Does $\pi$ contain every real? One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $\pi$, and then taking every $n$th digit.
$\pi = 3.141592653589793238462643383279502884197169399...$
for instance, for $m=3$, $n=5$, this will be
3.141592653589793238462643383279502884197169399...
giving
$0.469363019...$
I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.
So I guess this method can't create any real. But why?
 A: You basically present a mapping from $\mathbb N^2$ to $\mathbb R$ and ask if the mapping is surjective. Since $\mathbb N^2$ is countable, and $\mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.
In particular, the argument would go like so:


*

*Let $F$ be the mapping you construct going from $\mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $\pi$, and then taking every $n$th digit.

*Let $f:\mathbb N\to\mathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

*Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}\dots$

*Let $$b_i=\begin{cases}1 & \text{if } a_i^{(i)} = 0\\ 0& \text{if } a_i^{(i)} \neq 0\end{cases}$$

*Let $b=0.b_1b_2b_3\dots$. Clearly, for all $n\in\mathbb N$, we see that $F(f(n))\neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

*From $5$, it follows that the number $b$ is not in the domain of $F\circ f$.

*Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.

