4
$\begingroup$

Is it consistent with $\mathsf{ZF}$ (without $\mathsf{AC}$) that there is an infinite set $X$ and a subset $S\subseteq\mathcal P(X)$ of the same cardinality as $\mathcal P(X)$ with the property that whenever $s\neq t\in S$ we have $|s\cap t| \leq 1$?

Note that it is consistent that there is such a pair of sets $X,S$, with $|S|>|X|$, but no such $X$ can be well-orderable. See here and here.

Improve this question
$\endgroup$
20
  • $\begingroup$ Perhaps it is worth pointing out that it is definitely consistent that $|X|<|S|$. Is "cset" a typo? $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 3, 2015 at 16:41
  • $\begingroup$ The statement is clearly false when $S$ is finite (or empty), and you haven't imposed any requirement preventing that. So could you clarify the precise statement about which you are asking? It doesn't make sense to ask for consistency when $X$ and $S$ are already-fixed parameters, so I imagine you have in mind a statement quantifying over $X$ and $S$. The statement is also false when $S$ simply consists of all the singleton sets. Perhaps you are asking whether it is consistent that there is any instance of this at all? $\endgroup$
    – Joel David Hamkins
    Commented Mar 3, 2015 at 17:47
  • $\begingroup$ Hi Joel, Yes, the statement being asked is whether it is consistent to have any instances of this situation. $\endgroup$
    – Andrés E. Caicedo
    Commented Mar 3, 2015 at 18:10
  • $\begingroup$ In that case, the question should be edited. To start out by saying, "Let $X$ be an infinite set..." suggests that one has a universal quantifier, but really the question is: Is it consistent with ZF that there is a set $X$ and a collection $S$, such that...? $\endgroup$
    – Joel David Hamkins
    Commented Mar 3, 2015 at 18:19
  • 1
    $\begingroup$ More clearly: is it consistent with ZF that there is an infinite set $X$ and an injection $f:P(X)\to P(X)$ such that $|f(A)\cap f(B)|\leq 1$ whenever $A\neq B$? $\endgroup$
    – Joel David Hamkins
    Commented Mar 3, 2015 at 19:26

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .