2
$\begingroup$

Let $A$ be a set of cardinals. Prove that there is a cardinal that that is greater than every cardinal in $A$. Assume that there isn't such a cardinal. Then for any cardinal $x$ there is $y\in A$ such that $x\leq y$ and since each cardinal is an ordinal than $x\in y$ or $x=y$ and therefore the set $B=(\cup{A})\cup A$ contains each cardinal. By the axiom schema of separation there is a subset $C\subseteq B$ that contains exactly all the ordinals and that is a contradiction. Is my proof correct?

$\endgroup$
2
$\begingroup$

The proof is correct, but can be better executed. For example, you don't need the contradiction. Simply show that $\bigcup A$ is a cardinal, and then take $(\bigcup A)^+$, or its power set.

$\endgroup$

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.