Recent content by sephy

For Part A - I'm pretty sure F does exist: they give you that the intersection of A and B is non-empty, so it must contain at least one element, say \alpha\in A \bigcap B. Then define F={\alpha}. F is definitely in T, and the intersections of A, B and F are non-empty. For Part B - I...

I think you've pretty much done it - S is open means the boundary of S is a subset of Sc, so the boundary is not in S. Sc is open means boundary of Sc is a subset of S. Since you have shown that the boundary of Sc is equal to the boundary of S, this implies that the boundary of S is a...

The element is its own inverse, so the subgroup generated by that element would just be itself and the identity element. Maybe you are meant to assume that? Either that or the word "coset" is being mis-used- vela's right, it's not a coset unless it's made with a subgroup. Without the...

The right coset would be the set of all elements {\alpha} such that \alpha=((1 2),[1])(p,n) where p is an element of S3 and n is an element of Z2. I would list the elements of S3 X Z2 (which would be like ((1 2 3),[2]),((1 2 3),[2])...- there should be 12 elements I believe?), and operate ((1...