Reputation
613
Top tag
Next privilege 1,000 Rep.
See votes, expandable usercard
Badges
1 4 9
Newest
 Yearling
Impact
~6k people reached

  • 0 posts edited
  • 0 helpful flags
  • 42 votes cast
Jan
17
comment The Hales-Jewett Theorem for an infinite alphabet
One infinite version I know is the following: Given A countable such that $A=\bigcup A_n$ each $A_n$ finite, every finite coloring of $A^{<\omega}$ there exists an infinite sequence $\langle x_i: i\in \omega\rangle\subset A[x]^{<\omega}$ (x is the variable) such that $\{x_{n_0}[\lambda_0]^\frown \cdots ^\frown x_{n_k}[\lambda_k]: n_0<\cdots < n_k, \lambda_i\in A_{n_i}\}$ is homogeneous. But I must admit this is not quite what you ask for $HJT(\aleph_0)$..
Oct
27
awarded  Yearling
Oct
13
comment Idea behind the proof of consistency of club filter of $\omega_1$ is ultrafilter + ZF + DC
@MohammadGolshani : Well, the argument may not be applicable (if the measure sequence has a repeat point it preserves the measurability of $\kappa$.). Also that only concerns the Magidor product of the same poset in the ground model. Whereas the case in the problem, we will have to choose a new poset (i.e. the Radin poset formed by longer measure sequence containing repeat points), so if these posets satisfy some conditions (as in the proof of consistency of "$\kappa$ is both least strongly compact and least measurable), then there might not be a problem. But I have to think about it.
Oct
11
comment Idea behind the proof of consistency of club filter of $\omega_1$ is ultrafilter + ZF + DC
@MohammadGolshani I agree. I'm not quite sure what the corresponding support needs to be in the Mitchell forcing. Maybe full-support/Magidor iteration?
Oct
10
comment Idea behind the proof of consistency of club filter of $\omega_1$ is ultrafilter + ZF + DC
@AsafKaragila yes I did. I'm just not quite sure what would happen iterating Radin Forcing with sub measure sequences (if this is the right way to think about that).
Oct
10
asked Idea behind the proof of consistency of club filter of $\omega_1$ is ultrafilter + ZF + DC
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
comment 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
comment 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
comment A question regarding strong cardinals and measure sequence
Sorry you are right. I was confusing myself with the extenders of cardinal length.
Aug
21
comment 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
comment 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
comment 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
comment 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?