Jing Zhang
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 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