Uncountable, algebraically independent subset of $\mathbb{C}$?

Question

Does such a subset exist? I am interested in algebraic independence over $\mathbb{Q}$.

Could this be proven in an abstract way or would it be more appropriate to construct an explicit example?

Answer 1

Order the algebraically independent subsets of $\mathbb{C}$ by inclusion. The union of a chain of algebraically independent subsets of $\mathbb{C}$ is algebraically independent. Thus by Zorn's Lemma there is a maximal algebraically independent subset $I$ of $\mathbb{C}$.

The set $I$ cannot be countable, since the algebraic closure of a countable subset of $\mathbb{C}$ is countable.

One can give an alternate proof by well-ordering the reals as $r_{\alpha}$, where $\alpha$ ranges over the ordinals $\lt c$. Let $i_0$ be the first transcendental under this ordering. For any ordinal $\beta\gt 0$, let $i_{\beta}$ be the smallest real under the well ordering which is not algebraic over the $i_{\gamma}$ where $\gamma$ ranges over the ordinals $\lt \beta$.