Logic - 2018-11-15

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8:24 AM

@Holo That's different. Notice I didn't say countable submodel but merely countable model.

Also from the PDF you cited:

> As noted by Skolem himself, the condition that the countable model for T be an elementary submodel of M made an apparently essential use of the axiom of choice

Holo

@user21820 Oh! So for submodel we need choice but for arbitrary model with countable cardinality we don't

user21820

Well, the Lowenheim-Skolem theorem itself doesn't require "submodel", so countable model doesn't require AC. But as I said I'm not sure about arbitrary "cardinality" especially in the strong sense of bijection.

14 hours ago, by user21820

None of this requires AC. So the completeness theorem of a countable theory does not need AC. In general, you would get a model M such that C injects into M and M injects into N·C. I'm not sure whether that implies that M bijects with C.

That last bit in itself would be an interesting completely separate question.

Holo

I see, so I interpret it wrongly(I translated it to what I know to prove using choice without reason, i.e. submodel)

Btw @user21820 , an unrelated question: in PA- there is no induction so is it possible to prove that every number is either odd or even?(n even def: exists k such that n=2k, n odd def: it is one greater than an even number)

user21820

@Holo Excellent question. In fact, that is precisely why I mention PA− whenever talking about induction, because that very question is an easily explainable example of why induction is not superfluous:

A: Is induction something we take on faith?

Peter Smith's answer is correct, but it seems that you do not grasp logic well enough to understand it. So I'm going to explain it differently: Suppose you believe that there is such a thing as the counting number $1$ (whether as a concrete encoding in some physical medium such as a computer, ...

Some joker downvoted it, but the second section contains exactly that sentence (every element is even or odd) and explains why induction is necessary. I just edited it because I realized I used the same symbol "P" for different things...

Holo

8:50 AM

I see, the reason I thought about that is because few days ago I talked with Carl and Malice about the fact that for nonstandard model of PA and the fact that it is basically "omega+("around" every nonstandard element we can embed a copy of the integers) and then using the fact that there is even numbers in ever such copy we showed that the copies are dense without endpoints. I hoped we could do the same for PA-

user21820

@Holo Well the polynomial model P of PA− does not have any dense copies of Z. It's rather discrete.

Holo

@user21820 Yes, I didn't thought on that example :P

user21820

Grammar time! ( hadn't thought / didn't think ) about

Holo

@user21820 correct+correct=correct, so didn't thought is correct :D

6 hours later…

user21820

2:53 PM

@Holo Haha. If you want to stay undetected among native English speakers, you'd have to work on that a bit! But it's already quite good; that was the only mistake I noticed you making in a long while. =)

user21820

3:11 PM