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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 8:12 AM


quasi wrote: >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.
As regards the function f from the Cantor set C to [0,1] as described by baclesback, it seems, at first glance, that f satisfies the condition
x_n approaches x implies f(x_n) approaches f(x)
and if so, f is continuous.
But given the various blatantly flawed claims I made last night, I worry that may be missing something simple.
quasi



