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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 \( \mathbb{R}^1 \)?

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\( 2^{n-1} \) the the fifth line from proof while \( 2^{m-2} \) on the right bottom
Wait for @mukushla Ha ha ha ha..
@experimentX: Apostol?

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Other answers:

yes!! page no 55
Which Apostol is it?
Tom M. Apostle Mathematical Analysis Second Edition
I see.
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 \( I_k^{(1)}: -a \le x \le 0 \)
1 Attachment
\[-a\le x_k\le a\]
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..

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