Topology on Reals generated by Predicates What is the topology that the Real number has that is generated by the sets of the form $\{x\in \mathbb{R}:P(x)=T\}$ where P() is a predicate of the language of ordered fields (which is a predicate calculus with the constants $1,0$ the unary functions $-, $ and $()^{-1}$ defined only on non zero elements. The binary operations of addition and multiplication and the relation $<$) The sets clearly form a basis since the closure by finite conjunctions of predicates assures that the sets of the form mentioned above are closed by finite intersection. Each rational number can be expressed in this language so the set of rationals would have to be discrete. (The predicate $x=q$ is satisfied only by $q$ and so the singleton $\{q\}$ is open). By the same reasoning the algebraic numbers would need to be a discrete set in this topology. On the other hand the entire real line could not be discrete since that would imply an uncountable number of Predicates which cannot be the case since they are numerable. This topology would seem to be finer than the Euclidean one since one can define the sets of balls with rational radius and centered on a rational with a predicate which means this topology contains a basis of the euclidean one. What other kinds of sets would be open?
Edit: the topology must also be totally disconnected since the rationals are dense and the predicates of the form $x<q$ where $x$ is a variable and $q$ is a rational and the negation of such predicates mean that any pair of distinct reals cannot lie in the same connected part
 A: The theory of real closed fields admits quantifier elimination (over just the language of ordered rings, without the multiplicative inverse operation).  It follows that any definable (without parameters) subset of $\mathbb{R}$ is a finite Boolean combination of sets of the form $\{x:p(x)\geq 0\}$ where $p$ is a polynomial with integer coefficients.  Any such set is a finite union of intervals whose endpoints are algebraic numbers.  Since, as you have observed, every algebraic number is definable, it follows that the topology they generate is just the topology generated by open intervals and singletons whose elements are algebraic.  In other words, an open set in this topology is just a union of an open set in the usual topology and a subset of the real algebraic numbers.
This topology may seem pathological but it can in fact be embedded in $\mathbb{R}^2$.  Let $A$ be the set of real algebraic numbers and enumerate $A=\{a_n\}_{n\in\mathbb{Z}_+}$.  Consider the set $$X=(\mathbb{R}\setminus A)\times\{0\}\cup\{(a_n,1/n):n\in\mathbb{Z}_+\}\subset\mathbb{R}^2.$$  Then identifying $X$ with $\mathbb{R}$ via the first projection, the topology on $X$ as a subspace of $\mathbb{R}^2$ is the same as your topology on $\mathbb{R}$.
