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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):

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.

===================

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}##?

===================

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):

Does it go against the first paragraph written in post#80 or am I misinterpreting something:there is a model of ZFC in which every real number and indeed every set-theoretic object is 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.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.

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