Question

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

Answer 1

\( 2^{n-1} \) the the fifth line from proof while \( 2^{m-2} \) on the right bottom

Answer 2

Wait for @mukushla

Ha ha ha ha..

Answer 3

@experimentX: Apostol?

Answer 4

yes!! page no 55

Answer 5

Which Apostol is it?

Answer 6

Tom M. Apostle Mathematical Analysis Second Edition

Answer 7

I see.

Answer 8

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]).

Answer 9

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

Answer 10

right, but in the n dimensional case, the intervals in J_1 have length 2a.

Answer 11

Am I the only idiot here?

Answer 12

J_1 is defined as \( I_1^{(1)} \times I_2^{(1)} \times I_3^{(1)} \times ... I_n^{(1)}\)
where each \( I_k^{(1)}: -a \le x \le 0 \)

Answer 13

\[-a\le x_k\le a\]

Answer 14

i see i had been looking at the other page.

Answer 15

yeah, its a little confusing >.<

Answer 16

thanks for clearing up!!

Answer 17

@ParthKohli , No you are not alone here..

Answer 18

lol

Answer 19

Ha ha ha ha..

Even you know about Apostle..

I have heard the word first time in my Life..

Answer 20

He's just an author yaar

Answer 21

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..