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Continuous function from (0,1) onto [0,1]? 
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#1
Nov912, 10:03 PM

P: 126

I know that there does not exist a continuos function from [0,1] onto (0,1) because the image of a compact set for a continous function f must be compact, but isn't it also the case that the inverse image of a compact set must be compact? and a set in R is compact iff its closed and bounded right?



#2
Nov912, 11:25 PM

P: 1,622

First consider the continuous surjection [itex]f:(0,1) \rightarrow [0,1][/itex] defined as follows:
[tex] f(x) = \left\{ \begin{array}{lcl} 0 & : & x \in \left(0,\frac{1}{4}\right)\\[0.3em] 2x\frac{1}{2} & : & x \in \left[\frac{1}{4},\frac{3}{4}\right]\\[0.3em] 1 & : & x \in \left(\frac{3}{4},1\right) \end{array} \right. [/tex] Now to answer your other questions ... 


#3
Nov1012, 07:52 AM

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#4
Nov1212, 12:55 PM

P: 350

Continuous function from (0,1) onto [0,1]?
(1/2) (sin(2pi x)+1) maps (0,1) onto [0,1].
Clearly the problem is that this function is not injective. Same problem with the example by jgens. Does there exist an injective continuous function mapping (0,1) onto [0,1]? Assume there is, and suppose f(a)=0 and f(b)=1. WLOG assume b>a and let e>0 be small enough so that b+e<1. Since 1 is the max value of f, f(b+e) is strictly between 0 and 1. By the IVT, there exists c between a and b such that f(c)=f(b+e). So f can't be injective after all. More generally if f is injective and continuous from an interval of R into R, then it must be monotonic and its inverse must be continuous as well. 


#5
Nov1212, 05:52 PM

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P: 1,170

You can also use a result that a continuous bijection f: X>Y , between X compact
and Y Hausdorff , is a homeomorphism. And (0,1) and [0,1] are not homeomorphic for many reasons: [0,1] is compact and (0,1) is not, or (0,1) is 1connected and [0,1] is not  e.g., delete the endpoints of [0,1], and the space remains connected ( I think kconnectedness is also called the Euler number). This also shows there are no continuous bijections between (0,1) and [0,1) (because [0,1) is not 1connected; remove 0, and it remains connected.) 


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