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Topic:
Onto [0,1]
Replies:
40
Last Post:
Apr 29, 2013 10:16 PM



quasi
Posts:
12,067
Registered:
7/15/05


Re: Onto [0,1]
Posted:
Apr 25, 2013 5:31 AM


quasi wrote: >quasi wrote: >>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. > >I meant: > >Because f is _strictly_ increasing. > >>If X,Y are subsets of R, and f: X > Y is a monotonic function, > >I meant: > >If X,Y are subsets of R, and f: X > Y is a _strictly_ monotonic >function, > >>then f is continuous (with respect to the relative topologies on >>X and Y inherited from R).
Oops.
I wasn't thinking clearly  my general claim above fails badly.
For example, let
A = {n/(n+1)  n in N}
X = A U {1}
Y = A U {2}
and define f: X > Y by
f(x) = x if x =/= 1
f(1) = 2
Then f is increasing but not continuous.
Another second's thought yields the following even easier example
Let f: R > R be defined by
f(x) = x if x >= 0
f(x) = x1 if x < 0
Then f is increasing but not continuous.
I really don't know what I thinking.
quasi



