It is well known that cardinal numbers and the relations between them can be defined in ZF set theory (using the notion of "rank"), without the need of additional axioms. Can the following statement be proved from just the axioms of $\sf ZF$? " Let $k,\ell$ be cardinal numbers and let $k<\ell$. If $A$ is any set whose cardinal number is $k$, then there always exists a set $B$ whose cardinal number is $\ell$ and which contains $A$ as a subset".

  • $\begingroup$ This is borderline [axiom-of-choice]. If someone else thinks that this should be tagged as such, please make that edit. I'm still not sure. $\endgroup$ – Asaf Karagila Jul 7 '14 at 20:42


By the assumption that $k<\ell$ we essentially say that if $A$ is such that $|A|=k$ and $B'$ such that $|B'|=\ell$, then there is an injective function from $A$ into $B'$ (and there is no bijection between them, in this case). We can also assume that $A$ and $B'$ are disjoint, otherwise we replace $B'$ by $B'\times\{A\}$.

Fix any $f\colon A\to B'$ such injection, then the range of $f$, $A'=\{f(a)\mid a\in A\}$ is a subset of $B'$ and $f\colon A\to A'$ is a bijection. Define now $B=(B'\setminus A')\cup A$. It is easy to see why $|B|=\ell$ as well, and of course $A\subseteq B$.

  • $\begingroup$ When $\ell$ is infinite, $B = B' \cup A$ would work just as well, yes? $\endgroup$ – Nate Eldredge Jul 7 '14 at 20:46
  • 2
    $\begingroup$ @Nate: That depends. Are you assuming the axiom of choice? If not, then the answer is consistently no. $\endgroup$ – Asaf Karagila Jul 7 '14 at 20:47
  • $\begingroup$ Oh right, I need AC for that! $\endgroup$ – Nate Eldredge Jul 7 '14 at 20:48
  • $\begingroup$ @ Asaf Karaglia: Very nice. I could see how to prove that the answer was "YES" in the case that k and l were Alephs, but not in the more general case. $\endgroup$ – Garabed Gulbenkian Jul 8 '14 at 19:29
  • $\begingroup$ Although this proof works for most cases, there seems to be a problem if $B'\setminus A'$ and $A$ aren't disjoint. the work-around is, of course, to replace the original $B'$ with one that is disjoint from $A$. $\endgroup$ – Andreas Blass Aug 16 '14 at 6:04

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