619 reputation
528
bio website
location Unknow Galaxy
age
visits member for 2 years, 2 months
seen Dec 20 '13 at 18:33

A student in Philosophy but also curious in many fields.


I don't know,

But I want to know.


Jun
5
awarded  Yearling
May
20
accepted Which logic is stronger? SOL or $\frak{L}_{\infty,\infty}$?
May
8
awarded  Caucus
Apr
27
revised Which logic is stronger? SOL or $\frak{L}_{\infty,\infty}$?
clearify
Apr
27
asked Which logic is stronger? SOL or $\frak{L}_{\infty,\infty}$?
Apr
16
comment Can we conclude that $|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|=2^{|\alpha|}$ for all infinite ordinal $\alpha$ in $\mathsf{ZFC}$?
@Apostolos Thank you, that's all right.
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
Thank you for your suggestion.
Apr
16
asked Can we conclude that $|\prod_{\xi \in Ord, \xi=1}^{\xi<\alpha}\xi|=2^{|\alpha|}$ for all infinite ordinal $\alpha$ in $\mathsf{ZFC}$?
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
Thank you very much. Besides, a little tip: when $\alpha=\beta+1$ then $\Gamma(\aleph_\alpha)$ should be $2^{\aleph_0}\aleph_{\beta}^{|\beta|}$.
Apr
16
accepted Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
Apr
16
revised Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
correctify
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
I'm sorry that I have made another error: this product is the cardinal product of all positive cardinals smaller than $\aleph_\alpha$ whereas in the lemma 1 I had considered it as the cardinal product of all positive ordinals smaller than $\aleph_\alpha$.
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
I'm sorry about that...
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
@AsafKaragila Sorry, in that time I have mistook the definition of beth numbers (mistook that $2^{\aleph_\alpha}=\beth_{\alpha+1}$).
Apr
16
revised Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
fix
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
Thank you, it's amazing.
Apr
16
revised Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
fix
Apr
16
comment Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
@CliveNewstead Thanks, I'm going to fix it.
Apr
15
revised Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?
detail
Apr
15
asked Can we conclude $\prod_{\kappa \in Crd, \kappa=1}^{\kappa<\aleph_\alpha}\kappa=2^{\aleph_\alpha}$ in ZFC?