Question: Semi-algebraic geometry studies the solution sets of finitely many polynomial inequalities (in $\mathbb{R}^n$). What field (if any) could be considered the study of the solution sets of arbitrarily/infinitely many polynomial inequalities? Maybe real analytic geometry?
In (basic/elementary) algebraic geometry using $\mathbb{C}$/$\mathbb{P}^n(\mathbb{C})$ we have a Noetherian condition and the Nullstellensatz (I forget the precise reason) guaranteeing us that any projective variety (solution set of polynomial equations), even those defined in terms of infinitely many polynomial equations, can be written as the solution set of finitely many polynomial equations. I.e. there is effectively no difference between restricting to finitely many or allowing arbitrarily many defining conditions.
Polyhedra can be associated with finitely many linear inequalities in $\mathbb{R}^n$. If I understand/remember correctly, any (closed) convex set is (hypothetically) the intersection of the solutions to (arbitrarily) infinitely many linear inequalities in $\mathbb{R}^n$. (Where finite is a special case of infinite I guess.)
Since convex sets are much more general than polyhedra alone, we see that for this type of geometry, there is no analogous "Noetherian condition", and thus allowing infinitely many defining linear inequalities instead of only finitely many is a strict/substantive generalization.
Hence, one might expect that there is also a strict increase in generality when going from semi-algebraic geometry, i.e. finitely many polynomial inequalities, to infinitely many polynomial inequalities. (In particular, the restriction to only finitely many polynomial inequalities in the definition of semi-algebraic set seems motivated by the absence of a "Noetherian condition".)
Note that because $\mathbb{R}$ is an ordered field but $\mathbb{C}$ isn't, the notion of inequalities, and thus solutions sets of inequalities, makes sense in $\mathbb{R}^n$ but not in $\mathbb{C}^n$.
This answer mentions semi-analytic sets in $\mathbb{R}^n$, also noting the crucial feature of $\mathbb{R}$ being an ordered field. However, I don't really expect that semi-analytic geometry and/or real-analytic geometry is what results when we generalize from finitely many polynomial inequalities to infinitely many polynomial inequalities, for at least two reasons:
- We do have a Noetherian condition in the case of complex (projective) varieties (solutions to polynomial equations), and yet complex analytic geometry exists.
- The limit(s) involved in going from the intersection of finitely many sets of a certain class to infinitely many sets of the same class is qualitatively different from the type(s) of limit involved in constructing analytic functions from polynomials. In particular, when $n=1$, on any compact subset of $\mathbb{R}$ any continuous function can be approximated uniformly by polynomials by the Stone-Weierstrass theorem. I think a corollary of this is then that any inequality which, at least on a compact subset of $\mathbb{R}^n$, corresponds to a single continuous function, can be formed as the intersection of the solutions to infinitely many polynomial inequalities.
Of course, continuous functions are much, much more general than real-analytic functions. So perhaps a better suggestion for the "infinite limit" of semi-algebraic geometry would be the study of real topological manifolds? (Assuming these contain as special cases the solution sets of inequalities defined by continuous functions; I don't know much about topological manifold theory.)
Every subset of $\mathbb{R}^n$ can be defined by polynomial inequalities. Indeed, for any point $p=(p_1,\dots,p_n)$, the set $\mathbb{R}^n\setminus\{p\}$ is defined by the inequality $\sum(x_i-p_i)^2>0$. Given any $A\subseteq\mathbb{R}^n$, then, it is defined by the inequalities of this form for all $p\not\in A$.