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

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quasi

Posts: 10,208
Registered: 7/15/05
Re: Onto [0,1]
Posted: Apr 25, 2013 5:31 AM
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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.

quasi



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