Tuesday, October 04, 2016
A Second Order Statement true in (R,+) but not (Q,+)
In my last post I asked
Is there a first order statement true in (R,+) but false in (Q,+)
Is there a second order statement true in (R,+) but false in (Q,+)
First order: No.
One reader said it followed from the Lowenheim-Skolem Theorem. I can believe this but don't see how.
One reader said the theory of (Q,+) is complete. True, but would like a reference.
I would prove it by a Duplicator-Spoiler argument.
Second order: Yes
Some readers thought I had < in my language. I do not so I think that examples using < do not work- unless there is someway to express < in my language.
Some readers thought that second order meant you could quantify over functions. My intent was just to be able to quantify over sets.
Some had the right answers. The one I had in mind was
There exists A,B such that A,B are subgroups with at least two elements that only intersect at 0.
Here is a writeup.