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\def\header{{\hfill Set Theory of the Plane \hfill A. Miller \hfill \today
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\def\reals{{\mathbb R}}
\def\integers{{\mathbb Z}}
\def\rationals{{\mathbb Q}}
\def\sq{\subseteq}
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\def\rmor{\mbox{ or }}
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\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
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\newtheorem{question}[theorem]{Question}
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\begin{document}

\begin{flushright}
M873 Topics in Foundations\\
A.Miller\\
Spring 2005 \\
\end{flushright}

\bigskip

I plan to present the Jackson-Mauldin solution to the Steinhaus problem.
In the 1950s Steinhaus asked if there exists a subset S of the plane
which meets every isometric copy of the integer lattice, $\integers\times
\integers$, in exactly
one point.  I also plan to cover other results which involve the set
theoretical properties of the plane.  Davies has shown that the plane
can be partitioned into countably many pieces none of which contains two
pairs of points the same distance apart iff the continuum hypothesis is
true.  On the other hand Schmerl has shown that without any extra set
theoretical hypotheses the plane can always be partitioned into
countably many pieces so that no piece contains the vertices of an
isoceles triangle. 


\begin{theorem}
(R.O.Davies \cite{davies}, also in \cite{komjath}) 
CH is equivalent to:

There exists $S_k\sq \reals^2$ such that 
$$\reals^2=\bigcup_{k<\om}S_k$$
and for every $k$ if $x,y,u,v\in S_k$  and $x\neq y$ then
$$d(x,y)=d(u,v)\rmiff \{x,y\}=\{u,v\}$$
i.e., distances are unique in each $S_k$.
\end{theorem}

\begin{theorem}
(Kunen \cite{kunen}) CH implies that for each positive integer
$n$ there exists $S_k$ such that
$\reals^n=\bigcup_{k<\om}S_k$ and distances are unique in each $S_k$.
\end{theorem}

\begin{question}
(Kunen \cite{kunen}) For the $l_p$ norm $1<p<\infty$ 
does CH imply that $\reals^n$ can
be partitioned into countably many sets in which distances are unique?
\end{question}

Remarks \cite{kunen}.  It is false for $p=1$ or $p=\infty$ in the plane.  It is
also false for infinite dimensional Hilbert space $\ltwo$ by the following
proposition.  

\begin{prop}
(Posa see \cite{komjath}) There exists an uncountable 
${\mathcal A}\sq \ltwo$ such that 
$||a-b||$ is rational for any $a,b\in {\mathcal A}$. 
\end{prop}

\begin{question}
(Erdos) Does there exists a dense $D\sq\reals^2$ such that the
distance between any two points of $D$ is rational?
\end{question}

\begin{lemma}
If $V$ is a vector space over a countable field $F$, then there are $S_k$
$$V\sm \{\underline{0}\}=\bigcup_{k<\om} S_k$$
such that each $S_k$ contains no nontrivial solution to
$$\alpha x+\beta y=z$$
for $\alpha,\beta\in F$.  Trivial means
$$\alpha x+(1-\alpha)x=x.$$
\end{lemma}

\begin{theorem}
(Rado, in \cite{komjath}) $\reals^n$ can be partitioned into countably many
sets 
$$\reals^n=\bigcup_{k<\om}S_k$$
none of which contains the vertices
of a degenerate isoceles triangle (or equivalently a 3-term 
arithmetic progression):
 $${{p+r}\over 2}=q.$$
\end{theorem}

\begin{theorem}
(Ceder \cite{ceder} also in \cite{komjath}) 
$\reals^2$ can be partitioned into countably
many sets none of which contains the vertices of an equilateral triangle.
\end{theorem}

\begin{theorem}
(Ceder \cite{ceder}) If $\reals^n=\bigcup_{k<\om}S_k$ and each $S_k$ is Borel, then
there exists $k<\om$ such that for every finite $F\sq \reals^n$
there exists $p\in \reals^n$ and $\alpha>0$ such that
$p+\alpha F\sq S_k$.
\end{theorem}

\begin{question} (Erdos see \cite{svetic}) If $x_n$ is a sequence decreasing to
0, does there exist a  set of positive measure $C$ which avoids every similar
copy of it? What about $x_n={1\over 2^n}$? \end{question}

\begin{theorem}
(Gallai see \cite{graham}) Suppose $\reals^n=\bigcup_{k<N}S_k$ where
$N$ finite.  Then for every finite $F\sq \reals^n$
there exists $k<N$ and $p\in \reals^n$ and $\alpha>0$ such that
$p+\alpha F\sq S_k$
\end{theorem}

The Hales-Jewett Theorem is also in \cite{graham} as well as \cite{hj}.
Shelah's proof appears in the second edition 1990 of \cite{graham}
and there is an even better bound proved by Gowers \cite{gowers}
\cite{landman}.  For a version of Hales-Jewett with an infinite alphabet,
due to Miller and Prikry, see Carlson \cite{carlson} Theorem 15.

\begin{lemma}
(Tarski) The theory RCF (real-closed fields) eliminates quantifiers.
\end{lemma}
For a proof see Marker \cite{marker} or Shoenfield \cite{shoen}.

\begin{lemma}
(Implicit function theorem for real-analytic, see \cite{nara} 1.3.10)
\par Suppose $\Omega_1\sq\reals^n,\Omega_2\sq\reals$ are open, 
$f:\Omega_1\times\Omega_2\to\reals$ is real-analytic, 
$(a,b)\in \Omega_1\times\Omega_2$,
$$f(a,b)=0 \rmand {{\partial f}\over{\partial x_{n+1}}}(a,b)\neq 0.$$ 
Then there exists open $U_1\sq \Omega_1$ and $U_2\sq\Omega_2$ with
$(a,b)\in U_1\times U_2$ such that for every $x\in U_1$ there
exists a unique $y=y(x)\in U_2$ such that $f(x,y)=0$.  Furthermore
the function $x\mapsto y(x)$ is real-analytic.
\end{lemma}

\begin{theorem}
(Schmerl \cite{schmerl}) $\reals^n$ can be partitioned into countably
many sets none of which contains the vertices of an isoceles triangle.
\end{theorem}

The same partition has the property that the distance between distinct
points in any piece is irrational.  This yields a result of  Komjath
\cite{komjathdecomp}.  It also has the property that for the plane, no piece
contains the vertices of a triangle with rational area, a result proved by
Kunen (unpublished) assuming CH and generalized to $\reals^n$ by Komjath
\cite{komjathtetra},\cite{komjathmaster}. 

\begin{exercise}
Let $g:(\reals^n)^3\to \reals$ be defined by $g(a,b,c)=$ the area of
the triangle with vertices $a,b,c$.  Then $g$ is definable in
$(\reals,+,\cdot)$.  Prove or disprove: It is real-analytic.
\end{exercise}


\begin{theorem}
(Erdos-Kakutani \cite{ek}) CH is equivalent to $\reals\sm\{0\}=\cup_{n<\om}H_n$
where each $H_n$ is a Hamel base.
\end{theorem}

Remark. Assuming CH we can make the $H_n$ pairwise disjoint Hamel bases.

\begin{theorem}
(Beslagic \cite{beslagic}) $\neg$CH is equivalent to:
\par\noindent for
every partition ${\mathcal P}$ of the $\reals$ into countable sets there
exists a Hamel base $H$ such that $|p\cap H|\leq 1$ for every 
$p\in {\mathcal P}$.
\end{theorem}

\begin{theorem}
(Erdos, Komjath \cite{erdoskomjath}) CH is equivalent to 
there exists $S_k$ such that 
$$\reals^2=\bigcup_{k<\om}S_k$$
and no $S_k$ contains the vertices of a right triangle.
\end{theorem}

\begin{question}
(Erdos, Komjath \cite{erdoskomjath}) Is this result true for $\reals^n$
for every $n$?
\end{question}

\begin{theorem}
(Sierpinski, see Simms \cite{simms} for references and history)
The following are equivalent:
\begin{enumerate}
 \item CH
 \item $\reals^2=H\cup V$ where every horizontal line meets $H$ in a
 countable set and every vertical line meets $V$ in a countable set.
 \item $\reals^3=A_1\cup A_2\cup A_3$ where every line parallel
 to axis $x_i$ meets $A_i$ in a finite set.
 \item $\reals^3=B_1\cup B_2\cup B_3$ where every plane perpendicular
 to the $x_i$ axis meets $B_i$ in a countable set.
\end{enumerate}
\end{theorem}

\begin{theorem}
(Kuratowski, see Simms \cite{simms}) The following are equivalent:
\begin{enumerate}
 \item $|\reals|\leq \om_2$
 \item $\reals^3=A_1\cup A_2\cup A_3$ where every line parallel
 to axis $x_i$ meets $A_i$ in a countable set.
 \item $\reals^4=B_1\cup B_2\cup B_3\cup B_4$ where every 
 every line parallel
 to axis $x_i$ meets $B_i$ in a finite set.
\end{enumerate}
\end{theorem}
And the obvious generalizations to $|\reals|\leq \om_n$.

\begin{exercise}
Suppose that $\kappa\leq \aleph_m$ is equivalent to:
\par\noindent There exists $A_p$ such that
$$\kappa^n=\bigcup_{s\in [n]^k} A_s$$
and $|A_s(p)|<\aleph_l$ for each $s\in [n]^k$ and $p\in\kappa^s$
where
$$A_s(p)=\{(\alpha_i:i<n)\in \kappa^n\;\;:\;\; 
\forall i\in s\;\; \alpha_i=p(i)\}$$
Find $m=m(n,k,l)$ as a function of $n,k,l$.
\end{exercise}
Hint: $m=l+k-1$ for $k<n$ according to David Milovich.

\begin{question}
(Ristow, see Simms \cite{simms} 5.2) Characterize the sequences
$\sigma_i,\kappa_i$ for $i<n$ such that
$\om\leq \sigma_i\leq\kappa_i$ and there exists $E_i$ such
that 
$$\prod_{i<n}\kappa_i=\bigcup_{i<n}E_i$$
and $|l\cap E_i|<\sigma_i$ for each $i$ and line $l$ parallel to
the $i$th axis. 
\end{question}

\begin{theorem}
(Komjath \cite{komjathcloud}) CH is equivalent to $\reals^2$ is the
union of three clouds.  A cloud is set of points $C$ such that for some
point $p$ every line thru $p$ meets $C$ in a finite set.
The plane is the union of $n+2$ clouds if $|\reals|\leq \om_n$.
\end{theorem}

\begin{theorem}
(Bagemihl-Davies \cite{bag-plane}\cite{davies-fog}) CH is equivalent to
$\reals^2$ is the union of three fogs.  A fog $F\sq\reals^2$ is a set
with the property that for some slope $m$ every line with slope $m$
meets $F$ in a finite set.
\end{theorem}

\begin{theorem}
(Schmerl \cite{schmerlcloud}) CH implies that $\reals^2$ is the union
of three sprays.  A spray $S\sq\reals^2$ is a subset with the property
that for some $p\in\reals^2$ every circle centered at $p$ meets
at most finitely many elements of $S$.
\end{theorem}

\begin{question}
(Schmerl) If the plane is the union of three sprays, then does CH hold?
\end{question}

Two sprays cannot cover the plane.  Ditto for two fogs.
If $S^2$ can be covered by three ``sprays'' whose center lies
on a great circle, then CH is true.  To see this project the 
sphere to the plane of the great circle and get 2-1 map to 
cover the disk with three fogs.

\begin{exercise}
Define:
A set $W\sq\reals^2$ is a wringer iff there are two points
$p_1,p_2$ such that $C\cap W$ is finite for every circle containing $p_1$ and
$p_2$.  (The circles travel between the two points which wrings them dry.)
A set $P\sq\reals$ is a pinched wringer iff there is a line $l_0$ and point 
$p_0\in l_0$
such that $P\cap C$ is finite for every circle $C$ with $p_0\in C$ and 
center of $C$ on $l_0$.
A set $F\sq\reals$ is a flower iff there is a point $p_0$ and
radius $r$ such that
$F\cap C$ is finite for every circle $C$ with $p_0\in C$ and
radius $r$.
 
\par
Assume CH holds then
\begin{enumerate}

\item 
There exists three wringers $W_1,W_2,W_3$ determined by pairs of three points
$p_1,p_2,p_3$ which cover every point in $\reals^2$ not on the circle
containing $p_1,p_2,p_3$.

\item Three pinched wringers cover the plane.

\item Given a triangle the interior of
the triangle is covered by
three flowers at the vertices.

\end{enumerate}

Do any of these imply CH?
\end{exercise}

\begin{theorem}
(Komjath \cite{komjathcloud}) The plane is the union of countably many
starcircles.  A starcircle is a set of points $C$ such that for some
point $p$ every ray starting at $p$ meets $C$ in at most one point. 
\end{theorem}

\begin{theorem}
(Schmerl \cite{schmerlcloud}) The plane is not the union of finitely many
starcircles.  The plane is the union of four clouds iff $|\reals|\leq \om_2$.
Similarly, the union of $n+2$ clouds iff $|\reals|\leq \om_n$.
\end{theorem}

Analogous results holds for $n$ fogs. A fog $F$ can be thought of as the graph
of a finitely multi-valued function  which has been rotated. I.e., fix a line
$l$ orthogonal to the lines determining the fog, then every line orthogonal to
$l$ meets $F$ in a finite set.

\begin{theorem}
(Davies \cite{davies-curve}) There is a function $f:\reals\to\reals$ such
that the plane is the countable union of sets, each of which is isometric
to the graph of $f$.
\end{theorem}

Remark.  We can construct $f$ so that countably many rotations of its
graph cover the plane.  Erik Andrejko notes that in addition $f$ can
be made uniformly bounded, i.e., $|f(x)|\leq 1$ for all $x$.

\begin{question}
(Davies, see Simms \cite{simms} 5.1) Is the plane 
the countable union of sets congruent to the graph of a function
which is connected?  
\end{question}

\begin{theorem} 
(Mazurkiewicz 1933, see Simms \cite{simms} for reference)  The plane is not 
the finite union of curve-graphs. 
A curve graph is a subset of the plane which is isometric to the graph of some
function $f:\reals\to\reals$.  
\end{theorem}

This result follows easily from Gallai's Theorem.

\begin{theorem}
(Bagemihl \cite{bag-infdim}) There exists $S_k$ such that
$$\reals^\om=\bigcup_{n<\om} S_n$$
and $|S_n\cap l|\leq 1$ for every $n$ and line $l$ parallel to
$e_n$. 
\end{theorem}

\begin{theorem}
(Gruenhage \cite{gruen}) also in Komjath \cite{komjath}) 
There exists $F(l)\sq l$ finite for each line $l$ in the plane
and finite $G(p)\sq\{l:p\in l\}$ for each point $p\in\reals^2$ such
that for every line $l$ and point $p\in l$ either
$p\in F(l)$ or $l\in G(p)$.
\end{theorem}

\begin{theorem}
(Galvin) If $\reals\leq \om_n$ then 
\par (a) There are $F$ and $G$ as in Gruenhage's Theorem such that
$|F(l)|\leq n+1$ for every line $l$ and
\par (b) There are $F$ and $G$ as in Gruenhage's Theorem such that
$|G(p)|\leq n+1$ for every point $p$. 
\end{theorem}

\begin{question} In the Galvin result
\par (a) Is it consistent to have $|F(p)|\leq 1$ all $p$?
\par (b) Does $|F(p)|\leq 2$ all $p$ imply CH?
\par (c) Similarly for $|G(l)|$ in (a) and (b)?
\par (d) Is it consistent to have $|F(p)| + |G(l)|<n$
for all $p,l$?
\end{question}

I don't have a copy of the Gruenhage-Galvin unpublished manuscript
so perhaps these are known. Bergman-Hrushovski \cite{bh}
also has a proof of Galvin's Theorem as well as some questions. 

\begin{theorem}
(Davies \cite{davies-lines}) Suppose ${\mathcal L}_k$ for
$k<\omega$ is a partition of the lines of $\reals^2$. Then
there exists $S_k$ such that $\reals^2=\bigcup_k S_k$ and
for every $k$ and $S_k\cap l$ is
finite for every $l\in {\mathcal L}_k$.  In addition, if
$|\reals|\leq \om_n$ then we can find $S_k$ so that
$|S_k\cap l|\leq n+1$ for each $k<\om$ and $l\in {\mathcal L}_k$.
\end{theorem}

Remark.  Same proof for any $\reals^m$ replacing $\reals^2$.  This
result implies that the plane is the countable union of starcircles.

\begin{example}
(Davies) There exists ${\mathcal L}_k$ for
$k<\omega$ a partition of the lines of $\reals^2$ such that there
does not exists $S_k$ such that 
$|S_k\cap l|\leq 1$ for each $k<\om$ and $l\in {\mathcal L}_k$.
\end{example}

\begin{theorem}
(Erdos-Jackson-Mauldin \cite{ejm2}) Assume MA.
Suppose ${\mathcal L}_k$ for
$k<\omega$ is a partition of the lines of $\reals^2$. Then
there exists $S_k$ such that $\reals^2=\bigcup_k S_k$ and
for every $k$ and $|S_k\cap l|\leq 3$
for every $l\in {\mathcal L}_k$. 
\end{theorem}
So the ``3-point'' property does not imply any bound on the size of
the continuum.  
Later Schmerl showed that the  ``2-point'' property is just true
in ZFC.

\begin{theorem}
(Schmerl \cite{schmerl-2pt, schmerl-2pt2})
Suppose ${\mathcal L}_k$ for
$k<\omega$ is a partition of the lines of $\reals^2$. Then
there exists $S_k$ such that $\reals^2=\bigcup_k S_k$ and
for every $k$ and $|S_k\cap l|\leq 2$
for every $l\in {\mathcal L}_k$. 
\end{theorem}

\begin{theorem}
(Erdos-Jackson-Mauldin \cite{ejm1}) Assume $|\reals|\leq \om_{n-2}$.
Then
for every partition 
$$\bigcup_{k<n}{\mathcal L}_k=lines(\reals^2)$$
there exists $S_k$ such that $\reals^2=\bigcup_{k<n}S_k$ and
$|l\cap S_k|<\om$ for every $k<n$ and
$l\in {\mathcal L}_k$.
\end{theorem}

This results is also true for any $\reals^m$ in place of the plane.
It also implies that the plane is the union of $n$-clouds and
$n$-fogs.

\begin{theorem}
(Sierpinski see Komjath \cite{komjath}) There exists $A,S\sq\reals^2$ 
nontrivial so that for every $B$ an isometric copy of $A$  we
have $|A\cap S|=1$.
\end{theorem}

This answered a question of Steinhaus who also asked if $A$ could
be taken to be $\integers^2$.

\begin{theorem}
(Komjath \cite{komjath-stein1,komjath-stein2})
\par (a) There exists $S\sq \reals^2$ such that for every 
isometric copy $L\sq\reals^2$ of $\integers$ we have
that $|S\cap L|=1$.

\par (b) There exists $S\sq \reals^2$ such that for every 
isometric copy $L\sq\reals^2$ of $\rationals^2$ we have
that $|S\cap L|=1$.

\par (c) There exists $S\sq \reals^2$ such that for every 
isometric copy $L\sq\reals^2$ of $\rationals$ we have
that $|S\cap L|=1$.
\end{theorem}

Schmerl \cite{schmerl-stein} proves some generalizations of these 
for $\rationals^n\sq \reals^n$, and for isometric copies of
the $\integers$ and $\rationals$
in $\reals^n$.

\begin{prop}
(Schmerl \cite{schmerl-stein}) For $4\leq m<n$ there does not
exist $S\sq\reals^n$ such that for every
isometric copy $L\sq\reals^n$ of $\integers^m$ we have
that $|S\cap L|=1$.
\end{prop}

Similarly, for $\rationals^m$ and $\reals^n$.

\begin{prop}
(Jackson-Mauldin) Suppose $A\sq \reals^n$ and $S$ is Steinhaus for
$A$, i.e., $S$ meet every isometric copy of $A$ in exactly one point.
Then 
$$\{p+S:p\in A\}$$
is a partition of $\reals^n$.
\end{prop}

Hence if there $S$ in the plane meeting every isometric
copy of $\rationals$ in exactly one point, then the
plane can be partitioned into countably many such sets.
Similarly, if a Steinhaus set
for a countable $A$ has the Property of Baire (is Lebesgue measurable)
then it is nonmeager (positive measure).  If there is a Steinhaus
set for $A$ then for each positive $n<|A|$ there is a set $S$ meeting every 
isometric copy of $A$ in exactly $n$ points.

\begin{theorem}
(Jackson-Mauldin \cite{jm1,jm2,jm3}) 
There exists $S\sq \reals^2$ such that for every 
isometric copy $L\sq\reals^2$ of $\integers^2$ we have
that $|S\cap L|=1$.
\end{theorem}

\begin{lemma}
(Lemma B) Suppose $a_1,a_2,a_3$ are distinct points in the plane.
Suppose $r_1,r_2,r_3>0$ and $C_i$ is the circle centered at
$a_i$ with radius $r_i$.  Fix $\lambda_{i,j}>0$.  Suppose there
are infinitely many collinear triples 
$$(p_1,p_2,p_3)\in C_1\times C_2\times C_3$$
with $d(p_i,p_j)=\lambda_{i,j}$.  Then
$a_1,a_2,a_3$ are collinear, $r_1=r_2=r_3$, and 
$\lambda_{i,j}=d(a_i,a_j)$ for all $i,j\in 1,2,3$.
\end{lemma}

Lemma B is somewhere in between the original Komjath \cite{komjath-stein1}
version which used 5 circles and the stronger version in Jackson-Mauldin
\cite{jm2} which applies to noncollinear
triples.  The Maple proof for
our version is slightly simpler than the one given
in Jackson-Mauldin \cite{jm2} since it avoids Grobner basis. 

\begin{verbatim}
eq1:= (x+u*f-a)^2+ (y+u*g)^2  = r^2;
eq2:= (x+v*f-b)^2+(y+v*g-c)^2 = s^2;
eq1:= collect(expand(eq1),u):
eq2:= collect(expand(eq2),v):
eq1:= simplify(eq1,{x^2+y^2=1,f^2+g^2=1}):
eq2:= simplify(eq2,{x^2+y^2=1,f^2+g^2=1}):
eq1:=sort(collect(eq1,[x,y],'distributed'));
eq2:=sort(collect(eq2,[x,y],'distributed'));
sol:=solve({eq1,eq2},{x,y}): sol[1]; sol[2];
assign(sol):
z  := x^2+y^2-1:
bot:=denom(z):
bot:=factor(bot);
top:=numer(z):
top:=simplify(top,{g^2=1-f^2}): 
top:=sort(top,[f,g],'plex');
coeffs(expand(top),[f,g],'terms'): terms;
p:=lcoeff(top,[f,g],'t03'); t03;
q:=lcoeff(top,[g,f],'t12'); t12;
p:=factor(p);
q:=factor(q);
\end{verbatim}

\begin{theorem}
(Jackson-Mauldin \cite{jm3})
If $S\sq\reals^2$ has the property of Baire, then there exists
an isometry $\alpha$ of the plane such that
$|\alpha(\integers^2)\cap S|\neq 1$. 
\end{theorem}

This is open for Lebesgue measure, however 
Croft \cite{croft} (and independently Beck) has shown that there is no
Lebesgue measurable Steinhaus set which is measure zero outside a
bounded set.  It seems to be open if there can be a bounded
Steinhaus set.

Jackson and Mauldin also prove that there are Steinhaus sets for the lattice
$n\integers \times m\integers$ provided every prime which divides
$n$ or $m$ is 1 mod 4.  It seems to be open for $\integers \times 5\integers$.


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\end{thebibliography}

\end{document}
