Corollary

There does not exists a set that has every cardinal as an element.

If U were such a set, there would exist a set S, the sum of the family of sets $(X)_{X \in U}$, so that every cardinal is equipotent to a subset of S. In particular, let $\mathfrak{S} = \mathrm{Card}(S)$; since $2^\mathfrak{S}$ is a cardinal, we would have $2^\mathfrak{S} \leqslant \mathfrak{S}$, in contradiction to Theorem 2.