0
$\begingroup$

Let us call $f,g:\omega\to \omega$ almost totally distinct if $$|\{n\in \omega: f(n) = g(n)\}| < \aleph_0.\;\;\;\; (\star)$$ It is known that there are uncountable collections of almost totally distinct functions.

Question. Is the above statement still true if we replace $\aleph_0$ by $2$ in $(\star)$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

No. By pigeonhole principle any uncountable family of functions $f:\omega\to \omega$ contains two functions with the same pair $(f(1),f(2))$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ If this is so, why does the original statement hold (because the pigeonhole principle would say there is no uncountable family of almost totally distinct functions)? Gerhard "Bewildered By Infinitely Many Pigeons" Paseman, 2017.02.22. $\endgroup$
    – Gerhard Paseman
    Commented Feb 22, 2017 at 17:02
  • 1
    $\begingroup$ Ah. Even though I can find an uncountable family that agree on a given finite set, I can partition that family infinitely many times in a way agreeable to the definition of almost totally distinct. Once a finite bound on the size of the agreement set is put, I don't have the ability to split. OK. Gerhard "Returning You To Regular Programming" Paseman, 2017.02.22. $\endgroup$
    – Gerhard Paseman
    Commented Feb 22, 2017 at 17:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .