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swissgirl Group Title

Prove that the interval \(\{x \in \mathbb{R} \vert 0 \leq x \leq 1\}\) in uncountable. In other words show that no function \(f \mathbb{N} \to [0,1]\) can have one to one and onto correspondence.

  • one year ago
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  1. swissgirl Group Title
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    @KingGeorge Can you take a look?

    • one year ago
  2. KingGeorge Group Title
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    The classical way to do this, is by Cantor's diagonalization proof. Suppose you can list every number in [0,1]. So you have something like 1. 0.723462349786... 2. 0.473677967453... 3. 0.012384756293... . . . And so on.

    • one year ago
  3. swissgirl Group Title
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    The thing is we dont touch on Cantor's Diagonalization. We are learning abt limts and covergence

    • one year ago
  4. KingGeorge Group Title
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    So your teacher is looking for something that's not the diagonalization?

    • one year ago
  5. swissgirl Group Title
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    Yess

    • one year ago
  6. KingGeorge Group Title
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    Since you're talking about limits and convergence, I would then suppose that this has something to do with taking infinite sums that converge to real numbers, and somehow showing there are an uncountable number of these sums. Hmmm... I may have to think for a little bit.

    • one year ago
  7. swissgirl Group Title
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    Gonna think abt it toooo

    • one year ago
  8. swissgirl Group Title
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    Well actually it gives us the method we should be using

    • one year ago
  9. swissgirl Group Title
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    OK whatever its 3:30 am. I cant think anymore. i will have to put something together tomorrow

    • one year ago
  10. swissgirl Group Title
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    I am gonna head to bed now. I will come back tomorrow to check if you found some direction but otherwise don't stress over it. I will hopefully figure something out. Thanks King George :) Good Night

    • one year ago
  11. KingGeorge Group Title
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    For (a), perhaps a proof by induction will work. Base: n=1. Let f(1)=\(x_1\in[0,1]\). Then you can choose \(a_1\in[0,1]\) such that \(a_1\) is minimal and less than \(x_1\). Similarly, let \(b_1\in[0,1]\) such that \(b_1\) is maximal and less than \(x_1\). If \(x_1=0\), choose any \(a_1,b_1\in(0,1]\). Clearly, \(x_1\notin[a_1,b_1]\). Now assume true up to some \(k\in \mathbb{N}\). Then we have two cases. Case 1: \(x_{k+1}\notin[a_k,b_k]\). Here, just choose \(a_{k+1}>a_k\) and \(b_{k+1}<b_k\) such that \([a_{k+1},b_{k+1}]\subset[a_k,b_k]\). Case 2: \(x_{k+1}\in[a_k,b_k]\). Here, do the same kind of process we did for the base case. Hence, by induction, we're done.

    • one year ago
  12. KingGeorge Group Title
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    To show that \(\{a_n\}\) converges, WLOG let \(n>m\), and note that \(|a_n-a_m|<|b_m-a_m|\) since \(a_n\in[a_m,b_m]\). Then, since the interval gets smaller on every repetition, we can say that \(|b_m-a_m|<\epsilon\) for any \(\epsilon>0\) for sufficiently large \(m\). Hence, \(\{a_n\}\) is Cauchy, and is therefore convergent, and converges to some point A.

    • one year ago
  13. KingGeorge Group Title
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    Then, we can see that A is in \([a_n,b_n]\) for all \(n\in\mathbb{N}\) since each interval is contained within the previous one, and since \(\{a_n\}\) is a strictly increasing sequence. If \(A\notin[a_n,b_n]\) for some n, then since the sequence is increasing, it must be that \(A>b_k\) for some \(k\). However, \(a_k<b_l\) for any choice of \(k\) or \(l\), so this is a contradiction. Hence, \(A\in[a_n,b_n]\) for all \(n\in\mathbb{N}\).

    • one year ago
  14. KingGeorge Group Title
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    To finish it off, notice that \(f(n)\neq A\) for any choice of \(n\) since \(A\in[a_n,b_n]\) for all \(n\), but \(f(n)\notin[a_n,b_n]\) for all \(n\). Thus, \(f\) is not a surjective function, and the reals are uncountable.

    • one year ago
  15. KingGeorge Group Title
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    Just fyi, I did reference this page a little bit, however, they take some of these steps for granted (such as part (a)), and have a very informal proof. http://boolesrings.org/scoskey/my-favorite-proof-that-r-is-uncountable/

    • one year ago
  16. swissgirl Group Title
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    I wasnt exactly sure what we are proving for case 2 of part (a) If \(x_{k+1}\) is an element of \([a_k,b_k]\) Then it must be an element of [0,1]

    • one year ago
  17. swissgirl Group Title
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    That was the onlt part that was confusing me

    • one year ago
  18. KingGeorge Group Title
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    In retrospect, I probably should have changed part (a) so that you chose the smallest \(a_1\) such that \(a_1>0\), and \(a_1<x_1\). This would make my proof of part (c) more technically correct. However, this is assuming that \(x_1>0\). If \(x_1=0\), we would not be able to choose any \(a_1\) smaller than \(x_1\). Case 2 deals with that.

    • one year ago
  19. KingGeorge Group Title
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    No wait. Let me read over that more carefully before I continue.

    • one year ago
  20. KingGeorge Group Title
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    Oh right. The second case deals with the possibility of \(x_k\) landing in the interval already chosen. This limits the new interval we have to create, since \(x_k\) can not lie in the new interval. So I was lazy, and said to make the new interval in the same kind of fashion we chose the very first interval.

    • one year ago
  21. KingGeorge Group Title
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    That make more sense?

    • one year ago
  22. swissgirl Group Title
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    Ohhhh ok get it

    • one year ago
  23. swissgirl Group Title
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    Yaaaaaa

    • one year ago
  24. swissgirl Group Title
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    Hmmm I am also dont follow the part where u prove that \(A \in [a_n,b_n]\) for all \( n \in \mathbb{N}\)

    • one year ago
  25. KingGeorge Group Title
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    So we created the sequence \(\{a_k\}\) as a strictly increasing sequence (at least with the new revisions I just gave a few posts above). Thus, \(A>a_k\) for any \(k\in\mathbb{N}\). Therefore, we now suppose that \(A\notin[a_k,b_k]\) for some \(k\in\mathbb{N}\). So \(A>b_k\). However, since \(\{a_k\}\) converges to \(A\), we can choose some \(a_j\) such that \(b_k<a_j<A\). This is a contradiction, since we have always chosen our \(a_k\) to be less than \(b_k\). Therefore \(A\) is in the interval.

    • one year ago
  26. swissgirl Group Title
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    OHHHH I GET IT NOWWW

    • one year ago
  27. swissgirl Group Title
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    Thanks KG :)

    • one year ago
  28. swissgirl Group Title
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    Its really clear and and to the point. You always have this sleek way of proving

    • one year ago
  29. KingGeorge Group Title
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    Thanks :)

    • one year ago
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