Square of Menger groups | המחלקה למתמטיקה

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 - המחלקה למתמטיקה 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.