How would I prove that the interval $[0, 1]\setminus\mathbb{Q}$ is uncountable? I am studying measure theory and looking for how to prove this. I would like to achieve this without the use of measure theory. 
 A: First, we show that $\mathbb{Q}$ is countable and that the interval $[0,1]$ is uncountable. 
Once we have shown these facts, then we assume that $[0, 1] \setminus \mathbb{Q}$ is countable. 
But then $$[0, 1] = \left( [0, 1] \cap \mathbb{Q} \right) \cup \left( [0, 1] \setminus \mathbb{Q} \right).$$ 
Finally, we show that the union of any two (or finitely many!) countable sets is countable, and so is any (infinite) subset of a countable set, thus arriving at a contradiction. 
A: We can find an uncountable subset of $[0,1]\backslash \mathbb{Q}$. Here's a suggestion. Collect all numbers with specified decimal expansion:
For any $0,1$ sequence $a=(a_n)$, define 
$$
b_{a}=0.b_1 b_2 b_3 \cdots b_n \cdots,
$$
where $b_i=0$ if $i\neq  n!$, $i\neq n!+1$, $n\geq 2$ and 
$b_{n!}=1$, $b_{n!+1}=a_n$, $n\geq 2$. 
Collection of all such numbers is uncountable because it contains a copy of $\{0,1\}^{\mathbb{N}}$. 
Added: This set also provides an example of a perfect set consisted entirely of irrational numbers. 
