Is it consistent with ZF that there is no uncountable set of algebraically independent reals?

Question

Exactly what the title asks. This question was inspired by this one which looks for a countable such set. It is fairly easy to construct a size $\mathfrak{c}$ algebraically independent set of reals using a diagonalization of size $\mathfrak{c}$ if you have AC. What about without it though?

Answer 1

John von Neumann proved that $A_r = \sum_{n=0}^\infty \frac{2^{2^{[nr]}}}{2^{2^{n^2}}}$ are algebraically independent in

von Neumann, J., A system of algebraically independent numbers., Math. Ann. 99, 134-141 (1928). ZBL54.0096.02.

So the answer is no. It is always provable there is a set of size continuum of algebraically independent real numbers.

(Source: François G. Dorais on MathOverflow)