathbb{P}$ is proper if there exists a $\lambda>2^{|\mathbb{P}|}$ and a
club of countable $M\prec H_\lambda$ containing $\mathbb{P}$, such that for
every $p\in\mathbb{P}\cap M$, there is an $M$-generic condition below $p$.
\end{theorem}

Countably closed posets and c.c.c.\ posets are proper and all proper
posets preserve $\omega_1$ \cite{shelah:proper}.

\begin{definition}
\emph{The Proper Forcing Axiom} ({\rm PFA}) is the assertion that
for every proper poset $\mathbb{P}$ and every collection $\mathcal D$ of at
most $\omega_1$ many dense subsets of $\mathbb{P}$, there is a filter on
$\mathbb{P}$ that meets all of them.
\end{definition}

The Proper Forcing Axiom decides the size of the continuum. It was
shown in \cite{veli:pfa} that under {\rm PFA}, continuum is
$\omega_2$.

\section{Proper and Piecewise Proper Families}\label{sec:forcing}

Let $\mathfrak{X}$ be a family of reals. Define the poset $\mathfrak{X}/\mathrm{Fin}$ to consist
of the infinite sets in $\mathfrak{X}$ under the ordering of almost inclusion.
That is, for infinite $A$ and $B$ in $\mathfrak{X}$, we say that $A\leq B$ if
and only if $A\subseteq_\mathrm{Fin} B$. Observe that $\mathfrak{X}/\mathrm{Fin}$ is forcing
equivalent to forcing with the Boolean algebra $\mathfrak{X}$ modulo the ideal
of finite sets. A familiar and thoroughly studied instance of this
poset is $\mathcal{P}(\omega)/\mathrm{Fin}$.  For a property of posets $\mathscr
P$, if $\mathfrak{X}$ is an arithmetically closed family of reals and
$\mathfrak{X}/\mathrm{Fin}$ has $\mathscr P$, I will simply say that \emph{$\mathfrak{X}$ has
property $\mathscr P$}. An important point to be noted here is that
whenever a family $\mathfrak{X}$ is discussed as a poset, I will always be
assuming that \emph{it is arithmetically closed}. Recall that the
reason for this is the need for arithmetic closure of $\mathfrak{X}$ in the
constructions with models of {\rm PA} in which $\mathfrak{X}/\mathrm{Fin}$ is used.

The easiest way to show that a poset is proper to show that it is
c.c.c.\ or countably closed. Thus, every countable arithmetically
closed family is proper since it is c.c.c.\ and $\mathcal{P}(\mathbb {N})$ is
proper since it is countably closed. Unfortunately these two
conditions do not give us any other instances of proper families.

\begin{theorem}\label{th:wrong} Every c.c.c.\ family of reals is countable.
\end{theorem}
\begin{proof}
Let $\mathfrak{X}$ be an arithmetically closed family of reals.  If $x$ is a
finite subset of $\mathbb {N}$, let $\ulcorner x \urcorner$ denote the code
of $x$ using G\"{o}del's coding. For every $A\in\mathfrak{X}$, define an
associated $A'=\{\ulcorner A\cap n\urcorner\mid n\in\mathbb {N}\}$. Clearly
$A'$ is definable from $A$, and hence in $\mathfrak{X}$. Observe that if
$A\neq B$, then $|A'\cap B'|<\omega$. Hence if $A\neq B$, we get
that $A'$ and $B'$ are incompatible in $\mathfrak{X}/\mathrm{Fin}$. It follows that
$\mathscr A=\{A'\mid A\in\mathfrak{X}\}$ is an antichain of $\mathfrak{X}/\mathrm{Fin}$ of size
$|\mathfrak{X}|$.  This shows that $\mathfrak{X}/\mathrm{Fin}$ always has antichains as large as
the whole poset.
\end{proof}

Thus, the poset $\mathfrak{X}/\mathrm{Fin}$ has the worst possible chain condition,
namely $|\mathfrak{X}|^+$-c.c..

\begin{theorem} \label{th:countclosed} Every countably closed family
of reals is $\mathcal{P}(\mathbb {N})$.
\end{theorem}
\begin{proof}
I will show that every $A\subseteq \mathbb {N}$ is in $\mathfrak{X}$. Define a sequence
of subsets $\langle B_n\mid n\in\omega\rangle$ by $B_n=\{m\in\mathbb {N}\mid
(m)_n=\chi_A(n)\}$ where $\chi_A$ is the characteristic function of
$A$. Let $A_m=\cap_{n\leq m}B_n$ and observe that each $A_m$ is
infinite. Thus, $A_0\geq A_1\geq\cdots \geq A_m\geq\ldots$ are
elements of $\mathfrak{X}/\mathrm{Fin}$. By countable closure, there exists $C\in
\mathfrak{X}/\mathrm{Fin}$ such that $C\subseteq_{\mathrm{Fin}}A_m$ for all $m\in \mathbb {N}$. Thus,
$C\subseteq_\mathrm{Fin} B_n$ for all $n\in \mathbb {N}$. It follows that
$A=\{n\in\mathbb {N}\mid\exists m\,\forall k\in C\,\text{ if }k>m, \text{
then } (k)_n=1\}$. This shows that $A$ is arithmetic in $C$, and
hence $A\in\mathfrak{X}$ by arithmetic closure. Since $A$ was arbitrary, this
concludes the proof that $\mathfrak{X}=\mathcal{P}(\mathbb {N})$.
\end{proof}
The assumption that $\mathfrak{X}$ is arithmetically closed is not necessary
for Theorem \ref{th:countclosed}. Any family of reals $\mathfrak{X}$ such that
$\mathfrak{X}/\mathrm{Fin}$ is countably closed must be arithmetically closed (see
\cite{gitman:scott}) .

Enayat showed in \cite{enayat:endextensions} that {\rm ZFC} proves
the existence of an arithmetically closed family of size $\omega_1$
which collapses $\omega_1$, and hence is \emph{not} proper.

Later Enayat and Shelah showed in \cite{shelah:borel} that there is
a \emph{Borel} arithmetically closed family of size $\omega_1$ which
is \emph{not} proper as well.

I will show below that it is consistent with {\rm ZFC} that there
are continuum many proper families of size $\omega_1$ and it is
consistent with {\rm ZFC} that there are continuum many piecewise
proper families of size $\omega_2$. But first I will consider the
question of when does forcing to add new reals preserve the
properness of the reals of the ground model. I will show that if
{\rm CH} holds, forcing with a c.c.c.\ poset preserves the
properness of the reals of the ground model.

\begin{lemma}\label{le:intersect}
Let $\mathfrak{X}_0\subseteq\mathfrak{X}_1\subseteq\cdots\subseteq
\mathfrak{X}_\xi\subseteq\cdots$ for $\xi<\omega_1$ be a continuous chain of
countable families of reals and let $\mathfrak{X}=\cup_{\xi<\omega_1} \mathfrak{X}_\xi$.
If $M$ is a countable elementary substructure of some $H_\lambda$
and $\langle \mathfrak{X}_\xi\mid\xi<\omega_1\rangle\in M$, then $M\cap \mathfrak{X}=\mathfrak{X}_\alpha$
where $\alpha=\text{\emph{Ord}}^M\cap\omega_1$.
\end{lemma}
\begin{proof}
Let $\alpha=\text{Ord}^M\cap\omega_1$. Suppose $\xi\in\alpha$, then
$\xi\in M$, and hence $\mathfrak{X}_\xi\in M$. Since $\mathfrak{X}_\xi$ is countable, it
follows that $\mathfrak{X}_\xi\subseteq M$. Thus, $\mathfrak{X}_\alpha\subseteq\mathfrak{X}\cap
M$. Now suppose $A\in \mathfrak{X}\cap M$, then the least $\xi$ such that
$A\in\mathfrak{X}_\xi$ is definable in $H_\lambda$. It follows that $\xi\in
M$, and hence $\xi\in\alpha$. Thus, $\mathfrak{X}\cap M\subseteq \mathfrak{X}_\alpha$.
\end{proof}
\begin{lemma}\label{le:count}
 Suppose $\mathbb{P}$ is a c.c.c.\ poset and $G\subseteq \mathbb{Q}$ is $V$-generic
for a countably closed poset $\mathbb{Q}$. Then  $\mathbb{P}$ remains c.c.c.\ in
$V[G]$.
\end{lemma}
\begin{proof}
Suppose $\mathbb{P}$ does not remain c.c.c.\ in $V[G]$.  Fix a $\mathbb{Q}$-name
$\dot{A}$ and $r\in \mathbb{Q}$ such that $r\Vdash ``\dot{A}$ is a maximal
antichain of $\check{\mathbb{P}}\text{ of size }\omega_1"$. Choose $q_0\leq
r$ and $a_0\in \mathbb{P}$ such that $q_0\Vdash \check{a_0}\in \dot{A}$.
Suppose that we have defined $q_0\geq q_1\geq\cdots\geq
q_\xi\geq\cdots$ for $\xi<\beta$ where $\beta$ is some countable
ordinal, together with a corresponding sequence $\langle a_\xi\mid
\xi<\beta\rangle$ of elements of $\mathbb{P}$ such that $q_\xi\Vdash
\check{a}_\xi\in \dot{A}$ and $a_{\xi_1}\neq a_{\xi_2}$ for all
$\xi_1<\xi_2$. By countable closure of $\mathbb{Q}$, we can find $q\in\mathbb{Q}$
such that $q\leq q_\xi$ for all $\xi<\beta$.  Let $q_\beta\leq q$
and $a_\beta\in \mathbb{P}$ such that $q_\beta\Vdash \check{a}_\beta\in
\dot{A}$ and $a_\beta\neq a_\xi$ for all $\xi<\beta$.  Such
$a_\beta$ must exist since we assumed $r\Vdash ``\dot{A} \text{ is a
maximal antichain of }\check{\mathbb{Q}}\text{ of size }\omega_1"$ and
$q\leq r$. Thus, we can build a descending sequence $\langle
q_\xi\mid \xi<\omega_1\rangle$ of elements of $\mathbb{Q}$ and a
corresponding sequence $\langle a_\xi\mid \xi<\omega_1\rangle$ of
elements of $\mathbb{P}$ such that $q_\xi\Vdash \check{a}_\xi\in\dot{A}$.
But clearly $\langle a_\xi\mid \xi<\omega_1\rangle$ is an antichain
in $V$ of size $\omega_1$, which contradicts the assumption that
$\mathbb{P}$ was c.c.c..
\end{proof}
\begin{theorem}\label{th:oldreals}
If {\rm CH} holds, then $\mathcal{P}^V(\mathbb {N})/\mathrm{Fin}$ remains proper in any
generic extension by a c.c.c.\ poset.
\end{theorem}
\begin{proof}
Let $\mathbb{P}$ be a c.c.c.\ poset and fix a $V$-generic $g\subseteq \mathbb{P}$.
In $V$, let $\mathcal{P}(\mathbb {N})=\cup_{\xi<\omega_1} \mathfrak{X}_\xi$ where each
$\mathfrak{X}_\xi$ is countable and
$\mathfrak{X}_0\subseteq\mathfrak{X}_1\subseteq\cdots\subseteq\mathfrak{X}_\xi\subseteq\cdots$ is
a continuous chain. For sufficiently large cardinals $\lambda$, it
is easy to see that $H_\lambda^{V[g]}=H_\lambda[g]$. The countable
elementary substructures of $H_\lambda[g]$ of the form $M[g]$ where
$M\subseteq V$ and $\langle \mathfrak{X}_\xi\mid \xi<\omega_1\rangle, \mathbb{P}\in M[g]$ form
a club. So by Theorem \ref{th:equivproper}, it suffices to find
generic conditions only for such elementary substructures. Fix a
countable $M[g]\prec H_\lambda[g]$ in $V[g]$ such that $\langle
\mathfrak{X}_\xi\mid\xi<\omega_1\rangle, \mathbb{P}\in M[g]$ and $M\subseteq V$. We need
to prove that for every $B\in M[g]\cap \mathcal{P}(\mathbb {N})^V/\mathrm{Fin}$, there
exists $A\in \mathcal{P}(\mathbb {N})^V/\mathrm{Fin}$ such that $A\subseteq_\mathrm{Fin} B$ and
$A$ is $M[g]$-generic in $V[g]$. By Lemma \ref{le:intersect},
$M[g]\cap \mathcal{P}^V(\mathbb {N})=\mathfrak{X}_\alpha$ where $\alpha=\text{Ord}^M\cap
\omega_1$. Let $\mathcal D=\{\mathscr D\cap \mathfrak{X}_\alpha\mid \mathscr
D\in M\text{ and }\mathscr D\text{ dense in }\mathcal{P}^V(\mathbb {N})/\mathrm{Fin}\}$.
Observe that $\mathcal D\subseteq V$ and $|\mathcal D|=\omega$.
Since $\mathbb{P}$ is c.c.c., we can show that there is $\mathcal
D'\supseteq \mathcal D$ of size $\omega$ in $V$. In $V$, use
$\mathcal D'$ and $\mathfrak{X}_\alpha$ to define $\mathcal E=\{\mathscr D\in
\mathcal D'\mid \mathscr D\text{ dense in }\mathfrak{X}_\alpha\}$. It is clear
that $\mathcal D\subseteq \mathcal E$. By the countable closure of
$\mathcal{P}(\mathbb {N})^V/\mathrm{Fin}$ in $V$, we can find an infinite $A\subseteq_\mathrm{Fin}
B$ such that every $\mathscr D\in \mathcal E$ contains some $C$
above $A$. It follows that $A$ is $M$-generic. In fact, I will show
that $A$ is $M[g]$-generic. To verify this, we need to check that
whenever $A\in G$ and $G\subseteq \mathcal{P}(\mathbb {N})^V/\mathrm{Fin}$ is
$V[g]$-generic, then $M[g][G]\cap \text{Ord}=M[g]\cap \text{Ord}$.
Since we are forcing with $\mathbb{P}\times\mathcal{P}(\mathbb {N})/\mathrm{Fin}$, we have
$M[g][G]=M[G][g]$. It is clear that $M\cap \text{Ord}=M[g]\cap
\text{Ord}$, and so it remains to show that $M[G][g]\cap
\text{Ord}=M\cap \text{Ord}$. Since $A\in G$ and $A$ is $M$-generic,
we have that $M[G]\cap \text{Ord}=M\cap \text{Ord}$. The poset $\mathbb{P}$
remains c.c.c.\ in $V[G]$ by Lemma \ref{le:count} since
$\mathcal{P}(\mathbb {N})^V/\mathrm{Fin}$ is countably closed. Also we have $M[G]\prec
H_\lambda[G]$, even though $M[G]$ itself may not be an element of
$V[G]$. Let $\mathscr A$ be a maximal antichain of $\mathbb{P}$ in $M[G]$,
then $\mathscr A\in H_\lambda[G]$, and hence $\mathscr A$ has size
$\omega$. It follows that $\mathscr A\subseteq M[G]$. Since $g$ is
$V[G]$-generic, it must meet $\mathscr A$. So $g$ is $M[G]$-generic,
and hence $M[G][g]\cap \text{Ord}=M[G]\cap \text{Ord}$.
\end{proof}
It follows that it is consistent that there are uncountable proper
families other than $\mathcal{P}(\mathbb {N})$. Start in any universe satisfying
{\rm CH} and force to add a Cohen real. In the resulting generic
extension, the reals of $V$ will be an uncountable proper family.

Next, I will show how to force the existence of many proper
families. I will begin by looking at what properness translates into
in this specific context.
\begin{prop}\label{prop:proper} Suppose $\mathfrak{X}$ is a family of reals and
$\mathscr A$ is a countable antichain of $\mathfrak{X}/\mathrm{Fin}$.  Then  for $B\in
\mathfrak{X}$:
\begin{itemize}
\item[(1)] Every $V$-generic filter $G\subseteq \mathfrak{X}/\mathrm{Fin}$ containing $B$ meets
$\mathscr A$.
\item[(2)] There exists a finite list $A_0,\ldots, A_n\in \mathscr A$
such that $B\subseteq_\mathrm{Fin} A_0\cup\ldots\cup A_n$.
\end{itemize}
\end{prop}
\begin{proof}$\,$\\
(2)$\Longrightarrow$(1): Suppose $B\subseteq_\mathrm{Fin} A_0\cup\ldots\cup
A_n$ for some $A_0,\ldots, A_n\in \mathscr A$. Since a $V$-generic
filter $G$ is an ultrafilter, one of the $A_i$ must be in $G$.\\
(1)$\Longrightarrow$(2): Assume that every $V$-generic filter $G$
containing $B$ meets $\mathscr A$ and suppose toward a contradiction
that $(2)$ does not hold. Enumerate $\mathscr A=\{A_0,A_1,\ldots,
A_n,\ldots\}$. It follows that for all $n\in\mathbb {N}$, the intersection
$B\cap (\mathbb {N}-A_0)\cap\cdots\cap(\mathbb {N}-A_n)$ is infinite. Define
$C=\{c_n\mid n\in\mathbb {N}\}$ such that $c_0$ is the least element of
$B\cap(\mathbb {N}-A_0)$ and $c_{n+1}$ is the least element of
$B\cap(\mathbb {N}-A_0)\cap\cdots\cap (\mathbb {N}-A_{n+1})$ greater than $c_n$.
Clearly $C\subseteq B$ and $C\subseteq_\mathrm{Fin} (\mathbb {N}-A_n)$ for all $n\in
\mathbb {N}$.  Let $G$ be a $V$-generic filter containing $C$, then $B\in G$
and $(\mathbb {N}-A_n)\in G$ for all $n\in\mathbb {N}$. But this contradicts our
assumption that $G$ meets $\mathscr A$.
\end{proof}
\begin{corollary}\label{cor:properequiv}
A family of reals $\mathfrak{X}$ is proper if and only if there exists
$\lambda>2^{|\mathfrak{X}|}$ such that for every countable $M\prec H_\lambda$
containing $\mathfrak{X}$, whenever $C\in M\cap \mathfrak{X}/\mathrm{Fin}$, then there is
$B\subseteq_\mathrm{Fin} C$ in $\mathfrak{X}/\mathrm{Fin}$ such that for every maximal
antichain $\mathscr A\in M$ of $\mathfrak{X}/\mathrm{Fin}$, there are $A_0,\ldots,
A_n\in\mathscr A\cap M$ with $B\subseteq_\mathrm{Fin} A_0\cup\cdots\cup
A_n$.
\end{corollary}
\begin{proof}$\,$\\
($\Longrightarrow$): Suppose $\mathfrak{X}$ is proper. Then there is
$\lambda>2^{|\mathfrak{X}|}$ such that for every countable $M\prec H_\lambda$
containing $\mathfrak{X}$ and every $C\in M\cap \mathfrak{X}/\mathrm{Fin}$, there is an
$M$-generic $B\subseteq_\mathrm{Fin} C$ in $\mathfrak{X}/\mathrm{Fin}$. Fix a countable
$M\prec H_\lambda$ containing $\mathfrak{X}$ and $C\in M\cap \mathfrak{X}/\mathrm{Fin}$. Let
$B\subseteq_\mathrm{Fin} C$ be $M$-generic. Thus, every $V$-generic filter
containing $B$ must meet $\mathscr A\cap M$ for every maximal
antichain $\mathscr A\in M$ of $\mathfrak{X}/\mathrm{Fin}$. But since $\mathscr A\cap
M$ is countable, by Proposition \ref{prop:proper}, there exist
$A_0,\ldots,A_n\in \mathscr A\cap M$
such that $B\subseteq_\mathrm{Fin} A_0\cup\cdots\cup A_n$.\\
($\Longleftarrow$): Suppose that there is $\lambda>2^{|\mathfrak{X}|}$ such
that for every countable \hbox{$M\prec H_\lambda$} containing $\mathfrak{X}$,
whenever $C\in M\cap \mathfrak{X}/\mathrm{Fin}$, then there is $B\subseteq_\mathrm{Fin} C$ in
$\mathfrak{X}/\mathrm{Fin}$ such that for every maximal antichain $\mathscr A\in M$ of
$\mathfrak{X}/\mathrm{Fin}$, there are $A_0,\ldots, A_n\in\mathscr A\cap M$ with
$B\subseteq_\mathrm{Fin} A_0\cup\cdots\cup A_n$. Fix a countable $M\prec
H_\lambda$ with $\mathfrak{X}\in M$ and $C\in M\cap \mathfrak{X}/\mathrm{Fin}$. Let
$B\subseteq_\mathrm{Fin} C$ be as above. By Proposition \ref{prop:proper},
every $V$-generic filter $G$ containing $B$ must meet $\mathscr
A\cap M$ for every maximal antichain $\mathscr A\in M$. Thus, $B$ is
$M$-generic. Since $M$ was arbitrary, we can conclude that $\mathfrak{X}$ is
proper.
\end{proof}
The hypothesis of Corollary \ref{cor:properequiv} can be weakened,
by Theorem \ref{th:equivproper}, to finding for some $H_\lambda$,
only a club of countable $M$ having the desired property.

The next definition is key to all the remaining arguments in the
paper.
\begin{definition}
Let $\mathfrak{X}$ be a countable family of reals, let $\mathcal D$ be some
collection of dense subsets of $\mathfrak{X}/\mathrm{Fin}$, and let $B\in \mathfrak{X}$. We say
that an infinite set $A\subseteq\mathbb {N}$ is $\langle \mathfrak{X},\mathcal
D\rangle$\emph{-generic below} $B$ if $A\subseteq_\mathrm{Fin} B$ and for
every $\mathscr D\in\mathcal D$, there is $C\in \mathscr D$ such
that $A\subseteq_\mathrm{Fin} C$.
\end{definition}

Here one should think of the context of having some large family
$\mathfrak{Y}\in M\prec H_\lambda$ for a countable $M$, $\mathfrak{X}=\mathfrak{Y}\cap M$, and
$\mathcal D=\{\mathscr D\cap M\mid \mathscr D\in M\text{ and
}\mathscr D\text{ dense in }\mathfrak{Y}/\mathrm{Fin}\}$. We think of $A$ as coming
from the large family $\mathfrak{Y}$ and the requirement for $A$ to be
$\langle \mathfrak{X},\mathcal D\rangle$-generic is a strengthening of the
requirement to be $M$-generic.
\begin{lemma}\label{f:gen_elt}
Let $\mathfrak{X}$ be a countable family.  Assume that $B\in\mathfrak{X}/\mathrm{Fin}$ and
$G\subseteq \mathfrak{X}/\mathrm{Fin}$ is a $V$-generic filter containing $B$. Then in
$V[G]$, there is an infinite $A\subseteq\mathbb {N}$ such that
$A\subseteq_\mathrm{Fin} C$ for all $C\in G$. Furthermore, if $\mathcal D$
is the collection of dense subsets of $\mathfrak{X}/\mathrm{Fin}$ of $V$, then such an
$A$ is $\langle \mathfrak{X},\mathcal D\rangle$-generic below $B$.
\end{lemma}
\begin{proof}
Since $G$ is countable and directed in $V[G]$, there exists an
infinite $A\subseteq\mathbb {N}$ such that $A\subseteq_\mathrm{Fin} C$ for all $C\in
G$. For the ``furthermore" part, fix a dense subset $\mathscr D$ of
$\mathfrak{X}/\mathrm{Fin}$ in $V$. Since there is $C\in G\cap \mathscr D$, we have
$A\subseteq_\mathrm{Fin} C$. It is clear that $A\subseteq_\mathrm{Fin} B$ since
$B\in G$.
\end{proof}
\begin{lemma}\label{le:gen_elt}
Let $\mathfrak{X}_0\subseteq\mathfrak{X}_1\subseteq\cdots\subseteq
\mathfrak{X}_\xi\subseteq\cdots$ for $\xi<\omega_1$ be a continuous chain of
countable families of reals and let $\mathfrak{X}=\cup_{\xi<\omega_1} \mathfrak{X}_\xi$.
Assume that for every $\xi<\omega_1$, if $B\in \mathfrak{X}_\xi$ and $\mathcal
D$ is a countable collection of dense subsets of $\mathfrak{X}_\xi$, there is
$A\in \mathfrak{X}/\mathrm{Fin}$ that is $\langle \mathfrak{X}_\xi,\mathcal D\rangle$-generic
below $B$. Then $\mathfrak{X}$ is proper.
\end{lemma}
\begin{proof}
Fix a countable $M\prec H_\lambda$ such that $\langle \mathfrak{X}_\xi\mid
\xi<\omega_1\rangle\in M$. It suffices to show that generic
conditions exist for such $M$ since these form a club. By Lemma
\ref{le:intersect}, $\mathfrak{X}\cap M=\mathfrak{X}_\alpha$ where
$\alpha=\text{Ord}^M\cap \omega_1$. Fix $B\in\mathfrak{X}_\alpha$ and let
$\mathcal D=\{\mathscr D\cap M\mid \mathscr D\in M \text{ and
}\mathscr D \text{ dense in }\mathfrak{X}/\mathrm{Fin}\}$. By hypothesis, there is
$A\in\mathfrak{X}/\mathrm{Fin}$ that is $\langle\mathfrak{X}_\alpha,\mathcal D\rangle$-generic
below $B$. Clearly $A$ is $M$-generic.  Thus, we were able to find
an $M$-generic element below every $B\in M\cap \mathfrak{X}/\mathrm{Fin}$.
\end{proof}
We are finally ready to show how to force the existence of a proper
family of size $\omega_1$.
\begin{theorem}\label{th:forcing}
There is a generic extension of $V$ by a c.c.c.\ poset that
satisfies $\neg{\rm CH}$ and contains a proper family of reals of
size $\omega_1$.
\end{theorem}

\begin{proof}
First, note that we can assume without loss of generality that
$V\models \neg{\rm CH}$ since this is forceable by a c.c.c.\
forcing.

The forcing to add a proper family of reals will be a c.c.c.\ finite
support iteration $\mathbb{P}$ of length $\omega_1$. The iteration $\mathbb{P}$ will
add, step-by-step, a continuous chain
$\mathfrak{X}_0\subseteq\mathfrak{X}_1\subseteq\cdots\subseteq\mathfrak{X}_\xi\subseteq\cdots$ for
$\xi<\omega_1$ of countable arithmetically closed families such that
$\cup_{\xi<\omega_1} \mathfrak{X}_\xi$ will have the property of Lemma
\ref{le:gen_elt}. The idea will be to obtain generic elements for
$\mathfrak{X}_\xi$, as in Lemma \ref{f:gen_elt}, by adding generic filters.
Once $\mathfrak{X}_\xi$ has been constructed, I will force over $\mathfrak{X}_\xi/\mathrm{Fin}$
below every one of its elements cofinally often before the iteration
is over.  Every time such a forcing is done, I will obtain a generic
element for a new collection of dense sets. This element will be
added to $\mathfrak{X}_{\delta+1}$ where $\delta$ is the stage at which the
forcing was done.

Fix a bookkeeping function $f$ mapping $\omega_1$ onto
$\omega_1\times\omega$, having the properties that every pair
$\langle \alpha, n\rangle$ appears cofinally often in the range and
if $f(\xi)=\langle \alpha, n\rangle$, then $\alpha\leq \xi$.  Let
$\mathfrak{X}_0$ be any countable arithmetically closed family and fix an
enumeration $\mathfrak{X}_0=\{B_0^0,B_1^0,\ldots, B_n^0\ldots\}$. Each
subsequent $\mathfrak{X}_\xi$ will be created in $V^{\mathbb{P}_\xi}$. Suppose
$\lambda$ is a limit and $G_\lambda$ is generic for $\mathbb{P}_\lambda$. In
$V[G_\lambda]$, define $\mathfrak{X}_\lambda=\cup_{\xi<\lambda} \mathfrak{X}_\xi$ and
fix an enumeration
$\mathfrak{X}_\lambda=\{B_0^\lambda,B_1^\lambda,\ldots,B_n^\lambda,\ldots\}$.
Consult $f(\lambda)=\langle \xi,n\rangle$ and define
$\dot{\mathbb{Q}}_\lambda=\mathfrak{X}_\xi/\mathrm{Fin}$ below $B_n^\xi$. Suppose
$\delta=\beta+1$, then $\mathbb{P}_\delta=\mathbb{P}_\beta*\dot{\mathbb{Q}}_\beta$ where
$\dot{\mathbb{Q}}_\beta$ is $\mathfrak{X}_\xi/\mathrm{Fin}$ for some $\xi\leq\beta$ below one
of its elements.  In $V[G_\delta]=V[G_\beta][H]$, let
$A\subseteq_\mathrm{Fin} B$ for all $B\in H$ and define $\mathfrak{X}_\delta$ to be
the arithmetic closure of $\mathfrak{X}_\beta$ and $A$.  Also in
$V[G_\delta]$, fix an enumeration
$\mathfrak{X}_\delta=\{B_0^\delta,B_1^\delta,\ldots,B_n^\delta,\ldots\}$.
Consult $f(\delta)=\langle \xi,n\rangle$ and define
$\dot{\mathbb{Q}}_\delta=\mathfrak{X}_\xi/\mathrm{Fin}$ below $B_n^\xi$. At limits, use finite
support.

The poset $\mathbb{P}$ is c.c.c.\ since it is a finite support iteration of
c.c.c.\ posets (see \cite{jech:settheory}, p.\ 271).  Let $G$ be
$V$-generic for $\mathbb{P}$. It should be clear that we can use $G$ in
$V[G]$ to construct an arithmetically closed Scott set
$\mathfrak{X}=\cup_{\xi<\omega_1}\mathfrak{X}_\xi$. A standard nice name counting
argument shows that $(2^{\omega})^V=(2^{\omega})^{V[G]}$. Since we
assumed at the beginning that $V\models \neg{\rm CH}$, it follows
that $V[G]\models\neg{\rm CH}$.

Finally, we must see that $\mathfrak{X}$ satisfies the hypothesis of Lemma
\ref{le:gen_elt} in $V[G]$. Fix $\mathfrak{X}_\xi$, a set $B\in X_\xi$, and a
countable collection $\mathcal D$ of dense subsets of $\mathfrak{X}_\xi/\mathrm{Fin}$.
Since the poset $\mathbb{P}$ is a finite support c.c.c.\ iteration and all
elements of $\mathcal D$ are countable, they must appear at some
stage $\alpha$ below $\omega_1$. Since we force with $\mathfrak{X}_\xi/\mathrm{Fin}$
below $B$ cofinally often, we have added a $\langle \mathfrak{X}_\xi, \mathcal
D\rangle$-generic condition below $B$ at some stage above $\alpha$.
\end{proof}
\begin{corollary}
There is a generic extension of $V$ that satisfies  {\rm CH}  and
contains a proper family of reals of size $\omega_1$ other than
$\mathcal{P}(\mathbb {N})$.
\end{corollary}
\begin{proof}
As before, we can assume without loss of generality that $V\models
\neg{\rm CH}$. Force with $\mathbb{P}*\dot{\mathbb{Q}}$ where $\mathbb{P}$ is the forcing
iteration from Theorem \ref{th:forcing} and $\mathbb{Q}$ is the poset which
adds a subset to $\omega_1$ with countable conditions. Let $G*H$ be
$V$-generic for $\mathbb{P}*\dot{\mathbb{Q}}$, then clearly {\rm CH} holds in
$V[G][H]$. Also the family $\mathfrak{X}$ created from $G$ remains proper in
$V[G][H]$ since $\mathbb{Q}$ is a countably closed forcing, and therefore
cannot affect the properness of a family of reals.
\end{proof}
We can push this argument further to show that it is consistent with
{\rm ZFC} that there are \emph{continuum} many proper families of
reals of size $\omega_1$.

\begin{theorem}\label{th:continuum}
There is a generic extension of $V$ by a c.c.c.\ poset that
satisfies $\neg{\rm CH}$ and contains continuum many proper families
of reals of size $\omega_1$.
\end{theorem}
\begin{proof}
We start by  forcing ${\rm MA}+\neg{\rm CH}$. Since this can be done
by a c.c.c.\ forcing notion (\cite{jech:settheory}, \hbox{p. 272}),
we can assume without loss of generality that $V\models{\rm
MA}+\neg{\rm CH}$.

Define a finite support product $\mathbb{Q}=\Pi_{\xi<2^\omega}\mathbb{P}^\xi$ where
every $\mathbb{P}^\xi$ is an iteration of length $\omega_1$ as described in
Theorem \ref{th:forcing}. Since Martin's Axiom implies that finite
support products of c.c.c.\ posets are c.c.c.\ (see
\cite{jech:settheory}, \hbox{p. 277}), the product poset $\mathbb{Q}$ is
c.c.c.. Let $G\subseteq \mathbb{Q}$ be $V$-generic, then each
$G^\xi=G\upharpoonright \mathbb{P}^\xi$ together with $\mathbb{P}^\xi$ can be used
to build an arithmetically closed family $\mathfrak{X}^\xi$ as described in
Theorem \ref{th:forcing}.  Each such $\mathfrak{X}^\xi$ will be the union of
an increasing chain of countable arithmetically closed families
$\mathfrak{X}^\xi_\gamma$ for $\gamma<\omega_1$. First, I claim that all
$\mathfrak{X}^\xi$ are distinct. Fixing $\alpha<\beta$, I will show that
$\mathfrak{X}^\alpha\neq\mathfrak{X}^\beta$. Consider $V[G\upharpoonright
\beta+1]=V[G\upharpoonright \beta][G^\beta]$ a generic extension by
$(\mathbb{Q}\upharpoonright\beta)\times \mathbb{P}^\beta$. Observe that $\mathfrak{X}^\alpha$
already exists in $V[G\upharpoonright\beta]$.  Recall that to build
$\mathfrak{X}^\beta$, we start with an arithmetically closed countable family
$\mathfrak{X}_0^\beta$ and let the first poset in the iteration $\mathbb{P}^\beta$ be
$\mathfrak{X}_0^\beta/\mathrm{Fin}$. Let $g$ be the generic filter for
$\mathfrak{X}_0^\beta/\mathrm{Fin}$ definable from $G^\beta$. The next step in
constructing $\mathfrak{X}^\beta$ is to pick $A\subseteq\mathbb {N}$ such that
$A\subseteq_\mathrm{Fin} B$ for all $B\in g$ and define $\mathfrak{X}_1^\beta$ to be
the arithmetic closure of $\mathfrak{X}_0^\beta$ and $A$. It should be clear
that $g$ is definable from $A$ and $\mathfrak{X}_0^\beta$. Since $g$ is
$V[G\upharpoonright\beta]$-generic, it follows that $g\notin
V[G\upharpoonright\beta]$.  Thus, $A\notin
V[G\upharpoonright\beta]$, and hence $\mathfrak{X}^\beta\neq\mathfrak{X}^\alpha$. It
remains to show that each $\mathfrak{X}^\alpha$ is proper in $V[G]$.  Fix
$\alpha<2^\omega$ and let
$V[G]=V[G\upharpoonright\alpha][G^\alpha][G_\text{tail}]$ where
$G_\text{tail}$ is the generic for $\mathbb{Q}$ above $\alpha$.  By the
commutativity of products,
$V[G\upharpoonright\alpha][G^\alpha][G_\text{tail}]=
V[G\upharpoonright\alpha][G_\text{tail}][G^\alpha]$ and $G^\alpha$
is $V[G\upharpoonright\alpha][G_\text{tail}]$-generic.  Fix a
countable $M\prec H_\lambda^{V[G]}$ containing the sequence $\langle
\mathfrak{X}_\xi^\alpha\mid \xi<\omega_1\rangle$ as an element. By Lemma
\ref{le:intersect}, $M\cap \mathfrak{X}^\alpha$ is some $\mathfrak{X}_\gamma^\alpha$.
This is the key step of the proof since it allows us to know exactly
what $M\cap \mathfrak{X}^\alpha$ is, even though we know nothing about $M$.
Let $G_\xi^\alpha=G^\alpha\upharpoonright \mathbb{P}_\xi^\alpha$ for
$\xi<\omega_1$. Let $\mathcal D=\{\mathscr D\cap M\mid \mathscr D\in
M\text{ and }\mathscr D \text{ dense in }\mathfrak{X}^\alpha/\mathrm{Fin}\}$.  There
must be some $\beta<\omega_1$ such that $\mathcal D\in
V[G\upharpoonright\alpha][G_\text{tail}][G_\beta^\alpha]$.  By
construction, there must be some stage $\delta>\beta$ at which we
forced with $\mathfrak{X}_\gamma^\alpha/\mathrm{Fin}$ and added a set $A$ such that
$A\subseteq_\mathrm{Fin} B$ for all $B\in H$ where
$G_{\delta+1}^\alpha=G_\delta^\alpha
*H$.  Now observe that $H$ is $V[G\upharpoonright\alpha][G_\text{tail}]
[G^\alpha_\delta]$-generic for $\mathfrak{X}_\gamma^\alpha/\mathrm{Fin}$. Therefore
$H$ meets all the sets in $\mathcal D$.  So we can conclude that $A$
is $M$-generic.

A standard nice name counting argument will again show that
$(2^\omega)^V=(2^\omega)^{V[G]}$.  Thus, $V[G]$ satisfies $\neg{\rm
CH}$ and contains continuum many proper families of reals of size
$\omega_1$.
\end{proof}

Similar techniques allow us to force the existence of a piecewise
proper family of reals of size $\omega_2$.

\begin{lemma}\label{le:piecewise}
Let $\mathfrak{X}_0\subseteq\mathfrak{X}_1\subseteq\cdots\subseteq
\mathfrak{X}_\xi\subseteq\cdots$ for $\xi<\omega_1$ be a continuous chain of
countable families of reals and let $\mathfrak{X}=\cup_{\xi<\omega_1} \mathfrak{X}_\xi$.
Assume that for every $\xi<\omega_1$, if $B\in \mathfrak{X}_\xi$ and $\mathcal
D$ is a countable collection of dense subsets of $\mathfrak{X}_\xi$, there is
$A\in \mathfrak{X}/\mathrm{Fin}$ that is $\langle \mathfrak{X}_\xi,\mathcal D\rangle$-generic
below $B$. Then $\mathfrak{X}$ is proper and $\mathfrak{X}$ remains proper after forcing
with any absolutely c.c.c.\ poset.
\end{lemma}
\begin{proof}
The proof is a straightforward modification of the proof of Theorem
\ref{th:oldreals}. Let $\mathbb{P}$ be an absolutely c.c.c.\ poset and
$g\subseteq \mathbb{P}$ be $V$-generic. We need to show that $\mathfrak{X}$ is proper
in $V[g]$. Fix a countable $M[g]\prec H_\lambda [g]$ in $V[g]$ such
that $\langle \mathfrak{X}_\xi:\xi<\omega_1\rangle, \mathbb{P}\in M[g]$ and $M\subseteq V$.
Let $\mathfrak{X}_\alpha=M\cap \mathfrak{X}$ and let $\mathcal D=\{\mathscr D\cap
\mathfrak{X}_\xi\mid\mathscr D\in M\text{ and }\mathscr D\text{ dense in
}\mathfrak{X}\}$. Observe that $\mathcal D\subseteq V$ and $|\mathcal
D|=\omega$. Define $\mathcal E$ as in proof of Theorem
\ref{th:oldreals}. Now choose $A\in \mathfrak{X}$ that is $\langle
\mathfrak{X}_\alpha,\mathcal E\rangle$-generic in $V$. It follows that $A$ is
$M$-generic. Next proceed exactly as in the proof of Theorem
\ref{th:oldreals}, using the fact that $\mathbb{P}$ is absolutely c.c.c.\ in
the final stage of the argument.
\end{proof}
\begin{theorem}\label{th:piecewise}
There is a generic extension of $V$ by a c.c.c.\ poset which
contains a piecewise proper family of reals of size $\omega_2$.
\end{theorem}
\begin{proof}
We will define a c.c.c.\ forcing iteration $\mathbb{Q}_{\omega_2}$ of length
$\omega_2$ to accomplish this. Let $\mathbb{Q}_0$ be the forcing to add a
proper family $\mathfrak{X}_0$ of size $\omega_1$ (Theorem \ref{th:forcing}).
At the $\alpha^{\text{th}}$-stage, force with the poset to add a
proper family $\mathfrak{X}_{\alpha}\supseteq \cup_{\beta<\alpha}\mathfrak{X}_\beta$.
Observe here, that the poset from Theorem \ref{th:forcing} can be
very easily modified to the poset which adds a proper family
extending any family of reals from the ground model. Let $G\subseteq
\mathbb{Q}_{\omega_2}$ be $V$-generic. I claim each $\mathfrak{X}_\alpha$ remains
proper in $V[G]$. Fix $\mathfrak{X}_\alpha$ and factor the forcing
$\mathbb{Q}_{\omega_2}=\mathbb{Q}_\alpha*\mathbb{Q}_\text{tail}$. The family $\mathfrak{X}_\alpha$ is proper
in $V[G_\alpha]$ and $\mathbb{Q}_\text{tail}$ is absolutely c.c.c.\ in
$V[G_\alpha]$. The poset $\mathbb{Q}_\text{tail}$ is absolutely c.c.c\ since the
forcing to add a proper family is a finite support iteration of
countable posets. Thus, by Lemma \ref{le:piecewise}, $\mathfrak{X}_\alpha$
remains proper in $V[G_\alpha][G_\text{tail}]$. Thus,
$\mathfrak{X}=\cup_{\alpha<\omega_2} \mathfrak{X}_\alpha$ is clearly piecewise proper.
\end{proof}
By exactly following the proof of Theorem \ref{th:continuum}, we can
extend Theorem \ref{th:piecewise} to obtain:
\begin{theorem}
There is a generic extension of $V$ by a c.c.c.\ poset which
contains continuum many piecewise proper families of reals of size
$\omega_2$.
\end{theorem}

By Enayat's \cite{enayat:endextensions} example of a non-proper
arithmetically closed family of size $\omega_1$, we know that there
are piecewise proper families that are not proper. This follows by
recalling that arithmetically closed families of size $\omega_1$ are
trivially piecewise proper. It is not clear whether every proper
family has to be piecewise proper. In particular, it is not known
whether $\mathcal{P}(\mathbb {N})$ is piecewise proper. It follows that
$\mathcal{P}(\mathbb {N})$ \emph{can be} piecewise proper from the proof of
Theorem \ref{th:piecewise} since we can modify the construction to
end up with $\mathfrak{X}=\mathcal{P}(\mathbb {N})$.

Finally, I will discuss a possible construction for proper families
under {\rm PFA}. The idea is, in some sense, to mimic the forcing
iteration like that of Theorem \ref{th:forcing} in the ground model.
Unfortunately, the main problem with the construction is that it is
not clear whether we are getting the whole $\mathcal{P}(\mathbb {N})$. This
problem never arose in the forcing construction since we were
building families of size $\omega_1$ and knew that the continuum was
larger than $\omega_1$.  I will describe the construction and a
possible way of ensuring that the resulting family is not
$\mathcal{P}(\mathbb {N})$.

Fix an enumeration $\{\langle A_\xi,B_\xi\rangle\mid \xi<\omega_2\}$
of $\mathcal{P}(\omega)\times \mathcal{P}(\omega)$. Also fix a bookkeeping
function $f$ from $\omega_2$ onto $\omega_2$ such that each element
appears cofinally in the range. I will build a family $\mathfrak{X}$ of size
$\omega_2$ as the union of an increasing chain of arithmetically
closed families $\mathfrak{X}_\xi$ for $\xi<\omega_2$. Start with any
arithmetically closed family $\mathfrak{X}_0$ of size $\omega_1$. Suppose we
have constructed $\mathfrak{X}_\beta$ for $\beta\leq\alpha$ and we need to
construct $\mathfrak{X}_{\alpha+1}$. Consult $f(\alpha)=\gamma$ and consider
the pair $\langle A_\gamma,B_\gamma\rangle$ in the enumeration of
$\mathcal{P}(\omega)\times \mathcal{P}(\omega)$. First, suppose that
$A_\gamma$ codes a countable family $\mathfrak{Y}\subseteq \mathfrak{X}_\alpha$ and
$B_\gamma$ codes a countable collection $\mathcal D$ of dense
subsets of $\mathfrak{Y}$. Let $G$ be some filter on $\mathfrak{Y}$ meeting all sets in
$\mathcal D$ and let $A\subseteq_\mathrm{Fin} C$ for all $C\in G$. Define
$\mathfrak{X}_{\alpha+1}$ to be the arithmetic closure of $\mathfrak{X}_\alpha$ and $A$.
If the pair $\langle A_\gamma,B_\gamma\rangle $ does not code such
information, let $\mathfrak{X}_{\alpha+1}=\mathfrak{X}_\alpha$.  At limit stages take
unions.

I claim that $\mathfrak{X}$ is proper.  Fix some countable $M\prec H_\lambda$
containing $\mathfrak{X}$. Let $\mathfrak{Y}=M\cap \mathfrak{X}$ and let $\mathcal D=\{\mathscr
D\cap M\mid \mathscr D\in M\text{ and }\mathscr D \text{ dense in
}\mathfrak{X}/\mathrm{Fin} \}$. There must be some $\gamma$ such that $\langle
A_\gamma, B_\gamma\rangle$ codes $\mathfrak{Y}$ and $\mathcal D$. Let $\delta$
such that $M\cap \mathfrak{X}$ is contained in $\mathfrak{X}_\delta$, then there must be
some $\alpha>\delta$ such that $f(\alpha)=\gamma$. Thus, at stage
$\alpha$ in the construction we considered the pair $\langle
A_\gamma,B_\gamma\rangle$. Since $\alpha>\delta$, we have $M\cap
\mathfrak{X}=M\cap \mathfrak{X}_\alpha$.  It follows that at stage $\alpha$ we added an
$M$-generic set $A$ to $\mathfrak{X}$.

A way to prove that $\mathfrak{X}\neq \mathcal{P}(\mathbb {N})$ would be to show that some
fixed set $C$ is not in $\mathfrak{X}$.  Suppose the following question had a
positive answer:
\begin{question}\label{con:proper}
Let $\mathfrak{X}$ be an arithmetically closed family such that $C\notin \mathfrak{X}$
and $\mathfrak{Y}\subseteq \mathfrak{X}$ be a countable family. Is there a
$\mathfrak{Y}/\mathrm{Fin}$-name $\dot{A}$ such that $1_{\mathfrak{Y}/\mathrm{Fin}}\Vdash ``
\dot{A}\subseteq_\mathrm{Fin} B\text{ for all }B\in \dot{G}\text{ and
}\check{C}$ is not in the arithmetic closure of $\dot{A}\text{ and
}\check{\mathfrak{X}}$"?
\end{question}

Assuming that the answer to Question \ref{con:proper} is positive,
let us construct a proper family $\mathfrak{X}$ in such a way that $C$ is not
in $\mathfrak{X}$.  We will carry out the above construction being careful in
our choice of the filters $G$ and elements $A$. Start with $\mathfrak{X}_0$
that does not contain $C$ and assume that $C\notin \mathfrak{X}_\alpha$.
Suppose the pair $\langle A_\gamma,B_\gamma\rangle$ considered at
stage $\alpha$ codes meaningful information.  That is, $A_\gamma$
codes a countable family $\mathfrak{Y}\subseteq \mathfrak{X}_\alpha$ and $B_\gamma$
codes a countable collection $\mathcal D$ of dense subsets of $\mathfrak{Y}$.
Choose some transitive $N\prec H_{\omega_2}$ of size $\omega_1$ such
that $\mathfrak{X}_\alpha$, $\mathfrak{Y}$, and $\mathcal D$ are elements of $N$. Since
we assumed a positive answer to Question \ref{con:proper},
$H_{\omega_2}$ satisfies that there exists a $\mathfrak{Y}/\mathrm{Fin}$-name
$\dot{A}$ such that $1_{\mathfrak{Y}/\mathrm{Fin}}\Vdash ``\dot{A}\subseteq_\mathrm{Fin}
B\text{ for all }B\in \dot{G}\text{ and }\check{C}$ is not in the
arithmetic closure of $\dot{A}\text{ and }\check{\mathfrak{X}_\alpha}$". But
then $N$ satisfies the same statement by elementarity. Hence there
is $\dot{A}\in N$ such that $N$ satisfies $1_{\mathfrak{Y}/\mathrm{Fin}}\Vdash
``\dot{A}\subseteq_\mathrm{Fin} B\text{ for all }B\in \dot{G}\text{ and
}\check{C}$ is not in the arithmetic closure of $\dot{A}\text{ and}$
$\check{\mathfrak{X}_\alpha}$". Now use {\rm PFA} to find an $N$-generic
filter $G$ for $\mathfrak{Y}/\mathrm{Fin}$. Since $G$ is fully generic for the model
$N$, the model $N[G]$ will satisfy that $C$ is not in the arithmetic
closure of $\mathfrak{X}_\alpha$ and $A=\dot{A}_G$. Thus, it is really true
that $C$ is not in the arithmetic closure of $\mathfrak{X}_\alpha$ and $A$.
Since $G$ also met all the dense sets in $\mathcal D$ and
$A\subseteq_\mathrm{Fin} B$ for all $B\in G$, we can let $\mathfrak{X}_{\alpha+1}$ be
the arithmetic closure of $\mathfrak{X}_\alpha$ and $A$.  Thus, $C\notin
\mathfrak{X}_{\alpha+1}$. We can conclude that $C\notin\mathfrak{X}$.
\section{Questions}
\begin{question}
Can {\rm ZFC} or {\rm ZFC} + {\rm PFA} prove the existence of an
uncountable proper family of reals other than $\mathcal{P}(\mathbb {N})$?
\end{question}
\begin{question}
Can {\rm ZFC} or {\rm ZFC} + {\rm PFA} prove the existence of a
piecewise proper family of size $\omega_2$?
\end{question}
\begin{question}
Is it consistent with {\rm ZFC} that there are proper families of
reals of size $\omega_2$ other than $\mathcal{P}(\mathbb {N})$?
\end{question}
\begin{question}
What is the answer to Question \ref{con:proper}?
\end{question}
\begin{question}
Can $\mathcal{P}(\mathbb {N})$ be non-piecewise proper?
\end{question}
\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


text:
Best Online Sportsbooks - Betting News - Steelers vs. Colts Betting NFL Week 12, BetUS Odds and Analysis
Steelers vs. Colts Betting NFL Week 12, BetUS Odds and Analysis
It's not exactly a can't-miss event, but Monday night, the Pittsburgh Steelers (3-7) visit Lucas Oil Stadium to oppose the 4-6-1 Indianapolis Colts. With both sides needing the game, at least there's some incentive, The sides are good defensively, but struggle mightily with the ball. It's the Colts, as 2½-point favorites (total of 39) according to BetUS sportsbook NFL odds.
Pittsburgh Steelers vs. Indianapolis Odds by BetUs
Steelers +2½ ( -110 ) 39 ( -110 ) +120
Colts -2½ ( -110 ) 39 ( -110 ) -140
Steelers Lost Despite Scoring
The Pittsburgh Steelers have, without question, one of the worst offenses in the NFL. So, they went out and scored 30 points last Sunday, and lost, anyway. The team's 37-30 defeat to the visiting Cincinnati Bengals was the third loss in their last four. However, if the usually-moribund offense can build off that performance, perhaps they can find a way against Indianapolis.
Pittsburgh did get reigning Defensive Player of the Year TJ Watt back last week. Watt immediately showed his value, harassing Cincinnati quarterback while adding a blink-or-you'll-miss-it interception. Force turnovers and providing a short field is never bad, especially with a pathetic offense.
Pittsburgh's bottom 10 in passing yards per game, rushing yards per game, points per game (17) and points allowed per game, Watt will likely change the defensive numbers, but quarterback Kenny Pickett had made rookie mistakes (three touchdowns, eight interceptions) along the way.
Colts Flipping Series Script?
This has been a one-sided rivalry in recent decades. Indianapolis hasn't defeated Pittsburgh in 14 years, 0-7 since 2008. In a series dating to 1950, Pittsburgh has a 26-6 advantage. Suggesting the Colts are overdue is a bit of an understatement.
The Colts are coming off of a late-game, 17-16 loss to the Philadelphia Eagles. It was a game they had won, leading until the final 80 seconds.
Indy doesn't have a great offense, either, the second-worst points per-game average (15.7) the NFL. The Colts have worked around injuries to star running back Johnathan Taylor, while quarterback Matt Ryan, reinserted after a coaching change, has not matched his Atlanta resume (1p touchdowns nine interceptions. 29 sacks). Former Colt standout Jeff Saturday was a winner at Las Vegas in debut, but the jury is still out on whether the surprise hire is an improvement over Frank Reich as coach.
If you're into predicting the balance of each team's season, Pittsburgh's final win total is projected at 5½, while the Colts are at 6½. Those per NFL win odds.
Steelers in Nail-Biter
Acton will likely be under as far as the total is concerned, as two of the worst offenses in the NFL are on display. However, as the Steelers find some rhythm with Watt on the defense, and the Colts consistently playing close games, expect this game to barely hit the over, since 39 is a very low total.
Pittsburgh isn't great by any means, but is taken to win a nail-biter. Defense is better, but will have to contain Jonathan Taylor in the backfield. Take Pittsburgh +2½ in this game.

redpajama_set_name: RedPajamaCommonCrawl


Description "In compliance with the North Carolina Environmental Policy Act."; "October 2014"--Green sheet.
Physical Characteristics 1 volume (various pagings) : color maps (folded) ; 28 cm.
Digital Characteristics-A 36.1 MB; 55 p.


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Hindi Dj Songs 2015 Mp3 du Latest Punjabi Songs Mp3 Bollywood. Dj-pendu.com is tracked by us since November, 2015. Over the time it has been ranked as high as 1 201 081 in the world, while most of its traffic comes from India, where it reached as high as 68 741 position.


text:
Liolaemus yatel est une espèce de sauriens de la famille des Liolaemidae.

Répartition
Cette espèce est endémique de la province de Santa Cruz en Argentine.

Étymologie
Le nom spécifique yatel vient du tehuelche yatel, sol rocailleux, en référence à l'habitat de cette espèce.

Publication originale
 Abdala, Procopio, Stellatelli, Travaini, Rodríguez & Ruiz Monachesi, 2014 : New Patagonian species of Liolaemus (Iguania: Liolaemidae) and novelty in the lepidosis of the southernmost lizard of the world: Liolaemus magellanicus. Zootaxa, , .

Liens externes

Notes et références

Saurien (nom scientifique)
Liolaemidae
Faune endémique d'Argentine

redpajama_set_name: RedPajamaWikipedia


text:
Articles » Biographies
Nelle Nichols Peters
Jason Roe
Kansas City Public Library
Nelle Nichols Peters is known as a pathbreaking female architect, designer of nearly 1,000 local buildings, and one of the most prolific architects in Kansas City during the 1920s. Despite the fact that many of these buildings still occupy prominent locations, especially near the Country Club Plaza, Nelle Peters remains a relatively obscure figure in Kansas City history.
Nelle Peters. Courtesy of the State Historical Society of Missouri.
Born Nelle Elizabeth Nichols on December 11, 1884, in Niagara, North Dakota, she spent her childhood on the prairie until the Nichols family moved to Storm Lake, Iowa, in 1899. There she attended Buena Vista College, where she studied mathematics and art. Combining these skills, she attained employment as a draftswoman at the Eisentraut, Colby and Pottenger architectural firm in Sioux City, Iowa. She began to learn about architectural design while earning three dollars per week at the firm.
Nichols came to Kansas City in 1909 when the firm transferred her. Few developers wanted to hire a female architect in the 1910s, but she did manage to take on some of her own architectural projects outside the firm. In these projects, she demonstrated a talent for designing apartment buildings. In 1911 she married William H. Peters, a Kansas City Terminal Railroad designer, and took the name Nelle Nichols Peters.
Nelle Peters's career blossomed after she connected with the Phillips Building Company in 1913. Charles E. Phillips, local real estate developer and owner of the company, hired Peters as architect on dozens of projects in the ensuing years. She specialized in apartment buildings and hotels, but also designed corporate offices and single-family residences.
Peters developed a reputation not just for the rarity of female architects, but for her talent. Her signature designs placed apartment complexes around a central courtyard. Her careful attention to detail brought her some national recognition within the architectural profession. She minimized the total area taken up by her designs while still giving them an open feeling and functional spaces inside, especially with her usable kitchenette designs.
In 1923, Nelle and William Peters divorced, but Nelle embarked on the high point of her career. Peters's designs were not overly ornamental, but they were simultaneously attractive and functional. Many still stand today. Among her most notable designs were the Luzier Cosmetics Building at 3216 Gillham Plaza, and the Ambassador Hotel at 3560 Broadway, which was the largest apartment-hotel in Kansas City when it was completed in 1924. Other prominent buildings that she designed included the Valentine Hotel (3724 Broadway), the Belleclair Apartments (401 E. Armour St), the Del Monte Apartments (200 W. Armour), the Melbourne Apartments (303 Brush Creek), the Robert Louis Stevenson Apartments (4804 Jefferson Street), the Mark Twain, Robert Browning, and Oliver Wendell Holmes apartments (on the Plaza), and Charles Phillips's personal home (5825 Overhill Road).
Unfortunately, the drop in apartment construction during the Great Depression and Second World War curtailed Nelle Peters's career in the 1930s and 1940s. She remained in the business, but had to work as a seamstress to augment her income. She finally retired in 1967, although by that year it appears that she had not had a contract in nearly 11 years. Little else is known of Nelle Peters, who was at once one of the most prolific architects and an obscure figure in Kansas City history. When she died on October 7, 1974, at the age of 90, virtually no one acknowledged the significance of her accomplishments. While Nelle Nichols Peters remains largely invisible, many of the buildings she designed still contribute to character of the city's landscape.
Show Me Missouri Women
Kansas City Women of Independent Minds
"Biography of Nelle Nichols Peters (1884-1974)"
"The Architectural Career of Nelle Peters"
This article has been adapted from an article published at KChistory.org.


Experience. It's what sets your business apart from the shop down the street. And experience is what makes Sikkens the world's premium refinishing system.
to North America in 1983, we've offered the industry's most technologically advanced and innovative products that ensure a glossy, durable finish the first time, every time.
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text:
Q: Send String to webserver in iOS? I want to send a String (a long String, it's about 600 characters) to webserver (PHP webserver). Can i use NSURLRequest and NSURLConnection to send it to my PHP webserver?

A: exactly what you are looking for
http://forums.macrumors.com/showthread.php?t=689884
NSString *params = [[NSString alloc] initWithFormat:@"foo=longstringdatagoeshere"];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:[params dataUsingEncoding:NSUTF8StringEncoding]];
[[NSURLConnection alloc] initWithRequest:request delegate:self];

remember to implement the delegate methods
http://developer.apple.com/mac/library/documentation/Cocoa/Reference/Foundation/Classes/NSURLConnection_Class/Reference/Reference.html#//apple_ref/doc/uid/20001697-DontLinkElementID_17

redpajama_set_name: RedPajamaStackExchange


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text:
We get the latest from Las Vegas as the city witnesses the worst mass shooting in recent US history. Meanwhile, will Catalonia attempt to declare independence from Spain? Plus: infighting in the UK's ruling party and the woman behind some of the world's top podcasts.
Las Vegas shooting
At least 50 people have been killed and hundreds injured in the worst mass shooting in recent US history. Casey Morell from Nevada Public Radio tells us more about the attack.
Share chapter 1
Andrew Mueller
News from around the world, analysis and debate.
The Briefing - latest episodes
Joe Biden meets with Japan's prime minister Fumio Kishida; Austria is set to make coronavirus vaccines compulsory; and we remember the life and music of Brazil's legendary samba singer Elza Soares.
Thursday 20 January
The US Supreme Court rejects a final attempt by Donald Trump to block access to documents relating to the Capitol riots. Plus: business news, the Latin American front pages and this week's Global Countdown.
US secretary of state Antony Blinken visits Kyiv over brewing tensions on Ukraine's border. Plus: The Turkish president's offensive against K-pop and Netflix, and the latest in business, with Bloomberg, and urbanism.
Tuesday 18 January
Germany's foreign minister Annalena Baerbock touches down in Moscow, the second day of the World Economic Forum gets underway and our Asia editor James Chambers brings us an update. Plus: Brazil's booming pet industry.
Monday 17 January
North Korea fires its fourth missile test in a month, Québec's premier wants to tax those who refuse a vaccine and Monocle's Chiara Rimella talks licence fees and Sotheby's IPO. Plus: the latest in business and the WEF.
UK prime minister Boris Johnson faces fresh calls to resign after reports of yet another two parties at 10 Downing Street. Plus: US president Joe Biden pushes for new voting-rights legislation, our weekly What We Learned…
What might the impasse in talks between Nato and Russia mean for Ukraine? Plus: economics and the French election, Monocle's Nolan Giles has an update from Pitti Uomo and we head to Brazil for The Global Countdown.
Monocle 24's health expert Chris Smith explains the WHO's concerns over a huge spike in Omicron infections. Plus: Boris Johnson and the Downing Street party scandal, Austria's communists and the week's design news.
We bring you the latest revelations from the "bring-your-own-booze" lockdown party for the UK government in Downing Street. Plus: an update on the scandal engulfing Denmark's spy agencies, a roundup of stories from Nicar…
What do the US and Russia want from diplomatic talks in Geneva? Plus: a court in Myanmar hands the country's ousted leader Aung San Suu Kyi another jail term, we explore President Macron's relationship with the French and…
Friday 7 January
Former diplomat David Warren and Open Democracy's Paul Rogers explain the significance of the alliance between Japan and the US. Plus: what the rest of the world can learn from Bristol's reappraisal of its slavery heritage…
Why has Moscow sent its elite troops to quell violence in Kazakhstan? Plus: Monocle's Chris Lord and Sasha Issenberg on the anniversary of the 6 January Capitol riot, the day's business headlines and Monocle's Lucinda…
Wednesday 5 January
We have the latest from Seoul as North Korea reportedly tests another ballistic missile. Plus: Israel considers lifting travel curbs, foreign policy under Germany's new chancellor Olaf Scholz and Monocle's Chiara Rimella…
We ask what's next for Hong Kong as another independent media outlet closes its doors. Plus: Sweden and Finland refuse to rule out joining Nato; the day's business headlines from Bloomberg; and Josh Cowls tells us about a…
Monday 3 January
Could Omicron help some nations to reach herd immunity? Plus: Turkey's currency crisis, Monocle's Guy De Launey on Croatia's attempts to lure back its expat community and space scientist David Whitehouse on the James Webb…


May 31, 2011 • The NBA Finals begin Tuesday night with the Miami Heat taking on the Dallas Mavericks. The last time the two teams met in the championship round, the Heat was victorious.
May 30, 2011 • Freelance reporter Tom Finn, in the Yemeni capital, Sanaa, discusses the latest deadly clashes between anti-government protesters and security forces. Warplanes reportedly bombed the southern cit of Zinjibar. And in the city of Taiz, security forces reportedly killed at least 15 protesters.
May 30, 2011 • In Vietnam, black sunglasses are now banned for its police officers, along with smoking and chatting on the job. And traffic cops have been order to stop hiding behind trees to catch errant drivers. Vietnamese officials hope the changes will improve the police force's notoriously bad reputation.
May 30, 2011 • At the Knoxville Zoo, Al the giant tortoise hadn't seen a female turtle in almost 30 years. It's been about 20 years for his 90-year-old zoomate, Tex. Last Wednesday, Al quickly mated with a visiting female named Patches. So far, Tex is taking it slow with his potential mates.
May 30, 2011 • In Afghanistan, the U.S. death toll this year has topped 150. One of those was Staff Sgt. Edward Mills Jr. He was one of six soldiers who died last Thursday in Kandahar province by an improvised explosive device. Mills, of New Castle, Pa., was a "natural-born leader," says his brother.
May 30, 2011 • The tiny Persian Gulf state of Qatar is hoping to go where others haven't gone before. This coming weekend, officials at the Doha Trade Fair are organizing an attempt to build the world's largest Monopoly game. The board will be well over 1,500 square feet.
May 30, 2011 • Debt inspectors from the International Monetary Fund, the European Central Bank and the European Commission will be checking this week to see whether Greece has met the criteria for the next loan installment from a huge bailout. The deal, worked out last year, is worth about $155 billion.
May 27, 2011 • A Democrat in western New York won election to Congress this week in a district that had been held by the Republican Party for four decades. The deciding issue may have been the GOP's plan to revamp Medicare. In addition, more names are surfacing as potential Republican presidential candidates.
May 26, 2011 • From the initial protests in Cairo and the celebrations in Tahrir Square to being one of the first journalists to enter Libya, NPR's Lourdes Garcia-Navarro has been reporting the stories of revolution sweeping North Africa and the Middle East. While on a break from being overseas, she discusses what she's witnessed.
May 25, 2011 • More tornadoes raked across the south and Midwest Tuesday night and early Wednesday. Oklahoma was particularly hard hit as a series of strong twisters skipped across the state.
May 24, 2011 • As his European trip continues, President and Mrs. Obama are in London. A dense ash cloud from an Icelandic volcano forced the president to cut short his visit to Ireland and arrive early in London. While they are serious working days ahead, this day is one of pomp and ceremony.
May 23, 2011 • The poet's latest collection, Fall Higher, was published just days after he received a life-saving heart transplant. Now, Young is on the mend, but his book recalls when he was staring down death.
May 16, 2011 • More than a dozen people were killed Sunday as Israeli troops clashed with Arab protesters along three hostile borders, including the frontier with Syria. The violence came amid a wave of demonstrations marking a Palestinian day of mourning for their defeat at Israel's hands in 1948.
May 16, 2011 • The Endeavour's earlier launch was delayed because of technical issues. After this mission, it will become a museum exhibit in Los Angeles. NASA has just one shuttle launch left — the last flight of Atlantis is scheduled for this summer.
May 16, 2011 • Over the weekend, Mike Huckabee — Fox News host and former governor of Arkansas – announced he won't be running for president. Meanwhile, former House Speaker Newt Gingrich is making a big push in Iowa, just days after announcing his bid. And former Utah Governor Jon Huntsman is testing the waters in New Hampshire.


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A recent attack against Morgan Stanley that exposed hundreds of thousands of customer accounts was an inside job, a threat experts say is nearly impossible to stop.
Companies spend billions of dollars each year to protect from determined hackers attacking from across the Internet, but experts warn they shouldn't ignore a closer threat they aren't even ready for: Inside jobs.
Morgan Stanley, one of the world's largest financial services firms, revealed Monday its customer information was breached. But it wasn't the result of determined hackers or sophisticated email attacks. Instead, Morgan Stanley said it was an employee who stole data from more than 350,000 customer accounts.
The move is a wake-up call to companies, which spent an estimated $71.1 billion in 2014 on cybersecurity, up nearly 8 percent from the year before. And while hackers have successfully attacked large companies like JPMorgan, Target and Home Depot, experts warn employees pose just as much a threat, whether they act intentionally or by accident.
While the cybersecurity industry is devising an ever growing list of technology to protect from intrusions, it turns out there's relatively little that can be done to stop an insider who already has access to a company's otherwise highly protected data.
"There's always going to be a way, just like with hacking, for insider attacks to happen," said Lucas Zaichkowsky, who used to manage computers at a major credit card processor and is now a security expert at Resolution1.
Attacks by insiders are often characterized in three ways: They're hard to detect and don't happen often. But when an attack does come from the inside, it can be devastating. Security researchers at the Ponemon Institute say 88 percent of IT pros surveyed say they struggle to identify insider attacks, and security consultants at SpectorSoft say less than half of companies are even capable of noticing.
Few companies publicly disclose these types of attacks, and when they do they rarely estimate the damage. SpectorSoft said insider attacks -- 35 percent of all those committed -- cost US companies $40 billion in 2013 alone.
Despite the challenges in detecting and blocking insider attacks, there are ways that can help companies reduce the risk of insider attacks.
First, experts recommend companies tighten restrictions on highly sensitive data, locking files behind passcodes and security systems only employees and trusted business partners who must have access actually do. One way to do that is with cryptographic computer code, which jumbles a file's contents using an algorithm that only those with the proper computer keys have.
If the information does have to be accessible on the company network, "try to come up with a data policy that segregates it," said Andrew Conway, a site and data breach expert for security company Cloudmark.
Companies also need to monitor the actions of the employees who do have access, to ensure the data isn't copied or destroyed without approval. Many attacks have been spotted by warning systems, but the alarms went unnoticed.
Other long-standing techniques include preventing files from being copied to USB by physically blocking USB ports with liquid cement, and removing cameras from the screens of laptop and desktop computers. Some companies even use "air-gapped" computers -- machines which are neither connected to the Internet nor to other computers.
The US government has gone to some lengths to ensure certain kinds of data are better protected than others. Nuclear facilities, for example, have used air-gapped computers for years. There are also laws governing how medical data is stored and accessed, requiring companies to keep extensive logs every time the files are read.
Another option for companies is to bring on a chief security officer, someone who understands and knows how to balance security with a company's day-to-day business, and sets rules for how files are stored and accessed. Ponemon backed up this assertion in a 2013 study which found the cost per record exposed in a breach goes down when a company has hired such a person.
Even if companies implement all those measures, experts say they can't entirely secure their systems from a determined insider.
Consider famous information leakers like Edward Snowden and Chelsea Manning, each of whom accessed and leaked thousands of classified government documents. In both cases, they were able to circumvent some of the US government's most secure computer systems by virtue of being on the inside.
"If the NSA can't prevent an insider breach, then how is an enterprise company going to stop one?" Conway said.


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text:
Humana Festival 2008: The Complete Plays
by Lee Blessing, Gina Gionfriddo, Michael Friedman, Jim Lewis, Steven Cosson, Marc Bamuthi Joseph, Carly Mensch, Jennifer Haley, Zakiyyah Alexander, Rolin Jones, Jon Spurney, Alice Tuan, Daryl Watson, Marisa Wegrzyn, Ken Weitzman, Mike Lew, Naomi Wallace, Elaine Jarvik, M. Thomas Cooper
Productions (0)
Drama/Comedy
Humana Festival 2008: The Complete Plays brings together all eleven scripts from the 2008 Humana Festival of New American Plays, the 32nd annual cycle of world premiere productions staged at Actors Theatre of Louisville. The seven full-length plays and four ten-minute plays represent an exceptional array of work by some of the most exciting new voices in American theater -- from a scathing comedy about a blind date gone awry (Becky Shaw); to a documentary drama about the evangelical phenomenon in Colorado Springs (This Beautiful City); to a suburban street being undermined by zombies from a video game (Neighborhood 3: Requisition of Doom). Humana Festival 2008: The Complete Plays features a diverse selection of theatrical works that challenge and captivate audiences through movement, song, and bold explorations of what it means to be human. Read More
Read For FREE Add Book(s) To Cart
Standard Edition $19.95
NOTE: This book contains 11 plays. To perform any of the plays, each must be licensed separately.
Humana Festival 2008: The Complete Plays brings together all eleven scripts from the 2008 Humana Festival of New American Plays, the 32nd annual cycle of world premiere productions staged at Actors Theatre of Louisville. The seven full-length plays and four ten-minute plays represent an exceptional array of work by some of the most exciting new voices in American theater -- from a scathing comedy about a blind date gone awry (Becky Shaw); to a documentary drama about the evangelical phenomenon in Colorado Springs (This Beautiful City); to a suburban street being undermined by zombies from a video game (Neighborhood 3: Requisition of Doom). Humana Festival 2008: The Complete Plays features a diverse selection of theatrical works that challenge and captivate audiences through movement, song, and bold explorations of what it means to be human.
To purchase this book of 11 plays, click "Order this play" above. To perform an individual play, click on its title below:
All Hail Hurricane Gordo by Carly Mensch
Becky Shaw by Gina Gionfriddo
Dead Right by Elaine Jarvik
Game On by Zakiyyah Alexander, Rolin Jones, Jon Spurney, Alice Tuan, Daryl Watson, Marisa Wegrzyn, and Ken Weitzman
Great Falls by Lee Blessing
In Paris You Will Find Many Baguettes but Only One True Love by Michael Lew
Neighborhood 3: Requisition of Doom by Jennifer Haley
One Short Sleepe by Naomi Wallace
the break/s by Marc Bamuthi Joseph
This Beautiful City by Steven Cosson, Michael Friedman, and Jim Lewis
Tongue, Tied by M. Thomas Cooper
"Humana is the most invigorating theatre weekend of the year."
"The festival has gained a reputation for annually assembling a group of first-rate new scripts that invariably make their way to college, community and regional theatres around the globe and often to the New York off-Broadway or Broadway stage as well."
"The schisms in American society, both macro and micro, were on vivid display in this year's Humana Festival of New American Plays. The divisions between the religious right and the secular left, the tech-fueled widening of the generation gap and the ever-relevant question of what makes a modern marriage function smoothly were among the themes explored."
Too Much Light Makes The Baby Go Blind (30 plays in 60 minutes)
by Greg Allen
The Bold, The Young, and The Murdered
Heddatron
by Elizabeth Meriwether
Holmes and Watson: The Game's Afoot
a mystery adapted by Jon Jory
by Sean Grennan
Kafka's Metamorphosis
adapted by Steve Moulds
Crazytown
Nightfall With Edgar Allan Poe
by Eric Coble
Sherlock Holmes and The Case of the Jersey Lily
by Katie Forgette


text:
Q: Java Bean not found in configuration with multiple module I having a configuration problem that cause the error below. I know there are a tons of similar problem but I read a lot of answer that stackoverflow suggest with my title but I don't find the right answer yet.
Description:

Field collaborateurJpaRepository in com...infrastructure.persistence.user.UserRepositoryImpl required a bean of type 'com...infrastructure.persistence.jurisdiction.CollaborateurJpaRepository' that could not be found.

Action:

Consider defining a bean of type 'com...infrastructure.persistence.jurisdiction.CollaborateurJpaRepository' in your configuration.

Before to give some details of the code, I think it could be nice that I describe you brievly the context.
I work on a client project which has 8 maven modules (DDD methods) because at the end of the year 4 modules will be separate in an other server.
Recently we create a second oracle database so I had to create multiple configuration to refer to the good dataSource.
I think the problem is due to bad datasource configuration that block the bean spring instanciation.
We are using spring boot version 1.5.7.release.
First Infrastructure configuration
@Configuration
@ComponentScan(basePackages = "com...jade")
@EnableTransactionManagement
@EnableJpaRepositories(entityManagerFactoryRef = "jadeEntityManager", transactionManagerRef = "jadeTransactionManager")
@EntityScan(basePackages = { "com...jade", "org.springframework.data.jpa.convert.threeten" })
@Profile({"ARA_JADE_CONF", "LOCAL_JADE_CONF"})
public class InfrastructureConfig {

    private final static int JDBC_FETCH_SIZE = 1000;

    @Autowired
    private Environment environment;

    @Bean
    public LocalContainerEntityManagerFactoryBean jadeEntityManager() throws SQLException {
        LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
        em.setDataSource(dataSourceJade());
        em.setPackagesToScan(new String[] {"com...jade"});
        JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
        em.setJpaVendorAdapter(vendorAdapter);
        em.setJpaProperties(additionalProperties());
        return em;
    }
    @Bean
    @Primary
    public OracleDataSource dataSourceJade() throws SQLException {
        OracleDataSource dataSource = new OracleDataSource();
        dataSource.setUser(environment.getProperty("db.datasource.jade.username"));
        dataSource.setPassword(environment.getProperty("db.datasource.jade.password"));
        dataSource.setURL(environment.getProperty("db.datasource.jade.url"));
        dataSource.setImplicitCachingEnabled(true);
        dataSource.setFastConnectionFailoverEnabled(true);
        return dataSource;
    }
    @Bean(name="jdbcTemplateJade")
    public JdbcTemplate jdbcTemplateJade() throws SQLException {
        JdbcTemplate jdbcTemplate = new JdbcTemplate(dataSourceJade());
        jdbcTemplate.setResultsMapCaseInsensitive(true);
        jdbcTemplate.setFetchSize(JDBC_FETCH_SIZE);
        return jdbcTemplate;
    }
    @Bean
    public PlatformTransactionManager transactionManager() throws SQLException{
        JpaTransactionManager transactionManager = new JpaTransactionManager();
        transactionManager.setEntityManagerFactory(jadeEntityManager().getObject());

        return transactionManager;
    }
    @Bean
    public PersistenceExceptionTranslationPostProcessor exceptionTranslation() {
        return new PersistenceExceptionTranslationPostProcessor();
    } ...

Seconde Infrastructure configuration
@Configuration
@ComponentScan(basePackages =  "com.bnpparibas.sit.risk.art")
@EnableTransactionManagement
@EnableJpaRepositories(entityManagerFactoryRef = "artEntityManager", transactionManagerRef = "artTransactionManager")
@EntityScan(basePackages = { "com...art", "org.springframework.data.jpa.convert.threeten" })
@Profile({"LOCAL_ART_CONF,'ARA_ART_CONF"})
public class ArtInfrastructureConfig {

    private final static int JDBC_FETCH_SIZE = 1000;

    @Autowired
    private Environment environment;

    @Bean
    @Primary
    public LocalContainerEntityManagerFactoryBean artEntityManager() throws SQLException {
        LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
        em.setDataSource(dataSourceArt());
        em.setPackagesToScan(new String[] {"com...art.infrastructure.persistence"});
        JpaVendorAdapter vendorAdapter = new HibernateJpaVendorAdapter();
        em.setJpaVendorAdapter(vendorAdapter);
        em.setJpaProperties(additionalProperties());
        return em;
    }

    @Bean
    @Primary
    public OracleDataSource dataSourceArt() throws SQLException {
        OracleDataSource dataSource = new OracleDataSource();
        dataSource.setUser(environment.getProperty("db.datasource.art.username"));
        dataSource.setPassword(environment.getProperty("db.datasource.art.password"));
        dataSource.setURL(environment.getProperty("db.datasource.art.url"));
        dataSource.setImplicitCachingEnabled(true);
        dataSource.setFastConnectionFailoverEnabled(true);
        return dataSource;
    }

    @Bean
    @Primary
    public PlatformTransactionManager artTransactionManager()throws SQLException{
        JpaTransactionManager transactionManager = new JpaTransactionManager();
        transactionManager.setEntityManagerFactory(artEntityManager().getObject());

        return transactionManager;
    }

    @Bean
    @Primary
    public PersistenceExceptionTranslationPostProcessor exceptionTranslation() {
        return new PersistenceExceptionTranslationPostProcessor();
    }...

Interface link to Entity.
@DDD.Repository
@Repository
public interface CollaborateurJpaRepository extends CrudRepository<Collaborateur,String>{

    @Query("FROM ART_PARAM.ART_USER u WHERE u.refogUID = :uid")
    Collaborateur findCollaborateurById(@Param("uid") String uid);
}

Class that contain the autowired interface to retrieve some data of the second database.
@DDD.RepositoryImpl
@Repository
@Primary
public class UserRepositoryImpl implements UserRepository {

    @Autowired
    private CollaborateurJpaRepository collaborateurJpaRepository;

    @Override
    //@Cacheable(cacheNames = CacheConfig.CACHE_JURIDICTION, key = CacheConfig.CACHE_KEY_UTILISATEUR_ID)
    @Transactional("entityManagerFactoryArt")
    public User retrieveUser(UserId userId) {
        if(userId.getValue() == null) {
            return null;
        } else {
            Collaborateur collaborateur = collaborateurJpaRepository.findOne(userId.getValue());

            return new User(collaborateur.getRefogUID(),collaborateur.getRefogFirstName(),collaborateur.getRefogLastName(),collaborateur.getProfile(),collaborateur.isActiveUser(),collaborateur.isBankingSecrecy());
        }
    }


I've made it safely home from Atlanta! The convention was a blast. Tom and I had a great time. I totally destroyed my legs walking 6 miles while totally confused and sleep deprived. I ended up in some kinda underground abandoned city thankful lacking in mole people. I got to help one of my best friends sell darn near everything he brought with him, got to talk to a lot of cool people. Met Doug Walker of Nostalgia Critic for a second time and got to meet his brother Rob Walker for the first time. Big fan of his, probably more than Doug as odd as that may seem to people who know who I'm even talking about! XD I really wanted to meet ProJared but all I did was run into him on the elevator once and I didn't want to be that guy who corners a guy in an elevator that just wants to get back to his room and eat some dinner! Man. Momocon was great. Lot of wonderful people, great locations, hope I can go again someday. If you want to see pictures of the con- I really dont have much. Tom and I mostly just took pictures of our sketches we did while at the table. I highly recommend going over to our Twitter pages and checking them out. There's a link on the top bar to mine and Tom's is – https://twitter.com/TwoKinds. Cya Wednesday!