Square of Menger groups

Seminar
Speaker
Jialiang He (BIU)
Date
24/08/2020 - 13:00 - 11:00Add to Calendar 2020-08-24 11:00:00 2020-08-24 13:00:00 Square of Menger groups     This work is cooperated with Yinhe Peng and Liuzhen Wu. I will present a few constructions for square of Menger group  problem in metrizable sense and generally sense in this talk.      Under cov(M)=c, for any n\geq 1, there is a subgroup G of Z^N such that G^n is Menger but G^{n+1} is not Menger.   Under cov(M)=d=cf(d)$,  for any n\geq 1, there is a subgroup G of R such that  G^n is Menger but G^{n+1} is not Menger.  For any n\geq 1, There is a Menger subgroup G of R^{\omega_1} such that G^n is Menger butG^{n+1} is not Lindelof in ZFC.   According to  the known result, product of  Menger subspace  of Z^N is Menger  in Miller model, this shows Menger group square problem is independent with ZFC in the metrizable sense. But for nonmetrizable topological group, the answer is no. Building 216, Room 132 אוניברסיטת בר-אילן - Department of Mathematics mathoffice@math.biu.ac.il Asia/Jerusalem public
Place
Building 216, Room 132
Abstract
 

 

This work is cooperated with Yinhe Peng and Liuzhen Wu.
I will present a few constructions for square of Menger group  problem in metrizable sense and generally sense in this talk.  

 

 Under cov(M)=c, for any n\geq 1, there is a subgroup G of Z^N such that G^n is Menger but G^{n+1} is not Menger. 
 Under cov(M)=d=cf(d)$,  for any n\geq 1, there is a subgroup G of R such that  G^n is Menger but G^{n+1} is not Menger.
 For any n\geq 1, There is a Menger subgroup G of R^{\omega_1} such that G^n is Menger butG^{n+1} is not Lindelof in ZFC.
 
According to  the known result, product of  Menger subspace  of Z^N
is Menger  in Miller model, this shows Menger group square problem is independent with ZFC in the metrizable sense. But for nonmetrizable topological group, the answer is no.

Last Updated Date : 19/08/2020