I suppose the answer is E but not quite sure why because I only ruled out the first four. Any help would be appreciated!
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$\begingroup$The answer is not (e):
The set of all subsets of $\Bbb R$ is of cardinality strictly greater than $\Bbb R$.
The set of all countable subsets of $\Bbb R$ is the same cardinality as $\Bbb R$. See Cardinality of the set of at most countable subsets of the real line?
Reiterating the argument made there, invoking the axiom of choice we associate a sequence of real numbers to each countable subset. We have then $|\Bbb R^\Bbb N|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot \aleph_0}=2^{\aleph_0}=|\Bbb R|$
From here, we recognize that the set of all subsets of $\Bbb R$ is equal to the union of the set of countable subsets and the set of uncountable subsets of $\Bbb R$.
It follows then that the set of uncountable subsets of $\Bbb R$ must be of cardinality strictly greater than $\Bbb R$.
The answer is not (c) either as there is an obvious bijection between the set of cofinite subsets with the set of finite subsets. You can list finite subsets of $\Bbb N$ just as easily as you can list finite subsets of $\Bbb Z$. List all who have the maximum absolute value equal to 1 (there are finitely many), followed by all who have maximum absolute value equal to 2 (there are finitely many), and so on... showing that (c) is countable.
(b) is countable by the same logic.
(a) is countable as there is a clear bijection with $\Bbb Z\times \Bbb Z$.
Finally as for (d), an ordinal is the set of all ordinals below it, so if the object described in (d) were a set and existed it would be the "biggest ordinal" and would clearly be greater in size than $\aleph_3$ and so bigger than the cardinality of the real numbers as well.
I am less familiar with these objects and suspect that the object described in (d) is not even a set in the first place but rather a proper class.
$\endgroup$ $\begingroup$I think we can also show that e) is not the answer without the Axiom of Choice:
Let $\mathbb{R}_{\rm unc}$ be the set of all uncountable subsets of the reals. Also let $\mathcal{P}(S)$ be the power set of set $S$.
Starting with $\mathbb{R}\equiv (0,1)$ we can form a corresponding bijection $f:\mathcal{P}(\mathbb{R})\longrightarrow \mathcal{P}((0,1))$.
Now consider the map $g: \mathcal{P}(\mathbb{R})\longrightarrow \mathbb{R}_{\rm unc}$ given by $g(S) = f(S)\cup (1,2)$. This is an injective map since $f$ is injective.
Also the inclusion map $\iota: \mathbb{R}_\rm {unc}\longrightarrow \mathcal{P}(\mathbb{R})$ is an injection.
So by the Cantor-Schroeder-Bernstein Theorem, $ \mathbb{R}_\rm {unc}$ and $\mathcal{P}(\mathbb{R})$ have the same cardinality.
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