- #1
linulysses
- 4
- 0
As well known, for any topological space (X,T), there is a smallest measurable space (X,M) such that T\subset M. We say that (X,M) is generated by (X,T). Right now, I was wondering whether the "reverse" is true: for any measurable space (X,M), there exists a finest topological space (X,T) such that the smallest measurable space generated by (X,T) is exactly (X,M).
Last edited: