"There is no set containing everything"? I was reading this question regarding codomains, and I found something interesting in User134824's answer:
"On the other hand, owing to the set-theoretic fact that "there is no set containing everything," it's not possible to pick a single universal codomain for functions."
Why is it impossible to have a set containing everything? Why can't we define $U=\mathbb{R} \cup \mathbb{C} \cup ....$(all possible sets)?
P.S. This is a soft-question, so I am looking for intuitive, non-technical answers; I do not know any set theory
 A: With your $U$, then $\mathcal{P}(U)\subset U$ but, $\# \mathcal{P}(U)=2^{\#U}$, absurd
A: The problem via Cantor's paradox has already been noted. It is also the case that the most common set theories prove the existence of "the set of all $x\in A$ such that $x\notin x$". If $A$ is the universe, then there is a set $R$ containing every set that is not a member of itself; but $R\in R \iff R\notin R$, which is a paradox (Russell's, specifically).
More trivially, common set theories accept the Axiom of Foundation, which implies that no set can be a member of itself. But a set containing every set must have itself as a member.
There are, as someone mentions in the post linked to in the comments, consistent set theories with universal sets, but these theories must reject each of Foundation, the existence of $\{x: x\in A \wedge x\notin x\}$ for all $A$, and Cantor's theorem that $A < \mathcal{P}(A)$. The consequences of axiom systems that disprove these in favor of the existence of a universe can be counterintuitive or cumbersome; Currying a binary function might not work in $\mathsf{NFU}$, or complementation might not work in $\mathsf{GPK}$.
