Re: Enderton problem
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Wed, 05 Dec 2007 06:08:46 -0600
On Tue, 4 Dec 2007 13:03:16 -0800 (PST), Gc <Gcut667@xxxxxxxxxxx>
wrote:
On 4 joulu, 02:34, dpo...@xxxxxxxxxxxxx wrote:
If you have Enderton's A Mathematical Introduction to Logic handy,
this problem comes from number 7 in chapter 2, section 6.
Consider a language with a two-place function predicate symbol <, and
let N = (N; <) be the structure consisting of the natural numbers with
their usual ordering. Show that there is some A elementarily
equivalent to N such that <A has a descending chain.
Okay, here's what I'm thinking. We let the domain of A be the set {1/
n : n in the natural numbers} and define (m, n) is in <A iff (n, m) is
in <N. So A appears to have a descending chain. Now I need to show
that A and N are elementarily equivalant. I can do this by showing A
is a model for ThN. But...how do I do this? Enderton suggests
applying the compactness theorem...but I'm not sure how this leads to
showing they are elementarily equivalent.
I don`t see why you can`t just take the non-positive integers M with
their usual ordering. That structure is isomorphic to the (N; <),
It is?
Hint: No, it's not.
Hint: It's also not elementarily equivalent to N, because for example
it does not have a least element, while N does.
Hint: A structure that _is_ isomporphic to N cannot possibly
be the answer to the question, since if it's isomorphic to N
it does not have a descending chain.
so
it is also elementarily equivalent, which is a weaker property than
isomorphism but it is implied by it.
************************
David C. Ullrich
.
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