Expand LA by adding a constant c to obtain the language LA∗ = (+,·,<,0,1,c). — alcontali

She's adding a constant to the LANGUAGE of the theory. The model is a model for the language. The model is not a language.

There is a language. Then there are two things.

(1) A model for the language.

(2) A theory in the language.

Then the model is a model of the theory iff (the theory is in the language, and the model is a model for that language, and every sentence in the theory is true in the model).

These infinite cardinals — alcontali

Yes, as I said, if a theory has an infinite model, then it has both countable and uncountable models.

Therefore, Gitman's "language" does not always have a countable number of sentences — alcontali

That's a non sequitur. The set of sentences is countable but, if there is an infinite model of the theory, then there are countable and uncountable models of the theory.

And Gitman goes on to discuss particular countable models.

The language, the theory, and the model are separate, but related things.

If you like, I can recommend introductory textbooks in this subject that will explain all this for you, step-by-step, in greater detail and context than I can give in posts.

She's adding a constant to the LANGUAGE of the theory. — GrandMinnow

The language, the theory, and the model are separate, but related things. — GrandMinnow

In model theory, a signature σ is often called a vocabulary, or identified with the (first-order) language L to which it provides the non-logical symbols. — Wikipedia on the term signature

Yes, I see now, what Gitman wrote, is an extension of the signature, and therefore of the language of the theory. If it is a countable model then the extension will also show up in its (CS) definable language, but that is clearly not possible for nonstandard models of PA.

If you like, I can recommend introductory textbooks in this subject that will explain all this for you, step-by-step, in greater detail and context than I can give in posts. — GrandMinnow

There does not seem to be much that can be downloaded, but if you have any links to good material with examples, feel free to post them. From the paucity of materials online, the subject does not seem to be that popular, actually.

what Gitman wrote, is an extension of the signature, and therefore of the language of the theory — alcontali

Exactly.

paucity of materials online — alcontali

I don't know about online, but here is a course of books I highly recommend, in order of study:

* Logic: Techniques of Formal Reasoning - Kalish, Montague, Mar

This tells you how to make sure that your mathematical arguments are within the first order predicate calculus. But if you feel that your mathematical reasoning is not prone to such things as mistakes with quantifiers and improper instantiations of variables, then you can skip this book. Though I still highly recommend it to make really sure you're always on firm ground.

* Elements Of Set Theory - Enderton

From the previous book, you will have a good grasp of proof in the first order predicate calculus you need for a rigorous study of set theory.

* A Mathematical Introduction To Logic - Enderton

From the previous book, you will know the set theory you need for a rigorous study of mathematical logic.

* ... * ... * ... — GrandMinnow

It looks like worthless dead-tree material. I don't do that. There is no longer a need to cut trees in order to publish information. There is also no need any longer to physically sit a room for that purpose. We do not need these dinosaurs. If they do not put it online, then someone will put something better online.

With a view on staying relevant, maybe you and your friends can start reading HTML for dummies?

Whatever may be the demerits of printed books, at least I an tell you that those are exceptionally great books and it is not certain that there will be better ones online.

But did you even look to see whether these might also be online?

you and your friends — alcontali

What friends? You said I don't have any.

What friends? You said I didn't have any. — GrandMinnow

In that case, read "HTML for dummies" alone.

Because I recommended some printed books (which you may check for yourself whether they are also online or not), you infer that I don't know enough about HTML and should read a book about it? That would prevent me from ever again grievously recommending books when I don't know whether they are also available online?

Because I recommended some printed books (which you may check for yourself whether they are also online or not), you infer that I don't know enough about HTML and should read a book about it? That would prevent me from ever again grievously recommending books when I don't know whether or not they are also available online? — GrandMinnow

Look, the academia are stuck in a highly inefficient way of transmitting information while charging an arm and a leg in the process. Now, this subject is not vocational at all. So, it is indeed fun as a hobby, but the only thing you can professionally do with it, is teaching in the academia, and then also get stuck in highly inefficient ways of doing things while further bankrupting an already corrupted youth. I only use online resources for my hobbies. In fact, I only use online resources for professional activities too. So, we are not going to get anywhere discussing these things any further, because we simply live in two different worlds, while I more or less despise yours. So, no, we are simply not going to do things your way, and there is no need to argue over that. We'd better agree to disagree.

You more or less despise "my world", not knowing anything about it other than it includes people who may recommended books without first knowing whether they are also available online.

notation: for a function f, let f’y= the x such that fx =y.

Let M be a model for the language L.

Let S be any set whatsoever with the same cardinality as the universe U for M. Doesn’t matter the cardinalities of the members of S. Let f be a bijection from U onto S.

Then, for example, suppose ‘H’ is a relation symbol in L, and suppose M maps ‘H’ to the relation R in UxU.

Let M* be the model with universe S, and, for example, M* maps the symbol ‘H’ to {<j k> | <f’j f’k> in R} ... and in that way for the rest of the signature of L.

So M and M* are isomorphic.

Or did I make a mistake?

Happily, public universities are free in Brazil, so it is possible here to obtain higher education without incurring in large debt!



I'm not sure what your point is. Can you clarify?

Happily, public universities are free in Brazil — Nagase

Well, they are not really "free". The government still has to pay for them. At first glance, that is not the individual's problem, but sooner or later, it will still be.

The government naturally collects money from people from whom it is relatively easy to do that: the wage slaves in Brazil. In theory, they would try to mostly collect it from who has the highest income, but those are exactly the people who will most easily adjust to whatever is needed to avoid paying. For example, they avoid wage slavery. They do not receive too much income through that kind of contracts.

The poor, on the other hand, do not have much income anyway. It does not even matter how they receive it. So, it is not the poor who can keep that system financially afloat.

So, a strategy of large public expenditure depends on having a large middle class of easily taxable wage slaves, which is exactly the demographic that tends to disappear: smart enough to make more money than the poor in society but not smart enough to prevent the government from taxing their income away.

so it is possible here to obtain higher education without incurring in large debt! — Nagase

In that case, it is the government which incurs the debt.

They will try to get it back from the middle class, but only if spending on university really leads to increasing the size of the middle class of wage slaves. If the students rather graduate into working part time jobs at Starbucks, this strategy will still fail.

The biggest problem is that university does not increase the graduate's productivity. In fact, it is bad business to waste 4+ years of your life memorizing phone books replete with unimportant trivia.

If the government's plan fails financially, there is a real risk that their debt will still somehow end up being your problem. The funny thing is that the more money you have, the less it will affect you, because in that case, you already have lots of workarounds for the problem of the government trying to collect money from you.

I mentioned (putting it in these terms now) that for any theory T and any cardinality C, if there is a model M of T, then there is a model M* of T such that the universe for M* has a member of cardinality C. (Though originally I forgot to mention the obvious qualification that this pertains when there is a model M of T.)

You asked me what I had in mind. So in my previous post, I gave the proof.

Let's recap the discussion. I mentioned that the universes of ZF-Inf are all infinite, and remarked that this easily followed from the Power Set axiom. You then replied that "for any theory, and for any cardinality, there is a model of the theory such that the universe of the model has a member of that cardinality." I was puzzled by this and offered two interpretations: (i) on one interpretation, you were mentioning the fact that every theory with an infinite model has a model in any cardinality or (ii) you were mentioning the fact that, for any model M, it is possible to construct another model M* whose domain is composed of whatever you want. Now, (ii) was irrelevant to my remark, because I was talking about the size of the model, not the size of its members. And (i) forgets the fact that some theories have only finite models (say T implies "there is a y such that for every x, x=y").

Here is what I responded to:

The universes of ZF-Inf are all infinite. This is clear from the fact that ZF-Inf has the power set axiom, so that there's no bound for the size of its sets. — Nagase

When you say "its sets", I take it you mean the universes's sets, i.e. the sets that are in the universe. Of course, for any particular universe, there is a bound on the sizes of the sets. So I took it that you meant that for any set size there are universes for the theory that have members of that set size. And I agree. My point is that this holds for any theory whatsover, regardless of its axioms or, in particular the power set axiom.

if a given model has cardinality n, then it has no members of cardinality n+1 — Nagase

If by 'members' you mean members of the universe of the model, then the above is incorrect. Trvially, let the universe of some model be {2}. The universe has cardinality n=1 but it has a member of cardinality n+1. And beyond the trivial, for infinite universes, it is not precluded that a universe is of cardinality C but the universe has a member of cardinality C+1, where C+1 is the successor cardinal of C.

In the remarks you quote, I had in mind transitive models. My bad for not making that clearer.

I had in mind transitive models. — Nagase

That works. Thanks.