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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

@KingGeorge Can you take a look?
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.
The thing is we dont touch on Cantor's Diagonalization. We are learning abt limts and covergence

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

So your teacher is looking for something that's not the diagonalization?
Yess
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.
Gonna think abt it toooo
Well actually it gives us the method we should be using
OK whatever its 3:30 am. I cant think anymore. i will have to put something together tomorrow
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
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}
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.
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
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.
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/
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]
That was the onlt part that was confusing me
In retrospect, I probably should have changed part (a) so that you chose the smallest \(a_1\) such that \(a_1>0\), and \(a_10\). If \(x_1=0\), we would not be able to choose any \(a_1\) smaller than \(x_1\). Case 2 deals with that.
No wait. Let me read over that more carefully before I continue.
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.
That make more sense?
Ohhhh ok get it
Yaaaaaa
Hmmm I am also dont follow the part where u prove that \(A \in [a_n,b_n]\) for all \( n \in \mathbb{N}\)
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
OHHHH I GET IT NOWWW
Thanks KG :)
Its really clear and and to the point. You always have this sleek way of proving
Thanks :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question