# A Baby Rudin Related Question

• Sep 21st 2010, 05:04 PM
DontKnoMaff
A Baby Rudin Related Question
See the attachment for my question.

I am confused on how to do this problem... I need a little bit of help getting started.

(a), (b), and (c) seem trivial. I can't think of doing anything other than actually going through this process a couple of times to show these are all true by induction. I however have no idea nor understand part (d). Parts (e) and (f) I think I can manage pretty easily after getting the others done. Do you guys think induction is appropriate for the first 3? and could someone help shed some light on part (d).

This whole problem seems pretty weird to me. I dont even see how you could pick a pair of to start with..
• Sep 21st 2010, 07:03 PM
tonio
Quote:

Originally Posted by DontKnoMaff
See the attachment for my question.

I am confused on how to do this problem... I need a little bit of help getting started.

(a), (b), and (c) seem trivial. I can't think of doing anything other than actually going through this process a couple of times to show these are all true by induction. I however have no idea nor understand part (d). Parts (e) and (f) I think I can manage pretty easily after getting the others done. Do you guys think induction is appropriate for the first 3? and could someone help shed some light on part (d).

This whole problem seems pretty weird to me. I dont even see how you could pick a pair of to start with..

This is not more and not less that a particular case of the great Cantor's Theorem for closed embedded intervals (in the real line with the usual euclidean topology), and its usual, and perhaps easier, proof uses another great theorem: Bolzano-Weierstrass's: every bounded infinite sequence has a partial limit <==> every bounded infinite sequence has a subsequence which converges to a limit.
I'm not aware of any use of induction to prove this.

As for part (d) it follows directly from the work above: x belongs to all and each of of the closed intervals, so if $\displaystyle x^m\neq y$ then either $\displaystyle x^m<y$ or $\displaystyle x^m>y$ , say $\displaystyle y-x^m=\epsilon>0$ , but then we get a contradiction by considering that $\displaystyle a_n\xrightarrow [n\to \infty]{}x$ and thus some $\displaystyle a_n$ will be between $\displaystyle x^m\,\,and\,\,y$ ....and something very similar for $\displaystyle x^m>y$

Tonio
• Sep 21st 2010, 07:37 PM
DontKnoMaff
re:
I am sorry, were you referring to part (c) or (f) in that response?
• Sep 22nd 2010, 10:57 AM
tonio
Quote:

Originally Posted by DontKnoMaff
I am sorry, were you referring to part (c) or (f) in that response?

(f) relying on (c), of course...ain't this clear from the context?

Tonio
• Sep 23rd 2010, 03:18 PM
DontKnoMaff
Attachment 19030
Attachment 19029
I only asked because I was hoing for some help on (d) specifically, in any case, I would like to post my work on this and see if anybody has a problem with it. I am very unsure on how to do this...
part (a)-(d) see my attachment

(e) I would like to just use the Bolzano Weierstrass Theorem here.... This is as far as I have done..