Is there a $\Sigma_0$ definition of a bijection from $\omega\times \omega \to \omega$ ? If there is, it would help me in my (long) journey to find a $\Sigma_0$ bijection from $(2^\omega)^\omega \to \omega^\omega$ (if you have one that's quick for this second thing, it would help as well)
$\Sigma_0$ definition of a bijection $\omega \times \omega \to \omega$
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elementary-set-theory
first-order-logic
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0$\Delta_0$ over the raw language of set theory? Do you allow taking power sets, unions etc while forming quantifier bounds? – 2017-01-14
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0The raw language of set theory as you put it. For simplicity, one can consider that $\underline{\omega}$ is a constant symbol, interpreted as $\omega$ (the object $\omega$ is absolute for transitive models). I'm trying to find a bijection that's absolute between models (i.e. if $M$ is an inner transitive model, then $j^M$ is a bijection between $\omega\times omega$ and $\omega$, where $j^M$ is the defined bijection, as computed in $M$ (the final goal is to have $j$ defined so that $j^M$ is a bijection $((2^\omega)^\omega)^M \to (\omega^\omega)^M$ ) – 2017-01-14
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0So you can only use sets in the transitive closure of $\omega$ (which is $\omega$ itself)? I doubt that's enough to define a pairing function -- it would be the same as working in $(\mathbb N,{<})$ with no arithmetic other than successors avaialble. – 2017-01-14
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1Whats wrong with $(a,b)\mapsto\frac{(a+b)(a+b+1)}{2}+b$ ? – 2017-01-14
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0Henning Makholm: Well until I prove it, I can only think it's difficult, but not know it... Rene Schipperus: I doubt it'd be $\Sigma_0$.. I'm not even sure addition of ordinals is $\Sigma_0$ (it either involves recursion which isn't $\Sigma_0$ if I'm not mistaken, or isomorphisms which clearly aren't $\Sigma_0$ – 2017-01-15