I just started learning about cardinality in my math fundamentals class, and in particular I've just learned that $\aleph _0$ can be thought of as an equivalence class as follows: any set with cardinality equal to it are in its class. For instance, $\mathbb{N} \in \aleph _0$, and likewise, $\mathbb{Q} \in \aleph _0$ (if I have understood correctly).
That has lead me to wonder: what is the cardinality of this equivalence class? Is it countable? Uncountable? Thanks!
Best Answer
Part $1$: The cardinality is uncountable, with a caveat:
The cardinality of this equivalence class is, at first glance, uncountable. For notation's sake, let us denote this class by $\overline{\aleph_0}$, to avoid confusion with the cardinal.
It is known that the cardinality of the reals - that of the continuum - is greater than that of $\aleph_0$ (well-known from Cantor's diagonal argument). Taking this and that $| \Bbb N | = \aleph_0$ as given, consider the sets $\Bbb N \cup \{x\}$ where $x \in \Bbb R \setminus \Bbb N$ is fixed. It is trivial to show that this new set has cardinality equal to that of $\aleph_0$.
Consider the set of all such sets, for variable $x$ a real number but not a natural. It should be clear that since you have a unique set for each real number, then the set is bijective with the reals minus the naturals (which is obviously still the same as the reals in terms of cardinality). You could define the bijection explicitly:
$$f : \Bbb R \setminus \Bbb N \to \bigcup_{x \in \Bbb R \setminus N} \left\{ \Bbb N \cup \{x\} \right\} \;\;\;\;\; x \mapsto \left\{ N \cup \{x\} \right\}$$
(Anecdote, this motivated the unnecessary restriction that $x$ is not a natural number, the simplicity and clearness that $f$ is a bijection. Just in case you thought that was weird.)
Thus, since the sets are bijective, they share cardinality, i.e.
$$\overline{\aleph_0} \geq \left| \bigcup_{x \in \Bbb R \setminus N} \left\{ \Bbb N \cup \{x\} \right\} \right| = |\Bbb R \setminus \Bbb N| = | \Bbb R | = c > \aleph_0$$
Thus, by construction, we have found a subset of $\overline{\aleph_0}$ which is uncountably infinite.
(Note that this does not mean that $|\overline{\aleph_0}| = c$, because it's a subset, and that need not hold. It doesn't make an assertion as to the specific)
Part $2$: The cardinality of $\overline{\aleph_0}$ is actually not well-defined:
The logical follow-up question is obvious: what is the cardinality of $\overline{\aleph_0}$ if not $\aleph_0$?
In fact, there can be no such cardinality assigned to the class.
Notice that our method above essentially allows us to construct a subset of $\overline{\aleph_0}$ of any cardinality. Let $S$ be a set of cardinality $|S|$. Then generalize our argument from before by just letting $x$ be from, not $\Bbb R \setminus \Bbb N$, but $S$ (minus the members of $\Bbb N$ in $S$ if you so choose). Redefining $f$ to account for the obvious changes in the previous argument then shows that $\overline{\aleph_0} \geq |S|$, contradicting the assignment of the cardinal for appropriately chosen $S$.
It is well known that there exists no "maximum cardinal" - that is, there is always a set with a bigger cardinality (always a bigger fish, as it were). You can't assign a value to $\overline{\aleph_0}$, because you have a set with a larger cardinality, from which you could construct subsets of $\overline{\aleph_0}$, and thus have a subset with greater cardinality than $\overline{\aleph_0}$.
It has the sort of flavor of Russel's paradox.
Part $3$: A conclusion:
The conclusion is - to talk about the cardinality of the set $\overline{\aleph_0}$ is mostly meaningless. If it existed, it would be uncountable - but to assign such a cardinality would be contradictory. $\overline{\aleph_0}$ is just that big.
Credits to bof from the comments of the question body for the idea which led to the construction of this argument.