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I come up with the following set-theoretic question that has the flavor of Maximal Almost Disjoijnt (M.A.D.) families, although it is a bit different than the usual setting:

Let $\kappa$ be an uncountable cardinal. I want to find $\lambda$ such that:

1) there is a family of functions $f_\alpha:\lambda \rightarrow \kappa$, for all $\alpha< \lambda$,

2) for every $\alpha,\beta<\lambda$, $f_\alpha\cap f_\beta$ is finite (or empty), i.e. there are only finitely many $x\in \lambda$ such that $f_\alpha(x)=f_\beta(x)$, and

3) $\lambda$ is the greatest possible such cardinal.

Alternatively, we can see this family as one function $f:\lambda\times\lambda \rightarrow \kappa$ and the difference is that $\lambda$ controls both the domain of the functions as well as the number of them.

Now, it is trivial to construct a family of size $\kappa$ that satisfies (1) and (2) and it is not too hard to prove that it can not be done for $\lambda=\kappa^{++}$. So we are left we only two possibilities. Either $\lambda=\kappa$ is the best possible, or $\lambda=\kappa^{+}$ is the best possible.

Under the additional assumption $\kappa^\omega=\kappa$, we can prove that $\lambda=\kappa$ is the best possible.

My question: Does anyone know any references that relate to this question?

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  • $\begingroup$ Some of us instinctively read "M.A.D." as either "median absolute deviation" or "mean absolute deviation". (But ignore this silly comment.) $\endgroup$
    – Michael Hardy
    Commented May 20, 2011 at 2:35
  • $\begingroup$ If MAD was "median absolute deviation", it would not be a MAD family. Right? $\endgroup$
    – Ioannis Souldatos
    Commented May 20, 2011 at 19:05

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The article:

"Mad families and their neighbors" (2007) by Andreas Blass, Tapani Hyttinen, Yi Zhang

(pdf available here: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.107.3300&rep=rep1&type=pdf)

talks about several related notions of MAD, and might be a decent starting point.

This is an intriguing construction. In passing I would like to point out that $\kappa^\omega = \kappa \implies \lambda = \kappa$, says that under $GCH$ (so in $L$ for example) the only places the assertion can fail is at a singular cardinal of countable cofinality. As such you might want to have a look at what happens at $\omega_\omega$ under $CH$ (where you have $\forall n > 0 ({\omega_n}^{\omega} = \omega_n)$.) Or more to the point, what happens when you have a sequence of cardinals $\kappa_n : n\in \omega$, such that for each $n\in\omega$, $\lambda_n = \kappa_n$, and $\kappa = \sup\{ \kappa_n: n\in\omega \}$.

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  • $\begingroup$ Given that $\kappa^\omega=\kappa\Rightarrow \lambda=\kappa$, my first naive idea is that $\lambda$ is the minimum of $\kappa^+$ and $\kappa^\omega$. In particular, can we have $\kappa^\omega>\kappa$ and still $\lambda=\kappa$? I am not sure how difficult these questions are, but I do not have any examples when $\lambda=\kappa^+$, except for $\kappa=\aleph_0$. The argument (by Hjorth) uses Descriptive Set Theory and I wanted to see if anyone else knows anything about it. $\endgroup$
    – Ioannis Souldatos
    Commented May 20, 2011 at 19:03
  • $\begingroup$ Well, what I was alluding too was trying to glue an \omega sequence of these families together, and see if the result is such a family. $\endgroup$
    – Not Mike
    Commented Jul 29, 2011 at 16:11

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