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Sep
17 |
awarded | Popular Question |
Sep
12 |
answered | A question regarding strong cardinals and measure sequence |
Aug
30 |
comment |
A question regarding strong cardinals and measure sequence
Isn't the induction proof of $u_j(\beta)=u_i(\beta)$ only valid when $\beta<length(u_i)$, and a priori we do not know if $length(u_j)=length(u_i)$? I mean this is equivalent to showing that $u_i\restriction \beta \in M'$. |
Aug
28 |
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A question regarding strong cardinals and measure sequence
@Golshani: I just realized that $E'\not \in Ult(V,E')$ as otherwise $i_{E'}(\kappa)$ will have same cardinality as $i_{E'}(\kappa^+)$ computed in $Ult(V,E')$ which is impossible. (Sorry I've been out of Internet) |
Aug
22 |
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A question regarding strong cardinals and measure sequence
Maybe I'm still missing something easy, but in showing $u_j\restriction \gamma=u_i\restriction \gamma$ we need $u_j\restriction \beta \in M'$ for all $\beta<\gamma$. How do you see that $u_j\restriction \beta \in M'$ or equivalently $u_i\restriction \beta \in N$ (the length of this derived sequence of ultrafilters using $i$ might be shorter?) or equivalently $u_j\restriction \beta \in range(k)$ where $k$ is the map from $M'$ to $M$? Thanks in advance! |
Aug
21 |
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A question regarding strong cardinals and measure sequence
Sorry you are right. I was confusing myself with the extenders of cardinal length. |
Aug
21 |
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A question regarding strong cardinals and measure sequence
If $Ult(V,E')$ contains $V_{\kappa+2}$ wouldn't it contain $E'\in V_{\kappa+2}^{\kappa}$ too as $|\kappa+1|=\kappa$ and by $\kappa$-closure? |
Aug
21 |
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A question regarding strong cardinals and measure sequence
Yeah but in this case isn't $u_j\restriction \beta\in Ult(V,E') \forall 0<\beta<\kappa+1$ so clearly includes $u_j(1)$ as well. |
Aug
21 |
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A question regarding strong cardinals and measure sequence
Then what goes wrong if we look at $E'=E\restriction_{[\kappa+1]^{<\omega}}$? If even $U(0)=\{X\subset \kappa: \kappa \in j(X)\}$ is in $Ult(V,E')$, we could recover $E'$ in $Ult(V,E')$ as follows: in $Ult(V,E')$, $\alpha\in [\kappa+1]^n$ and $\kappa\in \alpha$, $X\in U_\alpha \Leftrightarrow \alpha-\{\kappa\}\in \{\beta: \exists \gamma \geq \sup \beta \beta\cup\{\gamma\}\in X \} \wedge X_{max}=\{\gamma\in \kappa: \gamma \geq\alpha-\{\kappa\}\} \in U(0)$. So $E'\in Ult(V,E')$? |
Aug
21 |
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A question regarding strong cardinals and measure sequence
@MohammadGolshani Why do the restrictions of the new measure sequence have to lie in M'? |
Aug
19 |
revised |
A question regarding strong cardinals and measure sequence
added 149 characters in body |
Aug
19 |
asked | A question regarding strong cardinals and measure sequence |
Feb
15 |
accepted | Consequences of ZF+“all subsets of reals are Lebesgue measurable” |
Feb
4 |
answered | What are your favorite instructional counterexamples? |
Jan
22 |
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Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@AndresCaicedo: I did have in mind that the length of an interval equals its measure. |
Jan
22 |
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Consequences of ZF+“all subsets of reals are Lebesgue measurable”
@Burak: I mean ~CH is a consequence of ZFC + there is a measure on \mathbb{R}, and this measure is necessarily not Lebesgue. You are right over ZF they are not really orthogonal. |
Jan
22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
added 29 characters in body |
Jan
22 |
comment |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
I meant Lebesgue measure is total otherwise choice is not necessarily ruled out. Sorry for the ambiguity. |
Jan
22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
added 10 characters in body |
Jan
22 |
revised |
Consequences of ZF+“all subsets of reals are Lebesgue measurable”
added 16 characters in body |