William Elliot wrote: >baclesback wrote: >> >> More specifically, use the representation of x in C Cantor >> set in base 3 with only 0's and 2's in the expansion of 3, >> and map >> >> f: x=0.a1a2..... ---> 0.b1b2....... >> >> Wheref(bi)= 0 , if ai=0 , f(bi)=1 , if ai=2 . >> >> This f is continuous, but not absolutely continuous ( which >> preserves sets of measure zero. ) > >Why is f continuous?
Because f is increasing.
If X,Y are subsets of R, and f: X -> Y is a monotonic function, then f is continuous (with respect to the relative topologies on X and Y inherited from R).