Curve in $(\mathbb{R},<)$ going to infinity

Question

My question is the following: Given the structure $(\mathbb{R},<)$ and $t \in \mathbb{R}$, can I have a definable function $f$ over a finite set of parameters, with domain $(-\infty, t)$ and with $lim_{x \to t} f(x) = \infty$ ?

Thanks!

Answer 1

No. Suppose $f$ is a function $D \rightarrow \mathbb{R}$ (here $D\subseteq\mathbb{R}$ is the domain of $f$) definable in the language $\{<\}$. Then $D$ can be expressed as a union of finitely many intervals, such that on each of these intervals, $f$ is either constant or the identity function.

This follows from quantifier elimination for the theory of dense linear orders. If you'd like more details, I can write them out later.

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Edit: Based on your comment, I guess you'd like more details.

First, notice that $\text{Th}(\mathbb{R},<) = \text{DLO}$, the theory of dense linear orders without endpoints. I'll assume you're familiar with the fact that this theory has quantifier elimination.

What are the definable subsets of $\mathbb{R}$ in this language? You can easily check that the set defined by an atomic formula in one free variable looks like $\{a\}$, $(-\infty,a)$, $(a,\infty)$, $\mathbb{R}$, or $\emptyset$ (corresponding to $x = a$, $x < a$, $a < x$, $x = x$, and $x < x$). In the cases involving $a$, $a$ is a parameter in the atomic formula. This is what makes $(\mathbb{R},<)$ so much simpler than the general o-minimal case - if we have the structure of a field around, for example, we could form terms from $a$ and obtain intervals with different endpoints.

By quantifier elimination, every formula in one free variable is equivalent to a boolean combination of atomic formulas. Using disjunctive normal form, you can see that such a thing can always be expressed a finite union of points and open intervals. Moreover, the points and the endpoints of the intervals involved are all among the parameters in the formula!

What are the definable functions in this language? Suppose $\phi_f(x,y)$ is a formula, with hidden parameters $\overline{r}$, defining the function $f$ (so $\mathbb{R}\models \phi_f(a,b)$ iff $f(a) = b$). For any $a$ in the domain of $f$, $\phi_f(a,y)$ defines a subset of $\mathbb{R}$. This subset is just a singleton, $\{f(a)\}$, so $f(a)$ must be among the parameters of $\phi_f(a,y)$. That is, $f(a)$ must be $a$ or one of the parameters in $\overline{r}$.

We've seen that at each point $a$ in the domain of $f$, we have $f(x) = x$ or $f(x) = r_i$ one of finitely many constants $\overline{r}$. To finish, note that the formulas $\phi_f(x,x)$ and $\phi_f(x,r_i)$ pick out the subsets of $\mathbb{R}$ on which $f$ exhibits each of these behaviors. Each of these subsets is a finite union of intervals, so we can decompose the domain of $f$ into a finite union of intervals, on each of which $f$ is constant or the identity function.