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experimentX
 3 years 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
experimentX
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1Wait for @mukushla Ha ha ha ha..

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

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

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

joemath314159
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1Am I the only idiot here?

experimentX
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[a\le x_k\le a\]

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

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

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

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

waterineyes
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1He's just an author yaar

waterineyes
 3 years 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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