TITLE: Easiest way to prove that $2^{\aleph_0} = c$
QUESTION [17 upvotes]: $\aleph_0$ is the cardinality of the set of natural numbers, $\aleph_0 = |N|$. $c$ is the cardinality of the continuum, i.e. the set of real numbers $c = |R|$.
I know that $|P(A)| = 2^{|A|}$. This means that the cardinality of the power set of a set is 2 raised to the power of the cardinality of that set.
This basically means that to prove  $2^{\aleph_0} = c$, I need to prove $c = |P(N)|$. I've seen the Cantor diagonal argument, although I have no idea on how to use it to prove this. 
Could anyone suggest ideas of the simplest way to prove that $2^{\aleph_0} = c$?

REPLY [5 votes]: Another approach (though slightly more advanced) is as follows:

Note that $| \mathbb{R} | = | ( 0 , 1 ) |$, and since $\mathbb{Q}$ is countable while $\mathbb{R}$ is uncountable, it follows that $| ( 0 , 1 ) | = | ( 0 , 1 ) \setminus \mathbb{Q} |$.
Note that every $x \in ( 0 , 1 ) \setminus \mathbb{Q}$ has a unique infinite continued fraction representation as $[0; a_1 , a_2 , \ldots ]$ where each $a_i \in \mathbb{N} = \{ 1 , 2 , \ldots \}$.  (And every such infinite continued fraction represents an irrational number in $(0,1)$.)  This establishes a bijection $\mathbb{N}^{\mathbb{N}} \to (0,1) \setminus \mathbb{Q}$.
It is quite easy to show that $| \mathbb{N}^{\mathbb{N}} | = 2^{\aleph_0}$.