experimentX
  • experimentX
Balzano-Weirerstrass Theorem: Does any one know why there is \( 2^{m-2} \) at the \( \mathbb{R^n} \) while there is \( 2^{m-1} \) on \( \mathbb{R}^1 \)? http://i.stack.imgur.com/O14VI.jpg
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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experimentX
  • experimentX
\( 2^{n-1} \) the the fifth line from proof while \( 2^{m-2} \) on the right bottom
anonymous
  • anonymous
Wait for @mukushla Ha ha ha ha..
ParthKohli
  • ParthKohli
@experimentX: Apostol?

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experimentX
  • experimentX
yes!! page no 55
ParthKohli
  • ParthKohli
Which Apostol is it?
experimentX
  • experimentX
Tom M. Apostle Mathematical Analysis Second Edition
ParthKohli
  • ParthKohli
I see.
anonymous
  • anonymous
It looks like its because of the way they labeled the subintervals in their proof. In the one dimensional case, they didnt label the starting interval (the one of length 2a, [-a,a]), while in the n-dimensional case they did (J_1 is the n-dimensional rectangle with sides [-a,a]).
experimentX
  • experimentX
the first interval is labelled as [a1,b1] which is [-a, 0] or the other, and it's length is 'a' if we continue this we, we get [an, bn] = a/2^(n-1) for the case of R^1
anonymous
  • anonymous
right, but in the n dimensional case, the intervals in J_1 have length 2a.
ParthKohli
  • ParthKohli
Am I the only idiot here?
experimentX
  • experimentX
J_1 is defined as \( I_1^{(1)} \times I_2^{(1)} \times I_3^{(1)} \times ... I_n^{(1)}\) where each \( I_k^{(1)}: -a \le x \le 0 \)
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
\[-a\le x_k\le a\]
experimentX
  • experimentX
i see i had been looking at the other page.
anonymous
  • anonymous
yeah, its a little confusing >.<
experimentX
  • experimentX
thanks for clearing up!!
anonymous
  • anonymous
@ParthKohli , No you are not alone here..
ParthKohli
  • ParthKohli
lol
anonymous
  • anonymous
Ha ha ha ha.. Even you know about Apostle.. I have heard the word first time in my Life..
ParthKohli
  • ParthKohli
He's just an author yaar
anonymous
  • anonymous
Ha ha ha... I think that is certain set in Mathematics which is given that name.. Ha ha ha. So, I am more Idiot than you..

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