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Do I need a function to show a bijection between intervals?

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to prove that [0,1) and (0,1] have the same cardinality. My question is, do I have to define a function from [0,1) → (0,1] in order to show a correspondence or is there another method?


    Thanks.
     
  2. jcsd
  3. Sep 28, 2012 #2

    jbunniii

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    There's actually a very simple bijection in this case. (Hint: most of the points can simply be mapped to themselves.)

    More interesting is to find one between (0,1) and [0,1].
     
  4. Sep 28, 2012 #3
    Would f(x)=1/1+x work? How would I find a bijection for the intervals you mentioned?
     
  5. Sep 28, 2012 #4

    Ray Vickson

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    No, f(x) = 1/1+x = 1 + x would not work, but f(x) = 1/(1+x) would be OK. Use brackets!

    RGV
     
  6. Sep 28, 2012 #5
    Oh yeah, that's what I meant :smile:

    In regards to the interval (whether open or closed), how can I be sure that the function will works as a bijection?
     
  7. Sep 28, 2012 #6

    jbunniii

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    If you can find an inverse function, then it's a bijection. To do that, put y = 1 / (1 + x) and try solving for x in terms of y.

    There's also a much simpler solution if you consider the hint I mentioned above.

    [edit] Wait a minute. Who says 1 / (1 + x) will work? It maps [0, 1) to (1/2, 1], not (0, 1]. You'll have to modify it slightly to get (0,1] as the image.
     
  8. Sep 28, 2012 #7
    jbunniii- How did you know that 1/(1+x) maps [0,1) to (1/2,1] and not (0,1]?

    And is the function you were hinting about f(x)=x+1?
     
  9. Sep 28, 2012 #8

    jbunniii

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    If 0 <= x < 1, then 1 <= 1+x < 2, so 1/2 < 1/(1+x) <= 1.

    If 0 <= x < 1, then 1 <= x+1 < 2, so that won't work. If you put change it slightly by putting a negative sign in the right place, it will work.
     
    Last edited: Sep 28, 2012
  10. Sep 28, 2012 #9
    I understand now, and the function you were referring to is f(x)=-x+1.

    Thanks :)
     
  11. Sep 28, 2012 #10

    jbunniii

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    Yep. Here's an even easier one, the one I was thinking of initially:

    [tex]f(x) = \begin{cases}
    1 & \text{if }x = 0 \\
    x & \text{if }x \neq 0
    \end{cases}
    [/tex]
    By the way, if you want a challenge, I highly recommend trying to come up with a bijection between (0,1) and [0,1]. Hint: you know it can't be a continuous function because such a function would have to preserve open and closed sets.
     
  12. Sep 28, 2012 #11
    Cool, I didn't think of that. I am actually working on that now. I have to show that [0,1] has the same cardinality as (0,1).
    I have a mapping:
    0 to 1/2
    1 to 1/3
    1/2 to 1/4
    etc.
    So f(x)=1/2, if x=0
    f(x)=1/(x+2), if x=1/n for positive integers n >= 1
    and f(x)=x for all other x.
    Is it correct syntax if I write it as a piece wise function?

    f(x)={1/(x+2), if x is in N[itex]\cup[/itex]{0}
    x, for all other x
    I'm not sure how to word that last part (f(x)=x).

    Edit:
    This seems cleaner:

    f(x)={1/(x+2), for x in NU{0}
    x, for x in ℝ\ NU{0}

    Where N denotes the set of all positive integers.
    That seems cleaner.
     
    Last edited: Sep 28, 2012
  13. Sep 28, 2012 #12

    jbunniii

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    Careful with your formula. For example, if x = 1/3, your formula gives f(1/3) = 1/(1/3 + 2) = 1/(7/3) = 3/7, which is not what you want. Instead, you could write [itex]f(1/n) = 1/(n+2)[/itex] for all [itex]n \in \mathbb{N}[/itex], [itex]f(0) = 1/2[/itex], and [itex]f(x) = x[/itex] for all other [itex]x[/itex].

    No, your "if x is in [itex]\mathbb{N} \cup \{0\}[/itex]" is wrong. That set is [itex]\{0, 1, 2, 3, \ldots)[/itex]. But most of these points are outside your domain, which is [itex][0,1][/itex]

    Similar problem here.
     
  14. Sep 28, 2012 #13
    I see my mistakes. If I use f(1/n) and f(x), then I wouldn't write it in the format used for piecewise functions?

    Also, to show the bijection, would I need to go through proving surjectivity/injectivity?
     
  15. Sep 28, 2012 #14

    jbunniii

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    I think you will still need a piecewise definition. It's hard to imagine a single formula working for all x. If you want to write it all in terms of f(x), you could put something like
    [tex]
    f(x) = \begin{cases}
    \frac{1}{(1/x) + 2} & \text{if }x = 1/n \text{ for some }n \in \mathbb{N} \\
    1/2 & \text{if }x = 0 \\
    x & \text{otherwise}
    \end{cases}
    [/tex]
    Yes, you should prove this. You could do this by showing that [itex]f[/itex] has an inverse.
     
  16. Sep 29, 2012 #15
    Okay, I did it. I let g be the inverse of f, defined by:

    g: (0,1)→[0,1]


    g(x)= { 0, if x=1/2
    [itex]\frac{1}{(1/x)-2}[/itex] , if x=1/n for n[itex]\in[/itex] N
    x, for all other x

    Then I would prove that it's an inverse by composing g[itex]\circ[/itex]f and f[itex]\circ[/itex]g

    Does that look fine?

    Thanks.
     
  17. Sep 29, 2012 #16

    jbunniii

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    It looks almost right, except the middle case is only valid for n >= 3. The first case handles n = 2. And n = 1 is outside the function's domain.
     
  18. Sep 29, 2012 #17
    Oh yeah, I had that written down but completely missed it. Thank you for all your help :smile:
     
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