First-Order Topology

Question

Is it possible to formalise topology in first-order logic? When one associates a first-order structure to a topological space, what does this mean?

Answer 1

A topology is not first-order definable, since it is defined in terms of subsets. However, one can define a topological first-order structure as follows:

First recall that a first-order language $\cal L$ consists of a set $\cal R$ of relation symbols and a set $\cal F$ of function symbols, and associated to each element $r\in \cal R$ (resp. $f\in \cal F$) a positive (resp. non-negative) integer called the arity of $r$ (resp. $f$). For each integer $n$, we let $\cal R_n$ (resp. $\cal F_n$) be the set of relation (resp. function) symbols of arity $n$. The elements of $\cal F_0$ are called constants.

An $\cal L$-structure $S$ is a topological $\cal L$-structure if the set $S$ is equipped with a topology such that:

1. for each $f\in \cal F_n$ ($n > 0$), the mapping $(s_1,\ldots, s_n)\mapsto f(s_1,\ldots, s_n)$, from $S^n$ into $S$, is continuous, and
2. for each $r\in \cal R_n$ ($n > 0$), the subset $r$ of $S^n$ is closed.