A question about cardinal numbers in ZF set theory.

Question

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".

Answer 1

Yes.

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 $A\subseteq B$.