## experimentX 4 years ago 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

1. experimentX

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

2. anonymous

Wait for @mukushla Ha ha ha ha..

3. ParthKohli

@experimentX: Apostol?

4. experimentX

yes!! page no 55

5. ParthKohli

Which Apostol is it?

6. experimentX

Tom M. Apostle Mathematical Analysis Second Edition

7. ParthKohli

I see.

8. anonymous

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]).

9. experimentX

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

10. anonymous

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

11. ParthKohli

Am I the only idiot here?

12. experimentX

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

13. anonymous

14. anonymous

$-a\le x_k\le a$

15. experimentX

i see i had been looking at the other page.

16. anonymous

yeah, its a little confusing >.<

17. experimentX

thanks for clearing up!!

18. anonymous

@ParthKohli , No you are not alone here..

19. ParthKohli

lol

20. anonymous

Ha ha ha ha.. Even you know about Apostle.. I have heard the word first time in my Life..

21. ParthKohli

He's just an author yaar

22. anonymous

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..