Is there any subset of Complex numbers that is algebraically closed?

Question

That any polynomial that is allowed to have coefficients from that subset has also a root in that subset

Answer 1

The set of algebraic numbers is algebraically closed. It is countably infinite, so it is a very small subset of the set of complex numbers.

Note that an algebraic number is a zero of a non-constant polynomial with integer coefficients. The set of algebraic numbers is the smallest algebraically closed subset of $\mathbb{C}$.

There are many other proper subsets of the complex numbers that are algebraically closed.

Answer 2

the algebraic numbers, say $Q_0$ are the algebraic closure of the rationals. if $\alpha_1$ is transcendental then set $Q_1$ to be the algebraic closure of $Q(\alpha_1)$. this can be continued quite far, modulo the availability of a choice function