This is very much the same post as I posted at math.stackexchange.

I am following the definitions of an o-minimal system in "Tame Topology and O-minimal Structures" by Lou van den Dries.

It is immediate from the definition that the graph of $\sin(x)$ is not a tame set (intersect it with $y=0$). But what about a slightly rotated one? Or one which is both rotated and translated. To me they look to be tame (unless rotated by $\pi/4$). Is it correct that these sets are contained in some o-minimal system? And how can I easily 'recognize' tame sets? E.g. my intuition is that a collection of sets in $\mathbb{R}^2$ are tame if they do not invalidate the minimality axiom ($S_1$ contains exactly finite unions of points and open intervals). If so I can just complete with whatever sets needed in order for it to be a o-minimal structure.

I am familiar with the monotonicity theorem and how that may obstruct a cell decomposition of the aforementioned sets but I would really like to see from the very definition what goes wrong. It would be no problem if all lines in $\mathbb{R}^2$ were included but I cannot see how that is the case (all horizontal, vertical and $y=x$ are included by definition).


Let $B$ be a slight rotation of the graph of the sine function and lets say that $B \in S_2$. Then one should be able to show from the definitions that the set $A$ of reals where $B$ has a "local maximum" is in $S_1$. But the set $A$ is not a finite union of points and intervals.

Edit 1: Here are some details as to why $A$ has to be in $S_1$.

First write a first order formula (using only the symbols $B$ and $\lt$ as binary predicates in addition to logical symbols) defining the elements of $A$. For instance, note that $a \in A$ if and only if:

$ \exists b,c,d [B(a,b) \wedge (c \lt a \lt d) \wedge \forall e,f [B(e,f) \wedge c \lt e \lt d \rightarrow f \leq b]] $.

Now the formula should guide you through the axioms you need to use to show that $A \in S_1$. For example, you could start saying something like "$A$ is the projection onto the first coordinate of certain set, which is in $S_4$ because...". Just remember that universal quantifiers correspond to complements of projections of complements and other logical symbols correspond to boolean operations.

Edit 2: On the other hand, if $B$ is the graph of the sine function rotated by $\pi/4$ and translated (say upwards) so that it doesn´t intersect the line $x=y$, then $B$ can actually be an element of an o-minimal structure (although not one that extends the ring structure of the reals). The reason is the following fact:

If $f,g: \mathbb{R} \to \mathbb{R}$ are two continuous increasing bijections such that $x \lt f(x)$ and $x \lt g(x)$ for all $x \in \mathbb{R}$, then the structures $(\mathbb{R},\lt,f)$ and $(\mathbb{R}, \lt, g)$ are isomorphic.

So taking $f$ as the function whose graph is $B$ and taking $g(x)=x+1$, we get that $(\mathbb{R},\lt,B)$ is o-minimal since $(\mathbb{R}, \lt, g)$ is o-minimal (being a "reduct" of the real field).

  • 2
    $\begingroup$ I don't see any need to intersect with a straight line. You just need to have some definable subset of the structure that isn't a finite union of intervals. Ramiro de la Vega proposes the set of $x$-coordinates of local maxima of the rotated graph, and this looks correct to me. $\endgroup$ – Andreas Blass Sep 19 '12 at 18:16
  • 1
    $\begingroup$ Because the projection of the set of points where the local maxima occur onto the $x$-axis belong to the o-minimal structure, by definition of o-minimal structure. $\endgroup$ – Todd Trimble Sep 19 '12 at 18:57
  • 1
    $\begingroup$ @Magnus: Write down what it means for $x$ to be the projection of a local maximum of a function $f$, and observe that what you wrote is a first-order definition in the language of the real field plus $f$. $\endgroup$ – Andreas Blass Sep 19 '12 at 19:37
  • 1
    $\begingroup$ @Andreas: I believe in this context you don't actually have the field structure (in which case the answer to the original question would be a bit easier since any straight line would be definable) although most of the time people only consider o-minimal structures which extend the real field. $\endgroup$ – Ramiro de la Vega Sep 19 '12 at 21:08
  • 1
    $\begingroup$ It seems to me that all sines rotated $\pi/4$ and translated, say, $k$ along the y-axis can very well be included in an o-minimal structure? This can be viewed as the graph of a strictly increasing function and hence does not conflict the monotonicity theorem. Also, it has no local maximas and after more thought it does not seem possible to get inflection points using only < (one needs subtractiona and division?). $\endgroup$ – M.B. Sep 20 '12 at 14:15

Lines, and more generally affine subspaces are definable. Here is why. Addition is definable, multiplication is definable, combining these you deduce that any affine map is definable. The zero set of a definable map is definable. An affine subspace is the zero set of some affine map, and its zero set will be definable.

The image of a definable set via a definable map is a definable set. Linar maps such as rotations are definable so the image of the graph of $\sin x$ via a rotation cannot be definable. If it were, applying the opposite rotation we would deduce that the graph itself would be definable.

Here an alternate argument. The intersection of two definable sets is a definable set. In particular, the intersection of a definable set with a line will be a definable set on that line, i.e., a finite union of open intervals and points. Clearly by rotating the graph of $y=\sin x$ you can still find a line so that its intersection with the rotated graph is an infinite countable set of points.

  • 5
    $\begingroup$ No, addition and multiplication need not be definable in the general sense of o-minimal structures -- only if you are considering structures that are expansions of the reals with its ordered ring structure. Most of the time this is what we want of course, but if you read this discussion, you'll see that the OP doesn't assume this. $\endgroup$ – Todd Trimble Sep 20 '12 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.