# Set of Cardinals

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?

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.