2
$\begingroup$

It's well known that a universal set, $U$, in ZFC, leads to a number of problems; such as Russell's Paradox when combined with specification. Another popular issue, as cited by Wikipedia is that

Because this power set is a set of sets, it would necessarily be a subset of the set of all sets, provided that both exist. However, this conflicts with Cantor's theorem that the power set of any set (whether infinite or not) always has strictly higher cardinality than the set itself.

This makes immediate sense at first glance. However, this feels more like a proof that the cardinality of a universal set simply behaves unexpectedly. In particular, there is one subtlety of Cantor's Theorem which has raised a question for me: The proof that $\text{card}(X)<\text{card}({\mathcal{P}(X))}$ is a result relying on the assumption that some map $f:X\to\mathcal{P}(X)$ exists at all.

Thus my question is, can such a map really exist for a universal set? If not, then its cardinality cannot be compared with its power set, thus cannot be said to violate Cantor's Theorem.


My work so far:

Of course, I realized this concern of mine can be easily disproven by an example construction of any function $f:U\to \mathcal{P}(U)$. Actually, since $U$ is its own powerset by virtue of containing all of its subsets, we just have to construct some $f:U\to U$. I assume this could be done in theory using replacement, but precisely how is beyond my grasp; hence any insight would be greatly appreciated.

$\endgroup$
2
  • $\begingroup$ If you just have to construct some $f:U\to U$, what's wrong with $f(u)=u$? $\endgroup$
    – bof
    May 29 at 1:31
  • $\begingroup$ @bof my thoughts as well; I just hadn't the confidence there wasn't some nuance I was overlooking. $\endgroup$
    – Graviton
    May 29 at 2:03

1 Answer 1

4
$\begingroup$

Consider $x\mapsto\{x\}$. (Note that in theories like $\mathsf{NF(U)}$ which do have a universal set, this doesn't correspond to a function!)

Incidentally, this doesn't really use replacement at all - Separation is enough (applied to the putative set $U\times\mathcal{P}(U)$). More generally, $\mathsf{ZF}$ (or indeed much less) proves that every set injects into its powerset.

$\endgroup$
3
  • 1
    $\begingroup$ I did considered this, (for the sake of brevity let's call such a function $f$). But assuming that $f=\{(x,\{x\}):x\in U\}$, is it provable that $f$ exists? $\endgroup$
    – Graviton
    May 28 at 23:35
  • 1
    $\begingroup$ @Graviton Yes (assuming $U$ is a set of course); as I say in my answer, separation does the job. $\endgroup$ May 28 at 23:35
  • 1
    $\begingroup$ Ah yes I see how separation does the job. Had a feeling it would be as trivial. Thanks for the brisk and efficient answer to my curiosity. I'll wait a few minutes before accepting as is customary but this does satisfy me. $\endgroup$
    – Graviton
    May 28 at 23:40

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.