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\centerline{\bf CMSC 752 Homework 3} 
\centerline{\bf Morally Due Tue Feb 18, 2025} 
\centerline{\bf Dead Cat Feb 20, 2025}

\bigskip

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\begin{enumerate}

\item
(50 points) 
We proved in class that 

{\it If $f\colon \Z\times \Z \into \Z$ then 
there exists $D \subseteq \Z$ such that
$f$ restricted to $\D\times \D$ is NOT onto. }

But in that proof it was quite possible for $f$ restricted to $\D\times \D$
to just miss ONE element. So it could be ALMOST onto. 

Gary was very mad about that!

A function $f$ is {\it lazy} if there are an infinite number of elements in the co-domain
such that $f$ does not hit (so VERY not-onto). 

PROVE the following:

{\it If $f\colon \Z\times \Z \into \Z$ then 
there exists $D \subseteq \Z$ such that
$f$ restricted to $\D\times \D$ is lazy.}

\newpage 

\item
(50 points) 
Prove that, for all $f\colon \Z\times \Z\times \Z \into \Z$
there exists $\D \subseteq \Z$ such that
$f$ restricted to $\D\times\D\times\D$ is NOT onto. 

\newpage
\item
(0 points, Extra Credit)
\begin{enumerate}
\item
(this part gets you no credit) 
Give your name (this will not get you any extra credit, but since I
grade this one by putting the names of who got it right into a file,
this makes my life easier.)

\item 
Find some value $c$ and prove the following:

{\it For all $\COL \colon \binom{\omega^2}{2} \into [1,000,000]$
there exists a $c$-homog set.}

Some points about this.
\begin{itemize}
\item
Your answer had to be well written.

\item
The optimal answer is $c=4$ though I doubt you can obtain that.
I am really looking for a NOT-MESSY proof of a weaker-than-known result.

\item
The first step you probably all know: WITHIN each copy of $\omega$
2-ary Ramsey, and then for all of the copies of $\omega$ use 1-ary Ramsey.
So you can just ASSUME that within each copy of $\omega$, all of the 
edges are RED.
(No extra credit for getting just that far.) 

\item
Some people tried to use the Infinite Bipartite Ramsey Theorem on all pairs.
This won't work since that yields 3-homog which is not true. However, if you can
get something like this to work (I couldn't) that would be great.
\end{itemize} 

\end{enumerate} 

\end{enumerate} 

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