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If I understand correctly, the real numbers can be formalised in a first-order, like in ZF. However, such a formalisation is not strong enough to ensure that all models of the reals are isomorphic.

Why is being able to ensure isomorphism between models interesting? What's significant about that? In terms of the first-order formalisation of the reals, when does it fall short compared to a second-order theory?

Thanks!

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In the case of $\mathbb{N}$, and to a lesser extent $\mathbb{R}$, we have a very strong intuition that are dealing with a definite "real" object. We would like to characterize that object using axioms. Sadly, it is a theorem that we cannot do that in a first-order theory. As to first-order formalization of the reals, if we just use $+$ and $\times$, we get a complete theory, but one with very limited expressive power. If we do it in ZFC or some extensions, we get something powerful enough for existing mathematics. –  André Nicolas Aug 19 '11 at 17:41
    
We can prove categoricity of $\mathbb{R}$ in ZFC. However, (with the usual caveat about consistency) there are many models of ZFC, even countable ones. So the problem of categoricity just gets transferred upwards. But for most purposes, relative categoricity is good enough. –  André Nicolas Aug 19 '11 at 18:35
    
In first-order logic, the reals have the same properties as the hyperreals. –  Ben Crowell Aug 19 '11 at 19:14
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To take a stab at the other question, ensuring isomorphism between models amounts to having a characterization of the object in question and so demonstrating that it makes sense to talk about the object in question. For example, up to isomorphism there is only one countable dense linear order without endpoints: any such order is order-isomorphic to $\langle\mathbb{Q},\le\rangle$. Thus, we can reasonably talk about the dense linear order without endpoints, and we can resolve questions about it by looking specifically at $\langle\mathbb{Q},\le\rangle$. Whether this is interesting is perhaps a matter of taste, but it pretty clearly is significant, in that it facilitates understanding and further investigation of the object. It also makes it easier to recognize the object when it appears in an unfamiliar guise.

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So why does having isomorphism between models amount to having a characterisation of the object? –  kate.r Aug 20 '11 at 13:53
    
@kate.r: (Sorry; I meant to get back to this sooner.) I simply meant that if all models are isomorphic, there is in effect only one object to study. I didn’t mean to suggest that a characterization theorem would automatically result. –  Brian M. Scott Aug 22 '11 at 4:05
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