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Indeed, the statement is that for any list of axioms there exists a countable set of objects satisfying them. For example, you could write down axioms for the

Indeed, the statement is that for any list of axioms there exists a countable set of objects satisfying them. For example, you could write down axioms for the real numbers by specifying that there should be relations called + and x with the standard properties such as commutativity, as well as an ordering relation < such that for all elements x and y there is an element z for which: x < z < y. Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well.

pdonis 6y ago
> Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. You missed the crucial property that rules out the rationals (more precisely, the rationals with their standard ordering): one way of stating it is that every sequence that has an upper bound in the set, must have a least upper bound in the set. The rationals do not satisfy this property (for example, consider the sequence of successive decimal expansions, each one to one more decimal place, of sqrt(2)), but the reals do. The challenge for me is to understand how there can still be countable sets that also satisfy that property of the reals. (Obviously any countable set can be put into one-to-one correspondence with the rationals, but for a countable set that satisfies the least upper bound property of the reals, such a correspondence with the rationals would put an ordering on the rationals that was not the standard one.)
- zzless 6y ago
  In fact, he did not miss anything. Using the language he started with (variables range over 'numbers', and the relations are <, >, +, and *), the reals and the rationals indeed have the same properties (elementary theory as logicians would put it). The reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). If I remember right, the fact he was referring to was proved by Tarsky.
  - saithound 6y ago
    The reals and the rationals do not have the same elementary theory over the language (>,+,*).
    - zzless 6y ago
      You are correct, one would have to exclude the multiplication for that.
  - pdonis 6y ago
    > he reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). Doesn't that mean that you can't even define the reals using the language he started with? If your language doesn't even let you express the difference between the reals and the rationals, it seems to me that the thing to do is to extend your language until it can.
    - Kranar 6y ago
      You're right that it is not possible to define the reals using the language he started with, but it's worse than that. It's also not possible to define the natural numbers using any first order theory. There is no way to extend a first order theory so that it defines the natural numbers and only the natural numbers and furthermore there is no way to define a first order theory that defines the reals and only the reals. Having said that, you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason.
      - pdonis 6y ago
        > It's also not possible to define the natural numbers using any first order theory. Yes, agreed. > you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason. Can you elaborate?
Kranar 6y ago
>Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. This wouldn't be correct, it's never the case that the set of rational numbers can satisfy a theory of real numbers, it's more subtle than that. It's that for any theory of the real numbers, there exist subsets of the real numbers that are countable that satisfy that theory. For example the subset of all computable real numbers will satisfy any theory of real numbers despite it being countable. It's simply not possible to define a first order theory that describes the real numbers as a whole and only the real numbers as a whole. However, there will never be any theory of real numbers that can be satisfied by the set of rational numbers. At a minimum any theory of real numbers would imply theorems that require the existence of a number that when squared was equal to 2. The rational numbers can not satisfy such a theorem.

pdonis 6y ago

> Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. You missed the crucial property that rules out the rationals (more precisely, the rationals with their standard ordering): one way of stating it is that every sequence that has an upper bound in the set, must have a least upper bound in the set. The rationals do not satisfy this property (for example, consider the sequence of successive decimal expansions, each one to one more decimal place, of sqrt(2)), but the reals do. The challenge for me is to understand how there can still be countable sets that also satisfy that property of the reals. (Obviously any countable set can be put into one-to-one correspondence with the rationals, but for a countable set that satisfies the least upper bound property of the reals, such a correspondence with the rationals would put an ordering on the rationals that was not the standard one.)

zzless 6y ago
In fact, he did not miss anything. Using the language he started with (variables range over 'numbers', and the relations are <, >, +, and *), the reals and the rationals indeed have the same properties (elementary theory as logicians would put it). The reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). If I remember right, the fact he was referring to was proved by Tarsky.
- saithound 6y ago
  The reals and the rationals do not have the same elementary theory over the language (>,+,*).
  - zzless 6y ago
    You are correct, one would have to exclude the multiplication for that.
- pdonis 6y ago
  > he reason things like \sqrt2 present no problems is that it is simply impossible to define such 'sequences of numbers' in this theory (you are only allowed to 'refer' to numbers by your variables, not ordinary sets and the usual language for sets is missing). Doesn't that mean that you can't even define the reals using the language he started with? If your language doesn't even let you express the difference between the reals and the rationals, it seems to me that the thing to do is to extend your language until it can.
  - Kranar 6y ago
    You're right that it is not possible to define the reals using the language he started with, but it's worse than that. It's also not possible to define the natural numbers using any first order theory. There is no way to extend a first order theory so that it defines the natural numbers and only the natural numbers and furthermore there is no way to define a first order theory that defines the reals and only the reals. Having said that, you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason.
    - pdonis 6y ago
      > It's also not possible to define the natural numbers using any first order theory. Yes, agreed. > you were originally right that no theory of the reals can be satisfied by the rationals, but that's for a fairly unrelated reason. Can you elaborate?

Kranar 6y ago

>Clearly the real numbers are a model for these axioms. But as it turns out the countable set of rational numbers is a model as well. This wouldn't be correct, it's never the case that the set of rational numbers can satisfy a theory of real numbers, it's more subtle than that. It's that for any theory of the real numbers, there exist subsets of the real numbers that are countable that satisfy that theory. For example the subset of all computable real numbers will satisfy any theory of real numbers despite it being countable. It's simply not possible to define a first order theory that describes the real numbers as a whole and only the real numbers as a whole. However, there will never be any theory of real numbers that can be satisfied by the set of rational numbers. At a minimum any theory of real numbers would imply theorems that require the existence of a number that when squared was equal to 2. The rational numbers can not satisfy such a theorem.