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swissgirl
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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
 one year ago
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
 one year ago

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swissgirl Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge Can you take a look?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
The thing is we dont touch on Cantor's Diagonalization. We are learning abt limts and covergence
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
So your teacher is looking for something that's not the diagonalization?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Gonna think abt it toooo
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Well actually it gives us the method we should be using
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
OK whatever its 3:30 am. I cant think anymore. i will have to put something together tomorrow
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
To show that \(\{a_n\}\) converges, WLOG let \(n>m\), and note that \(a_na_m<b_ma_m\) since \(a_n\in[a_m,b_m]\). Then, since the interval gets smaller on every repetition, we can say that \(b_ma_m<\epsilon\) 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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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/myfavoriteproofthatrisuncountable/
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
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

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
That was the onlt part that was confusing me
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
No wait. Let me read over that more carefully before I continue.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
That make more sense?
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Ohhhh ok get it
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
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

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
OHHHH I GET IT NOWWW
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Thanks KG :)
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Its really clear and and to the point. You always have this sleek way of proving
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.3
Thanks :)
 one year ago
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