Intuition behind o-minimal structures. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:31:40Z http://mathoverflow.net/feeds/question/107588 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107588/intuition-behind-o-minimal-structures Intuition behind o-minimal structures. Magnus Botnan 2012-09-19T16:27:04Z 2012-09-20T23:22:08Z <p>This is very much the same post as I posted at <a href="http://math.stackexchange.com/questions/195203/some-intuition-behind-o-minimal-systems" rel="nofollow">math.stackexchange</a>. </p> <p>I am following the definitions of an o-minimal system in "Tame Topology and O-minimal Structures" by Lou van den Dries.</p> <p>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 <em>some</em> 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. </p> <p>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). </p> http://mathoverflow.net/questions/107588/intuition-behind-o-minimal-structures/107596#107596 Answer by Ramiro de la Vega for Intuition behind o-minimal structures. Ramiro de la Vega 2012-09-19T17:34:54Z 2012-09-20T23:22:08Z <p>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.</p> <p><strong>Edit 1:</strong> Here are some details as to why $A$ has to be in $S_1$.</p> <p>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:</p> <p>$\exists b,c,d [B(a,b) \wedge (c \lt a \lt d) \wedge \forall e,f [B(e,f) \wedge c \lt e \lt d \rightarrow f \leq b]]$.</p> <p>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.</p> <p><strong>Edit 2:</strong> 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:</p> <blockquote> <p>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.</p> </blockquote> <p>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).</p> http://mathoverflow.net/questions/107588/intuition-behind-o-minimal-structures/107667#107667 Answer by Liviu Nicolaescu for Intuition behind o-minimal structures. Liviu Nicolaescu 2012-09-20T09:43:08Z 2012-09-20T09:43:08Z <p>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.</p> <p>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.</p> <p>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.</p>