Re: Infinitely Many Complete 1-Types



On 2 Jun 2006 21:40:45 -0700, Blake Manner wrote:

I would really appreciate it if I could get someone to look at a proof
for me. Its regarding a qualifying exam that I took last semester. I
got the problem wrong, and I'm trying to see where I went wrong.

Instead of trying to make it all confusing looking where I put the
LaTex and gibberish looking things on here, I just put it on my web
page.

I would REALLY REALLY appreciate it if I cold get some feedback on this
proof.

The address is http://www.math.umd.edu/~cnglover/logic/probs.htm


I don't follow the step
"So T has only finitely many types realized."
Well, I am not certain what it means.

I think you need to show that \psi_i whitnesses that \Psi_i is principle.

In other words for any formula \theta, you have \theta is equivalent to
some boolean combination of the \psi_i

Personally I would do it differently: consider a non principle type p, and
try to construct a sequence of formulae \alpha_i contained in p so that for
each n we have

T |- \exists x \alpha_1 and ...\alpha_{n-1} and not \psi_n

Just a side note:
I have been posting a lot lately because I'm studying to take this exam
again, and so I don't want to be in this position again, so I want to
take every step to ensure I pass this exam next time around. Since the
problem in the past has always been that I had nowhere to ask questions
and get a better understanding of the concepts I get confused about,
I've been doing that a lot here. But I really hope that people are not
getting tired of my posts and/or thinking that I ask too many
elementary questions. I feel like I have a grasp on a lot of this
stuff, but there are certain sections that I am absolutely dumfounded
on, and so after a year of working on my own, I've decided to seek help
from people who know this stuff. But I really appreciate any help
and/or direction that I can get on this stuff.

Blake.

No problem, it is a while since I looked at alot of this stuff, but its the
stuff that got me hooked on model theory.

I wish you luck in your exam!
.



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