3
$\begingroup$

When it is said that almost every model of $\sf ZF + \neg AC$ do have paradoxical partitioning $\sf PP$, that is: there exists a set $X$ and a partition $P$ of $X$ that is strictly larger than $X$. Is it also provable that: $$|P|<|\mathcal P(X)| \text{ ?}$$

$\endgroup$
  • 1
    $\begingroup$ I am not sure what was unclear in my comment to you the other day that "PP" is what you called "IP" (and promptly went on to ignore my comment about it being called PP). How about we all start giving different names for things? What's the point of having a common language? The existence of a paradoxical partition is quite literally the negation of the weak partition principle (WPP). Using a term to denote both a statement and its negation is bad practice. So I am going to ask you again, and hopefully I won't make this sort of comments for a third time. $\endgroup$ – Asaf Karagila Oct 19 at 7:30
  • $\begingroup$ @AsafKaragila, but I saw written as paradoxical partition and not as the partition principle. This is not my naming. Actually that was the reason behind me writing $\sf IP$ in the last posting, because it resolves this confusion. $\endgroup$ – Zuhair Oct 19 at 7:32
  • 1
    $\begingroup$ Yes, I figured you read the Taylor–Wagon paper. $\endgroup$ – Asaf Karagila Oct 19 at 7:33
5
$\begingroup$

Yes. Note that trivially $|P|\leq|\mathcal{P}(X)|$ since $P\subseteq\mathcal{P}(X)$. If $|P|=|\mathcal{P}(X)|$, then we can define a surjection $X\to\mathcal{P}(X)$ by mapping each element of $X$ to the element of $P$ that contains it and then composing with a bijection $P\to\mathcal{P}(X)$. This is impossible by Cantor's theorem.

$\endgroup$
4
$\begingroup$

Yes. If $P$ is a partition of $X$, then $$|P|\lt|\mathcal P(P)|\le|\mathcal P(X)|$$ by virtue of Cantor's theorem and the obvious injection $\mathcal P(P)\to\mathcal P(X)$.

$\endgroup$
3
$\begingroup$

We can say a bit more.

Note that if there is a surjection $f:A\to B$ then $f^{-1}:\mathcal P(B)\to\mathcal P(A)$ is an injection. Also, if $g:C\to D$ is an injection and $C\ne\emptyset$, then mapping $D\smallsetminus g[C]$ to a fixed point of $C$ and everything else to their preimage by $g$ is a surjection from $D$ to $C$.

It follows that if $\sim$ is an equivalence relation on a set $X$ and $|X/{\sim}|>|X|$, then there are surjections from $X$ onto $X/{\sim}$ and also from $X/{\sim}$ onto $X$, and therefore there are injections from $\mathcal P(X)$ into $\mathcal P(X/{\sim})$ and vice versa. But then $|\mathcal P(X)|=|\mathcal P(X/{\sim})|$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.