Assume generating means that for all $k$ there exists $N_k$ such that after $N_k$ steps the $k$th cell on the tape consists of the $k$th digit of $\pi$'s binary expansion and will stay there forever.
Can such a machine exist? can it not?
EDIT
Is this true for every irrational number?
So this is true for $\pi$ since there exists a Turing machine to calculate the $k$th digit of $pi$ so we can construct a machine that calls the prior machine to calculate digit by digit.
But this can not be true for all irrational numbers since there are uncountably many irrational numbers and only countably many turing machines.
From this we can conclude that $\pi$ is among a special countable set of irrational numbers, call it $G$, that are generatable in the sense above.
Do we know what is the set $G = \{ g \in \mathbb{R} \setminus \mathbb{Q} : g$ is generatable $\}$ in a mathematical sense or a more informative way? We know $\pi \in G$ and we know $G$ is countable. Is there more to the picture?
