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seen Feb 26 at 5:32

Jul
22
accepted $H(\kappa)$-absoluteness of a formula
Jul
21
comment About singular $\beth_{\alpha}$ for limit ordinals $\alpha$
math.uni-bonn.de/people/logic/teaching/2013SS/… page 65, near the top, under "Case 2"
Jul
21
asked About singular $\beth_{\alpha}$ for limit ordinals $\alpha$
Jul
21
revised $H(\kappa)$-absoluteness of a formula
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Jul
21
awarded  Yearling
Jul
21
revised $H(\kappa)$-absoluteness of a formula
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Jul
21
comment $H(\kappa)$-absoluteness of a formula
I think I fixed everything now.
Jul
21
revised $H(\kappa)$-absoluteness of a formula
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Jul
21
comment $H(\kappa)$-absoluteness of a formula
Alright, thank you for your answers.
Jul
20
comment $H(\kappa)$-absoluteness of a formula
I see this shouldn't work because it would let ZFC prove its consistency, but I don't see the mistake in the argument at the moment.
Jul
20
comment $H(\kappa)$-absoluteness of a formula
I have a question regarding "cannot form elementary substructure of V", if you don't mind. Trying to form a Skolem hull for a set $X$, if there is $x\in V$ with $\varphi(x,y_1,..,y_k)$ for $y_i\in X$, choose $\alpha$ so that x is in $V_{\alpha}$. Since X is a set, there is $\alpha$ so that this works for all $\varphi$ and parameters in $X$ at the same time. Then wellorder $V_{\alpha}$ and choose the smallest element for each such $\varphi$, $y_i$. Take $X_1$ as the set of those elements. Repeat this, get sets $X_i\subset X_{i+1}$, the union ($i<\omega$) should be an elem. substructure of V.
Jul
20
revised $H(\kappa)$-absoluteness of a formula
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Jul
20
revised $H(\kappa)$-absoluteness of a formula
edited body
Jul
20
answered $H(\kappa)$-absoluteness of a formula
Jul
20
comment $H(\kappa)$-absoluteness of a formula
I just realized I made a mistake in the question. $\varphi$ should be absolute between models of ZF minus powerset (which includes $H(\kappa)$ for uncountable $\kappa$). Also, $\kappa$ should indeed be uncountable regular. Sorry for all that.
Jul
20
revised $H(\kappa)$-absoluteness of a formula
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Jul
20
comment $H(\kappa)$-absoluteness of a formula
No, $\kappa$ is arbitrary.
Jul
20
comment $H(\kappa)$-absoluteness of a formula
This means that, for any $y\in H(\kappa)$, $\psi(y)$ holds in V iff it holds in $H(\kappa)$. So in this case the task is to show that $x$ can be chosen in $H(\kappa)$ if such a x exists in V.
Jul
20
asked $H(\kappa)$-absoluteness of a formula
Jul
10
revised Is this formula $\Sigma_1^{ZF}$?
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