Does the Continuum hypothesis say anything about the cardinality of the set $3^{{\aleph}_0}$? I thought of this amazing method which produces a set of cardinality $3^n$ from a set of cardinality $n$:
If we take a set S of cardinality $n$ and produce ordered pairs of the form (A,B), where A and B are subsets of S such that $A\cap B =\emptyset$, then the set of all those ordered pairs has cardinality $3^n$. This is simply the multinomial theorem. I even tested this on the set {1,2}. I got $3^2=9$ ordered pairs: ($\emptyset$, $\emptyset$), ($\emptyset$, {1}), ($\emptyset$, {2}), ($\emptyset$, {1,2}), ({1}, $\emptyset$), ({2,} $\emptyset$), ({1,2}, $\emptyset$), ({1},{2}), and ({2},{1}).
Let's call this set of cardinality $3^n$, the superpower set of  S. I was wondering that, if we assume the continuum hypothesis to be true, can we derive the cardinality of the superpower set of Natural numbers? What implication does $2^{{\aleph}_0}={{\aleph}_1}$ have on $3^{{\aleph}_0}=?$. Does $3^{{\aleph}_0}$ equal ${{\aleph}_2}$ or something?
 A: In general, given $\kappa$ an infinite cardinal, we have
$$ 2^\kappa \leq 3^\kappa \leq (2^\kappa)^\kappa = 2^{\kappa\cdot\kappa}=2^\kappa$$
showing that $2^\kappa = 3^\kappa$. Thus, if $2^{\aleph_0} = \aleph_\alpha$, then $3^{\aleph_0} = \aleph_\alpha$.
(Alternate Solution) For another (easier) way of thinking of it, we can define an explicit injection of $3^\kappa$ into $2^\kappa$. For simplicity we will just do the case where $\kappa=\aleph_0$, but there isn't too much difficulty extending to arbitrary $\kappa$ (infinite cardinal). The idea is to think of $2^{\aleph_0}$ as being the set of countably-infinite sequences of $0$'s and $1$'s, $2^\mathbb{N}$. Likewise, $3^{\aleph_0}$ is the set of countably-infinite sequences of $0$'s, $1$'s, and $2$'s, $3^\mathbb{N}$. We can 'encode' a sequence in $3^{\aleph_0}$ by using the encoding 
$$0 \mapsto 00, 1 \mapsto 01, 2 \mapsto 11$$ 
In other words, we define $f: 3^{\mathbb{N}} \to 2^{\mathbb{N}}$ by sending a sequence $(x_n)_{n\in\mathbb{N}} \in 3^\mathbb{N}$ to the sequence $(y_n)_{n\in\mathbb{N}}$ where
$$y_{2n},y_{2n+1} = \begin{cases} 0,0 & \text{if $x_n = 0$} \\ 0,1 & \text{if $x_n=1$} \\ 1,1 & \text{if $x_n=2$} \end{cases}$$
$f$ is an injection, and there is another obvious injection $2^\mathbb{N} \to 3^\mathbb{N}$. Cantor-Schroeder-Bernstein then shows $2^{\aleph_0} = 3^{\aleph_0}$.
(Alternate Solution 2) For a fun proof, note that when you put the discrete topology $3=\{0,1,2\}$, $3^\mathbb{N}$ (with the product topology) is a compact metrizable topological space. But since compact metrizable spaces are second-countable, the cardinality of $3^\mathbb{N}$ is at most continuum, i.e. $2^{\aleph_0}$.
