An Elementary Observation Regarding Definability

[SSequence]

Few months ago there was a discussion in the topic(Complex numbers in QM) regarding the notion of definable real numbers. The discussion was in the first 3 or 4 pages of that topic.

Anyway, I thought of a reasonably interesting observation about it. Since the main theme of that topic seems quite different, I have posted this as a separate thread. I think I have gotten some idea of post#80. Below is one way I thought about it:

Consider, in the language of set-theory, we consider all well-formed-formulas that have one free variable. Since the formulas are easily counted we could use the notation $\phi_n(x)$ (where $n \in \mathbb{N}$) to denote $n$-th formula. $x$ is supposed to be the free variable in the formula. Now if we consider the set of reals $\mathbb{R}$ (say, as subset of naturals), then we say that a given formula $\phi_i(x)$ defines the real number $r$ iff $r$ is the only real number satisfying that formula.

Now suppose that we have a well-ordering of reals available to us and say its order-type is $\beta \geq \omega_1$. Now, using that well-ordering, for every formula $\phi_i(x)$ we can test whether it is satisfied by only one real or not. So, for example, we first test whether $\phi_0(x)$ is satisfied by a unique real or not ... and then $\phi_1(x)$, $\phi_2(x)$, $\phi_3(x)$, $\phi_4(x)$ etc. This way we can pick all those reals which are definable. Then we diagonalise through these definable real numbers giving us a new real number $R$ which is different from every definable real.

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It is quite informal, but it seems plausible to me that if there is a formula (in language of set-theory) for the well-ordering of $\mathbb{R}$ then it can be shown that there must also be a formula which is only satisfied by the real number $R$ above (and no other real number). But that would mean that $R$ is definable (which it isn't supposed to be since it was formed from diagonalisation of definable reals). Hence, it seems to me, that we also have to assume that there can't be any formula (in set-theory language) describing the well-order of $\mathbb{R}$?

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Does any of this sound remotely reasonable? There is one other thing, which I don't quite get. In this question it is mentioned (in the accepted answer):

there is a model of ZFC in which every real number and indeed every set-theoretic object is definable.

Does it go against the first paragraph written in post#80 or am I misinterpreting something:

As far as I know, nobody tries to do mathematics using only definable objects, because the usual mathematical axioms don't hold when restricted to definable objects. However, the set of reals is certainly definable.

Sorry I didn't want to link the above question (since it seems far too advanced for this particular discussion, where I am just trying to get a sense of things). Also, quoting a specific sentence is kind of nitpicking, but nevertheless, the two statements felt at odds to me, so it seemed reasonable to inquire.

 

[A. Neumaier]

The diagonalizatio you refer to happens on the objectlevel, wile the construction of the reals happens on the metalevel so the model of the reals is metacountable but intrinsically uncountable. There are not enough intrinsic maps to make the metbijection needed.

As to your other question, if a set is definable it does not mean that all its elements are individually definable.

 

[SSequence]

Below, denoting the language of (first-order) set theory as $L_S$.

As to your other question, if a set is definable it does not mean that all its elements are individually definable.

I assume that this is the kind of thing that's happening with $\mathbb{R}$. In general, knowing a diagonalisation through reals definable in $L_S$ doesn't mean that we can write down a formula for the diagonalised real [uniquely satisfied by that real alone in the set $\mathbb{R}$] in $L_S$. Right? At least that's what I gathered from discussion/posts in previous thread.

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Also, to re-iterate my question in the first half of OP: My question was that if we assume that there is a formula for a well-order of $\mathbb{R}$ in $L_S$, it seems plausible that we can conclude that there is also a (unique) formula for a real $R$ which is not definable (in $L_S$)? I am not familiar enough with logical constructions/machinery to know for sure.

Hence, by argument in prev. paragraph, can we conclude that there is no formula for a well-order of $\mathbb{R}$ in $L_S$? And furthermore is the line of reasoning in previous paragraph correct?

 

[A. Neumaier]

Hence, by argument in prev. paragraph, can we conclude that there is no formula for a well-order of $\mathbb{R}$ in $L_S$? And furthermore is the line of reasoning in previous paragraph correct?

In no ZFC model of the reals, there is a formula for well-ordering.

 

[SSequence]

I see. So it seems reasonable to assume that when people talk about formula for well-ordering of $\mathbb{R}$, they are talking about a language that is different (presumably richer?) from $L_S$ in some way.

 

[A. Neumaier]

I see. So it seems reasonable to assume that when people talk about formula for well-ordering of $\mathbb{R}$, they are talking about a language that is different (presumably richer?) from $L_S$ in some way.

In general, existence is nonconstructive, unlike formulas.

 

[SSequence]

@A. Neumaier
I think your post#4 may not be correct. No? Or maybe I misinterpreted.

At any rate, I think my post#1 isn't correct. I don't know where, but perhaps somewhere around the part about diagonalization.

Anyway, I guess it doesn't matter.

 

[A. Neumaier]

I think your post#4 may not be correct.

Why do you think so?

 

[SSequence]

For example:
https://mathoverflow.net/questions/6593
To be fair though, I know very little when it comes to more subtle matter related to set-theory. I can try to paraphrase my own understanding too (which is rather based on indirect directions) but I have tried to quote from a source [due to previous sentence].

 

[A. Neumaier]

For example:
https://mathoverflow.net/questions/6593
To be fair though, I know very little when it comes to more subtle matter related to set-theory. I can try to paraphrase my own understanding too (which is rather based on indirect directions) but I have tried to quote from a source [due to previous sentence].

The reference you quote does not give a formula in the set theoretic universe itself but a construction on the metalevel in which one can discuss the set theoretic universe. In particular, ''definable'' is a concept on the metalevel only.

You need to distinguish between object level and metalevel. A discussion about formulas happens on the metalevel, whereas the object level consists of the formulas and their formal derivations only.

 

[SSequence]

I am totally confused :p ... its the nature of the topic.

Though idk what you mean when you say meta-level or object-level. I think I kind of understand "little" about what you are trying to say though.

When you say "object level" perhaps you have pre-conceived idea of a universe in mind. And when you say "meta-level" perhaps you are talking about a specific model (which is always a set). Sorry for possibly distorted description of your view. But that's the best I can guess.

At any rate, I guess I misinterpreted post#4, since you seem to have fairly nuanced understanding on this.

Edit:
So perhaps when you wrote in post#4:
"In no ZFC model of the reals,"
you meant something like:
"In no ZFC model of the reals,"

But then one question perhaps maybe whether there is any model which has the reals in the first place at all (possibly even under additional axioms such as constructiblity etc.)? I don't have any idea about it.

 

[WWGD]

For example, Mathematical Logic uses a metalanguage to study Mathematics, the object being studied, which has its own language. Similar to the way in which the grammar associated to a language has its own metalanguage to study, say, English.I hope I understood your question correctly.

 

[SSequence]

Read from post#7 onwards. As I mentioned that conclusion drawn in post#1 isn't justified.

That's because, unlike what I mentioned there [when we have the assumption that the base axioms are congruous] there exists models M and N such that:
---- there exists a well-order of $\mathbb{R}$ in both M and N
---- In M there exists no formula for the well-order of $\mathbb{R}$
---- In N there does exist a formula for the well-order of $\mathbb{R}$

Last two points are technical and I don't know the details (though I know a little bit for second point). However, they are correct.

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Regarding post#11, I was just trying to guess post#10 (and perhaps my guess was off).

 

[SSequence]

I think I have gotten a very rough idea about few issues.

(1) If we have two (set) models M and N then the quantifications of a given formula will range over the appropriate set. So, for example, it could be possible that some formula is satisfied by a unique element (when applied to M) but there is no element which satisfies the very same formula (when applied to N).

That's because the quantifications in the same formula would be over the elements of M (OR over the elements of N) depending on which structure we are considering (M or N).

(2) Let's suppose that N is bigger than M. So, for example, if we could well-order the elements of M, then we could do a diagonalization like the one in OP I think. But for that diagonalization to work, it would have to be from the "outside" perspective of N. Since apparently, to do it, we would seemingly require the order-type (from the perspective of N) of a well-order of M. [and for example, once again, it would be impossible that M contains that order-type as its member ... or something like that :p]

[EDIT:] There is a further point that I think should be relevant. Whether the "diagonalized" real is uniquely described by a formula (from the perspective of N) or not is still not decided just because the construction is explicit. Few further conditions need to be satisfied for that I think. For one, the exact order-type (in N) which describes the well-order of elements of M is quite relevant.[END]

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In some sense I guess its a bit like the notion of a (countable) model M, where the bijection between $\omega$ and $\aleph^M_1$ exists but the bijection is not a member of M itself.

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The above points are meant to be qualitative. Obviously I would need much (much) more expediency to get more comfortable with these issues.

P.S.
There is one more thing that I think might be mentioned w.r.t. formulas $\phi_n(x)$ in OP. There are two kind of "reasonably natural" conditions we could put:
(i) A given $\phi_n(x)$ (for some specific $n$) is satisfied by one and only one set.
(ii) A given $\phi_n(x)$ (for some specific $n$) is satisfied only by one subset of $\omega$. That is, there might be some other sets which may or may not satisfy the given formula, but once we constrain ourselves to subsets of $\omega$ then there should only be one such set.

The first one, I guess, would be usually meant.