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?
Asked
Active
Viewed 144 times
6
-
4https://mathoverflow.net/questions/23202/explicit-big-linearly-independent-sets See the answer, it gives an algebraically independent sets. – Asaf Karagila Oct 15 '18 at 18:55
1 Answers
10
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.
Asaf Karagila
- 393,674
-
-
2
-
-
1
-
Thank you for the answer. And also for the comment pointing to the correct post on Math overflow. I must say I'm very surprised. I assumed the answer would be Yes. Given you can't always get a vector basis. I have much more thinking to do. – DRF Oct 16 '18 at 12:36
-
Transcendence basis and Hamel basis are very different from "uncountable sets of such and such". – Asaf Karagila Oct 16 '18 at 12:37