Here's the question you clicked on:
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
\( 2^{n-1} \) the the fifth line from proof while \( 2^{m-2} \) on the right bottom
Wait for @mukushla Ha ha ha ha..
@experimentX: Apostol?
Which Apostol is it?
Tom M. Apostle Mathematical Analysis Second Edition
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]).
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
right, but in the n dimensional case, the intervals in J_1 have length 2a.
Am I the only idiot here?
J_1 is defined as \( I_1^{(1)} \times I_2^{(1)} \times I_3^{(1)} \times ... I_n^{(1)}\) where each \( I_k^{(1)}: -a \le x \le 0 \)
\[-a\le x_k\le a\]
i see i had been looking at the other page.
yeah, its a little confusing >.<
thanks for clearing up!!
@ParthKohli , No you are not alone here..
Ha ha ha ha.. Even you know about Apostle.. I have heard the word first time in my Life..
He's just an author yaar
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..