Timeline for Undefinability of $\mathbb{Z}$ in the reals
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 16, 2019 at 18:01 | comment | added | user44143 | $(\forall x\, \exists y>x\, \phi(y))\rightarrow (\exists x\,\forall y>x\, \phi(y))$ | |
| Apr 25, 2017 at 8:32 | comment | added | Cubikova | @MattF. True, indeed this is some sort of "o-minimality near infinity": every cofinal definable set contains an interval of the form $(a,+\infty)$ for some $a\in \mathbb{R}$. This property implies of course that cofinal definable sets are uncountable, but also states the stronger condition of containing an interval. I guess the we can weaken then the property that implies the undefinability of $\mathbb{Z}$ to be just ''cofinal definable sets are uncountable''. | |
| Apr 25, 2017 at 7:18 | comment | added | Mikhail Katz | @MattF, it would be nice to have some parentheses in the formula you presented. Without them this is a bit tricky to read. | |
| Apr 24, 2017 at 20:34 | comment | added | user44143 | Or: from the similarly weak fact about Th(R) that $\forall x \exists y>x \,\phi(y) \rightarrow \exists x \forall y>x \,\phi(y)$ for any potential definition $\phi$. This might be easier to prove than quantifier elimination, without all the technicalities of Sturm's lemma, just focusing on the easy algebraic geometry near infinity. | |
| Apr 24, 2017 at 14:49 | history | answered | Cubikova | CC BY-SA 3.0 |