I'm trying to come up with a countable set of real numbers that are transcendental and algebraically independent that's concrete and easy to work with.
The idea I had was to come up with very sparse decimals whose sole digits are $0$ and $1$ (although the use of base ten is somewhat arbitrary) where the non-zero digits get further and further away very rapidly.
Let $b(p)$ denote $\sum_{k \ge 1} 10 ^ {-p^{k!}}$ for a prime $p$.
Is the set consisting of $b(p)$ for prime $p$ algebraically independent?
What follows is my attempt to solve the problem myself.
I have the following argument for why a specific $b(p)$ is transcendental. The argument is not complete, I don't know how to show that we can neglect carries.
Let $q(x) = \sum a_ix^i$ be a polynomial of degree $n$.
Let $\Delta_k(a)$ be the number of ways of making $a$ out of $k$ prime factorial powers.
We are looking at $q\left( \sum_{k \ge 1} \frac{1}{10^{p^{k!}}} \right)$.
$$ \sum_{i \ge 0} a_i \left(\sum_{k \ge 1} \frac{1}{10^ {p^{k!}}} \right)^i \\ \sum_{i \ge 0} a_i \sum_{s \ge 1} \frac{\Delta_i(s)}{10^s} \\ \sum_{i \ge 0, s \ge 1} \frac{a_i \Delta_i(s)}{10^s} $$
Let's look at $\Delta_n(p^{(n+1)!})$ where $n$ is the degree of $q(x)$. I claim that $\Delta_n(p^{(n+1)!})$ is $1$ because it would require $n+1$ copies of $p^{n!}$ multiplied together to make $p^{(n+1)!}$. Therefore the value of $\Delta_n(p^{(n+1)!})$ is exactly one, not accounting for caries.
By a similar argument so is $\Delta_n(p^{(n+2)!}), \Delta_n(p^{(n+3)!}) \cdots$.
Intuitively, the values of nearby digits are far too low for a cascading carry to propagate to these places, but I'm not sure how to prove that.