Consider the set $S$ that contains all sets $X$ such that $|X|\geq |\Bbb N|$.
The whole reason ZFC was made was because of Russel’s paradox right? And so we know that we can't have a set that contains itself. So because of this: $S \notin S$
And so we have that $|S| < |\Bbb N|$.(i.e: that there are only finitely many sets of infinite cardinality).
But clearly this is not the case! One way to see this is to note that all of:
$\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 2)}, $\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 3)}, $\{x \in \Bbb{Z}$ $s.t. x \equiv 1$ (mod 4)}, ... $\in S$
So clearly $|S| \geq |\Bbb N|$
This brings about a contradiction. And one would determine that it was irrational to consider the set $S$.
My question, essentially, is: What's going on here? Surely there is some problem with the logic above, as it should be rational to consider the collection of sets with infinite cardinality, right?
I'm afraid my ZFC knowledge isn't quite up to par (at all) so I'm not able to work out exactly what is wrong with my argument.
In my search for answers I came across some other posts on here, but none of them seemed to be directly comparable to my question.
Thank you in advance!!
The thing is that you define $S$ to be the "set" of $X$ that satisfy some property. If you do not tell where you pick $X$ from, the object $S$ you're creating might not even be a set. Some of the sets that are constructed this way and is axiomatized to be a set are the power set $P(A)=\{X\mid X\subseteq A\}$, and the union $\bigcup C=\{x\mid \exists A\in C : x\in A\}$. If $S$ is not a set, some properties we are familiar with might not be true.
See the term "class" for further information.