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 one year ago
BalzanoWeirerstrass Theorem: Does any one know why there is \( 2^{m2} \) at the \( \mathbb{R^n} \) while there is \( 2^{m1} \) on \( \mathbb{R}^1 \)?
http://i.stack.imgur.com/O14VI.jpg
 one year ago
BalzanoWeirerstrass Theorem: Does any one know why there is \( 2^{m2} \) at the \( \mathbb{R^n} \) while there is \( 2^{m1} \) on \( \mathbb{R}^1 \)? http://i.stack.imgur.com/O14VI.jpg

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experimentX
 one year ago
Best ResponseYou've already chosen the best response.1\( 2^{n1} \) the the fifth line from proof while \( 2^{m2} \) on the right bottom

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1Wait for @mukushla Ha ha ha ha..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1@experimentX: Apostol?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Which Apostol is it?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1Tom M. Apostle Mathematical Analysis Second Edition

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1It 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 ndimensional case they did (J_1 is the ndimensional rectangle with sides [a,a]).

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1the 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^(n1) for the case of R^1

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1right, but in the n dimensional case, the intervals in J_1 have length 2a.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Am I the only idiot here?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1J_1 is defined as \( I_1^{(1)} \times I_2^{(1)} \times I_3^{(1)} \times ... I_n^{(1)}\) where each \( I_k^{(1)}: a \le x \le 0 \)

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1\[a\le x_k\le a\]

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1i see i had been looking at the other page.

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.1yeah, its a little confusing >.<

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1thanks for clearing up!!

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1@ParthKohli , No you are not alone here..

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1Ha ha ha ha.. Even you know about Apostle.. I have heard the word first time in my Life..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1He's just an author yaar

waterineyes
 one year ago
Best ResponseYou've already chosen the best response.1Ha 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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