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Real analysis

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that there does not exist a continuous, bijective function ##f:[0,1)\to \mathbb{R}.##

    2. The attempt at a solution

    I am stumped on how to do this question. What I was thinking of doing was assume that there is a function and arrive at a contradiction, in doing so I know I need to use the IVT since ##f## is assumed to be continuous, but I can't create such a contradiction.
     
  2. jcsd
  3. Nov 19, 2013 #2
    Theorem:
    Let be [itex]f : A \subset E \longrightarrow F[/itex]
    The following properties are equivalent:
    1) f is continuos in [itex]A[/itex]
    2) If [itex]T[/itex] is a open set in (F,d'), then [itex]f^{-1}(T)[/itex] is open set in the subspace (A,d)
    3) If [itex]T[/itex] is a closed set in (F,d'), then [itex]f^{-1}(T)[/itex] is closed set in the subspace (A,d)
    4) For all set [itex]S[/itex] with [itex]S \subset A[/itex]:
    [itex]f(A \cap \overline{S} \subset \overline{f(S)}[/itex]

    I studied it in a course of "Introduction to Analysis" (It was roughly Metric Spaces Topology).
     
    Last edited: Nov 19, 2013
  4. Nov 19, 2013 #3

    Dick

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    Think back to one of your previous exercises and draw pictures of possibilities for the function. Using the IVT, isn't it true that since f is one-to-one then f([0,1/2])=[f(0),f(1/2)]? Now where are you going to put f((1/2,1))??
     
    Last edited: Nov 19, 2013
  5. Nov 19, 2013 #4
    Dick - So I can assume an ##x\in [0,1/2]## such that ##f(x)\le f(0)##?
     
    Last edited: Nov 19, 2013
  6. Nov 19, 2013 #5

    Dick

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    Well, no. I actually stated what I said a little wrong. I meant that f([0,1/2]) is equal to either [f(0),f(1/2)] or [f(1/2),f(0)] depending on whether f(0)<f(1/2) or vice versa. You remember the previous exercise, right?
     
  7. Nov 19, 2013 #6
    Hmm, can you elaborate a bit more? If I consider the case when ##f(0) <f(1/2)## then that must imply that ##[f(0),f(1/2]##?
     
  8. Nov 19, 2013 #7

    Dick

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  9. Nov 19, 2013 #8
    But in that proof, we considered ##x\in (a,b)## such that ##f(x) <f(a)## and you said I can't do that here.
     
  10. Nov 19, 2013 #9

    Dick

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    In that proof they assumed a<b because they wanted to write the interval [a,b] without having to say [a,b] if a<b or [b,a] if b<a. It's not important. The substance of the proof is. In this case I know 0<1/2. So I don't have to say 'assume 0<1/2'. f([0,1/2]) is either [f(0),f(1/2)] or [f(1/2),f(0)] depending. That's what I would like you to confirm that you still understand. I don't know what this x is you are speaking of here.
     
  11. Nov 19, 2013 #10
    So, since this is a bijection then for ##0<1/2## we have ##[f(0), f(1/2)]## and for ##1/2 < 1## we have ##[f(1/2),f(1)]##.
     
  12. Nov 19, 2013 #11

    Dick

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    No, you don't. I don't even think that means anything. f(1) isn't even necessarily defined. Can you just confirm that you understand why f([0,1/2]) is equal to either [f(0),f(1/2)] or [f(1/2),f(0)], you seemed comfortable with the proof yesterday.
     
  13. Nov 20, 2013 #12
    Yes, I can confirm that because we proved that yesterday.
     
  14. Nov 20, 2013 #13

    Dick

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    Alright! Let's assume for the moment that f([0,1/2])=[f(0),f(1/2)]. Then f(3/4) is either greater than f(1/2) or less than f(0), right? You should really be drawing pictures here.
     
  15. Nov 20, 2013 #14
    I drew a graph of a function ##f(x) = x## so it's just a diagonal line. Now for ##f(3/4)## why would it be less that ##f(0)##?
     
  16. Nov 20, 2013 #15

    Dick

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    f(x) is any bijective continuous function. It doesn't have to be f(x)=x. Draw a picture where you wiggle f(x) around anyway you want as long as it's continuous. I'm going to bed soon so I'll try and give you too many hints and see if you can figure it out. If f(3/4)>f(1/2), and since the function is onto, there must also be an x such that f(x)<f(0) and x must be greater than 1/2. Why? Try and draw such functions see why they can't be bijective. Using the IVT that would give you a contradiction with one-to-oneness. Just try it. I'll be back tomorrow.
     
    Last edited: Nov 20, 2013
  17. Nov 20, 2013 #16
    Okay, I will try it. Thanks!
     
  18. Nov 20, 2013 #17

    Dick

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    Actually trying to work your previous result into this isn't making it any simpler. Just think about f(0). Since f is onto, there must be an x in (0,1) such that f(x)=f(0)+1 and a y in (0,1) such that f(y)=f(0)-1. Just think about that.
     
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