Is there another way to proof that there can't be a bijection between reals and natural not using Cantor diagonal? Is there another way to proof that there can't be a bijection between reals and natural not using Cantor diagonal?
I was wondering about diagonal arguments in general and paradoxes that don't use diagonal arguments. Then I was puzzled because I couldn't think another way to show that the cardinality of the reals isn't the same as the cardinality of naturals.
 A: One way is through the use of measure theory. The measure of the unit interval $[0,1]$ is $1$, while the measure of any countable subset of the reals is $0$. This is enough to demonstrate that $[0,1]$ is uncountable, for if it were countable then its outer measure would have to be $0$.
The outer measure of a general subset $A \subset \mathbb R$ is defined as
$$m^*(A) = \inf \Big\{ \sum_i \ell(I_i) : A \subset \bigcup_i I_i\Big\}$$
where $(I_i)_i$ is any countable collection of intervals, $I_i = [a_i, b_i]$ for some reals $a_i < b_i$ for each $i$, and $\ell(I)$ is the length of the interval, defined as $\ell(a,b) = b - a$. What this definition means is: cover the set $A$ by (at most countably many) intervals, however you may like, and sum the lengths of those intervals. The (limiting) smallest possible total length is what the outer measure of $A$ is.
See here for a proof that a countable set has outer measure zero, and see here for a proof that the unit interval $[0,1]$ has outer measure $1$.
