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

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quasi

Posts: 10,403
Registered: 7/15/05
Re: Onto [0,1]
Posted: Apr 25, 2013 8:12 AM
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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) = x-1 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



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