Example of set of cardinality $\aleph_2$ I am looking for an example of a set of cardinality $\aleph_2$, such as the continuum is an example for cardinality $\aleph_1$.
 A: Assuming the generalized continuum hypothesis, $\aleph_2=2^{\aleph_1}=2^{2^{\aleph_0}}=\cal |P(P(\Bbb N))|=|P(\Bbb R)|$.
If you think about the cardinality of the real numbers as $\aleph_1$, then you might as well think about the cardinality of $\mathcal P(\Bbb R)$ as $\aleph_2$.
Of course, as Clive pointed out, it is provably unprovable that $|\Bbb R|=\aleph_1$ from the standard axioms of set theory (namely $\sf ZFC$), so it is also unprovable that $\mathcal P(\Bbb R)$ has size $\aleph_2$. In fact under some set theoretic axioms we can prove that $\Bbb R$ is a set of size $\aleph_2$.
Instead, the definition of $\aleph_1$ is the least cardinality larger than $\aleph_0$. Namely, the least size of an uncountable set. The definition of $\aleph_2$ is the least uncountable cardinal larger than $\aleph_1$. Of course, the way I wrote the definitions here require us to assume the axiom of choice, but this can be slightly modified to avoid this issue.
Using this definitions, we have really standard ways to define $\aleph_1$. Consider the set of all well-orderings of $\Bbb N$, namely $X=\{R\subseteq\Bbb{N\times N}\mid (\Bbb N,R)\text{ is a well order}\}$. Now consider $\equiv$ defined on $X$ as the order-isomorphism relation. Then $\aleph_1$ is defined as the cardinality of $X/\equiv$, the set of equivalence classes of well-orders of $\Bbb N$.
Similarly, if we consider $A$ to be a set of size $\aleph_1$, then $\aleph_2$ is defined as the cardinality of the set $\{R\subseteq A\times A\mid (A,R)\text{ is a well order}\}/\equiv$, where $\equiv$ as before is the order-isomorphic relation.
Now if we consider an ordinal as a set $T$ such that $(T,\in)$ is a well ordered set, and whenever $x\in T$, then $x\subseteq T$ (for example $\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}$ are all ordinals). Then $\aleph_1$ is the size of the least ordinal which is not countable, and $\aleph_2$ is the size of the least ordinal which is not countable or has size $\aleph_1$.
A: If you think of "the continuum as an example for cardinality $\aleph_1$" then I'm guessing that you (or your professor) are (at least implicitly) assuming the so-called Generalized Continuum Hypothesis. Nothing wrong with that; the GCH is a perfectly cromulent foundation for mathematics, even if the set-theory experts around here are more interested in various exotic alternatives.
In that case, the set of all functions $f:\mathbb R\to\mathbb R$ is a good example for cardinality $\aleph_2$. However, it is important to realize that we are considering arbitrary functions. If you only count continuous functions, there are only $\aleph_1$ of those. So most of those $\aleph_2$ functions are very ragged, and their graphs do not look like anything you would call a "curve".
You can also get $\aleph_2$ as the number of functions $f:\mathbb R\to\{0,1\}$ which is the same as the number of subsets of $\mathbb R$, since a subset $X$ can be coded by a function $f$ such that $f(x)=1$ if $x\in X$ and $f(x)=0$ if $x\notin X$.
A: The continuum isn't provably a set of cardinality $\aleph_1$ unless you assume the continuum hypothesis. The most canonical set of cardinality $\aleph_{\alpha}$ (for any ordinal number $\alpha$) is the ordinal $\omega_{\alpha}$.
...so for example
$$\omega_1 = \text{the set of countable ordinals}$$
has cardinality $\aleph_1$, and
$$\omega_2 = \text{the set of ordinals whose cardinality is} \le \aleph_1$$
has cardinality $\aleph_2$.
The continuum, i.e. the cardinality of $\mathbb{R}$ is $2^{\aleph_0}$, which is not provably equal to $\aleph_{\alpha}$ for any fixed value of $\alpha$.
