Second order logic implies categoricity in peano arithmetic. But why are the models isomorphic to the standard model of aritmhetic and not to another non standard model, for example?
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The point is that there is a single second-order sentence which completely characterizes (up to isomorphism) the structure $(\mathbb{N}, 0, 1, <, +, \times)$. Specifically, the standard model is characterized by the property that every element has only finitely many predecessors - and we can write this in second-order logic as $$\forall n\forall F([\forall m(m<n\implies F(m)<n) \wedge \forall m_1, m_2(m_1\not=m_2\implies F(m_1)\not=F(m_2))]$$ $$\implies \forall m<n\exists k<n(F(k)=m));$$ in English, this is just saying "for every $n$, every injective function from $\{m: m<n\}$ to $\{m: m<n\}$ is surjective." This sentence, when conjoined with the finitely many non-induction axioms of Peano arithmetic, characterizes the standard model up to isomorphism.
answered Nov 16, 2015 at 5:40
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$\begingroup$ It is said that there are infinite (in metamathematical intuitive sense) numbers, but the sets $ n i \in N $ and $ n<m $ implies this sets are finite. .Thanks for the answer. Now I have not clear my position about second order logic, maybe it is not only first order set theory in disguise. When I started to interest for theoretical basis for maths (my formation is scientific but not mathematical). But, why the mathematicals are so reticents to second order logic then? $\endgroup$ Nov 16, 2015 at 6:07
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1$\begingroup$ Well, it doesn't have anything like a decent proof system. That's pretty terrible. But we do use second- (and higher-)order logic informally, all the time. So you'll have to clarify what you mean when you say mathematicians are hesitant to use it. $\endgroup$ Nov 16, 2015 at 6:16