Is it possible to make the set of all sets of cardinality $\aleph_0$?

I know that in ZFC that some collections of objects cannot be gathered together into a set (for example, the "set of all sets") does not exist, nor is "the set that just contains itself."

Is it possible to construct the set of all sets with cardinality $\aleph_0$? Or does this set not exist?

Thanks!

Solution 1:

It does not exist. Suppose that it did, and call it $C$. The union axiom then ensures that the set $\bigcup C$ exists. But every set $x$ is an element of a countably infinite set, e.g., $\{x\}\cup\omega$, so for each set $x$ there is a $y\in C$ such that $x\in y$, and therefore $x\in\bigcup C$. That is, every set is an element of $\bigcup C$, which you already know is impossible. Thus, $C$ cannot exist.

Solution 2:

No.

The reason is simple:

For every set $A$, the set $\{ A \} \cup \mathbb N$ is countable. And the set of all such sets would contain the "set" of all sets as a subset.

Solution 3:

Suppose $x$ is a set, and consider corresponding set

$$X_x = \{(0, X), (1, X), (2, X), \ldots \}$$

Then each $X_x$ has cardinality $\aleph_0$, and there are as many of the $X_x$ as there are of sets $x$, i.e. too many to form a set.