First-order definition of "$f$ is continuous at $x$" using just $<$

Question

I need to show that the set $\{ a\in \mathbb{R}\ |\ f\ \text {is continuous at a}\}$ is definable in the structure $(\mathbb{R};<,\ f)$, where $f$ is just some function $f:\mathbb{R}\rightarrow\mathbb{R}$.

I'm having a really hard time getting started, given the pretty severe limitations of the language. Any help would be appreciated.

Answer 1

Here, $\mathbb{R}$ is just an ordered structure, with the order topology, which is the same as the usual topology. The algebraic operations aren't needed to express continuity:

$$ \begin{align} \varphi(x) \iff (\forall r_0, r_1)\,(&r_0 < f(x) < r_1\!\implies\! \\ &(\exists u_0,u_1)(u_0 < x < u_1\, \land \\ &\quad\qquad(\forall y)(u_0 < y < u_1 \!\implies\! r_0 < f(y) < r_1))) \\ \end{align} $$ That is, "for every open interval $(r_0, r_1)$ around $f(x)$, there is an open interval $(u_0, u_1)$ around $x$ such that $f$ maps $(u_0, u_1)$ into $(r_0, r_1)$". So this is a $\Pi^0_3$ predicate.