nonstandard topology? It is possible to have nonstandard models of PA where the natural numbers are different.
The definition of a topology requires a notion of finiteness.

What happens if we use a nonstandard model of arithmetic to say what finiteness means in the definition of topology?

 A: It turns out that we needn't use arithmetic to define finiteness at all. For example, a set $S$ of sets is finite (in the sense we're used to) if and only if each non-empty subset of the power set of $S$ has a minimal element with respect to inclusion ($\subseteq$). A general set $X$ (not necessarily a set of sets) is finite if and only if every injection $X\to X$ is a surjection. Alternately, $X$ is finite if and only if every inductive family of subsets of $X$ contains $X$. An inductive family of subsets of $X$ is a set $\mathcal{A}$ of subsets of $X$, such that $\emptyset\in \mathcal{A}$, and such that for all $A\in\mathcal{A}$ and all $x\in X$ we have $A\cup\{x\}\in\mathcal{A}$.
A: The definitions of topology require some notion of set theory. Therefore we need to work in a theory which is stronger than that of $\sf PA$.
Mostly we work in $\sf ZFC$, or some variation of it. The important part is that these theories prove the consistency of $\sf PA_2$, that is the second-order theory of $\sf PA$, which is categorical. So internally to a universe of $\sf ZFC$ there is no such thing as a non-standard integer.
Of course, there are many different ways of defining finiteness (especially in $\sf ZFC$), which do not make any appeal to the integers themselves. But the point is that topology is working within the universe of a theory that proves that all the integers are standard (again, internally, it might be the case that the model itself does have non-standard integers, but it cannot know about it from an internal point of view).
