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1d
comment Keisler Order: Saturated Ultrapowers
I've elaborated a little more, though it's been a while since I've done anything with ultrapowers and my notes are spotty, so I wouldn't trust myself to flesh out the details without creating an enormous mess. But I think what I've written out is (hopefully) correct, and gives some idea of this.
1d
revised Keisler Order: Saturated Ultrapowers
added 902 characters in body
1d
comment Keisler Order: Saturated Ultrapowers
Sorry; yes, you are exactly right - $f$ itself need not be multiplicative (as you pointed out, it's very easy for it not to be), but there is a refinement which is. I just fixed it, I hope it's correct now.
1d
revised Keisler Order: Saturated Ultrapowers
added 63 characters in body
1d
revised Keisler Order: Saturated Ultrapowers
added 538 characters in body
2d
revised Keisler Order: Saturated Ultrapowers
added 21 characters in body
Aug
17
answered Axiomatizability of finite Isomorphic Classes
Aug
17
answered Keisler Order: Saturated Ultrapowers
Jun
12
awarded  Yearling
May
26
comment first order logic models
The answer will (obviously) be 'no'. To see this, suppose $\varphi$ were such a sentence and either induct on complexity or put it in some sort of normal form ('prenex normal form' seems to be the phrase used on Wikipedia). The analysis may be tedious, but should come with relative ease.
Jan
5
comment Must vectors in $\mathbb{R}^n$ have their “tail” at origin?
When you think of $\mathbb{R}^n$ geometrically as a vector space, you are considering each vector as a direction arrow with a tail at the origin. So $(0, 10) - (0, 9)$ is an arrow $(0, 1)$ with a tail at the origin. Vectors here are only directions; the arrow you would want to draw from the head of one vector to the head of the other is not actually a vector unless you translate it back to the origin. If you try to define a vector space structure on $\mathbb{R}^n$ using the geometric interpretation of addition above, you'll run into problems (e.g. with additive identity).
Nov
21
answered is it possible for two fields with the same characteristic to not be isomorphic?
Nov
13
answered A subgroup $H$ is normal in $G$ if and only if…
Nov
6
revised Evaluate the limit without L'Hospital
fixed texing
Nov
6
suggested suggested edit on Evaluate the limit without L'Hospital
Oct
28
comment What is $R(\omega)$ (and where can I find definitions for similar common notation)?
Most books have an index of symbols at the back (my copy of Chang & Keisler does), which you can use to find where things are first defined. $R(\omega)$ is defined inductively on p. 45, and more clearly on p. 588.
Oct
28
revised $\omega$-categoricity of a theory
added 76 characters in body
Oct
28
answered $\omega$-categoricity of a theory
Oct
27
suggested suggested edit on Why $p \leftrightarrow q$ is equivalent to $(p \wedge q) \vee (\neg p \wedge \neg q)$? Without using the truth table
Oct
27
revised Is every theorem of PA true in the standard model of number theory $N$?
added 81 characters in body