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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
 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
 one year ago

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

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

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

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

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

joemath314159Best ResponseYou've already chosen the best response.1
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 ndimensional case they did (J_1 is the ndimensional rectangle with sides [a,a]).
 one year ago

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

joemath314159Best ResponseYou've already chosen the best response.1
right, but in the n dimensional case, the intervals in J_1 have length 2a.
 one year ago

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

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

joemath314159Best ResponseYou've already chosen the best response.1
\[a\le x_k\le a\]
 one year ago

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

joemath314159Best ResponseYou've already chosen the best response.1
yeah, its a little confusing >.<
 one year ago

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

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

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

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

waterineyesBest ResponseYou've already chosen the best response.1
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..
 one year ago
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