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 2006-10-09, 18:43 #1 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 10100000012 Posts S^oo ? I have a homework problem regarding $S^{\infty}$ and while I'm not looking for anyone to do my homework, I don't know how to define $S^{\infty}$. The only two definitions I know for $S^{n}$ are the points that are distance 1 from the origin in $\mathbb{R}^{n}$ and the boundary of $\mathbb{B}^{n}$, neither of which are helping me for the infinite case. Any info would be greatly appreciated P.S. Wow, that's an attractive avatar.
 2008-03-04, 16:14 #2 m_f_h     Feb 2007 24·33 Posts (to reactivate this subforum without news since beg. of December...) If you suggest the context of S^n in IR^n, I suppose S^oo is IR^IN, (vectors with oo number of components, a.k.a. sequences), probably restricted to a subspace with finite norm. But since not all norms are equivalent in oo-dimensional case, one must know which one is meant. It could be the \ell^2 norm (sqrt of the infinite sum of squared components which is required to converge) or the max norm or some ultrametric norm like limsup |x_n|^{w_n} where (w_n) can be any sequence of weights (e.g. w_n=1/log(n) gives sequences of at most polynomial growth, and that ultranorm gives (exp() of) the corresponding power of n). Last fiddled with by m_f_h on 2008-03-04 at 16:40
 2008-03-04, 16:40 #3 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 33·239 Posts I'd have thought S^\infty was the set of all vectors of real numbers the sums of whose squares is 1 ... what is it you need to prove about it?
2008-03-04, 17:45   #4
m_f_h

Feb 2007

24×33 Posts

Quote:
 Originally Posted by fivemack I'd have thought S^\infty was the set of all vectors of real numbers the sums of whose squares is 1 ... what is it you need to prove about it?
you mean: the set of all vectors of any length? well vectors of finite length can always be considered as sequences with only a finite number of nonzero components. so, it is essentially what I said. (with \ell^2 norm, where the sum of squared components of course converges if there is only a finite number of nonzero ones.).

 2008-03-05, 00:03 #5 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 64110 Posts Thanks for the timely response. So, if we take $S^{0}$ and two line segments, we can identify the boundaries of the line segments with $S^{0}$ and obtain $S^{1}$. Similarly, if we take $S^{1}$ and two discs, we can identify the boundaries of the two discs with $S^{1}$ and obtain $S^{2}$. We can carry this process on indefinitely, and define $S^{\infty}$ to be what we get in the limit Last fiddled with by Orgasmic Troll on 2008-03-05 at 00:04
2008-03-08, 20:20   #6
m_f_h

Feb 2007

43210 Posts

Quote:
 Originally Posted by Orgasmic Troll Thanks for the timely response. So, if we take $S^{0}$ and two line segments, we can identify the boundaries of the line segments with $S^{0}$ and obtain $S^{1}$. Similarly, if we take $S^{1}$ and two discs, we can identify the boundaries of the two discs with $S^{1}$ and obtain $S^{2}$. We can carry this process on indefinitely, and define $S^{\infty}$ to be what we get in the limit
That's complicated... and maybe a bit misleading, since after any finite number of steps, you are always still in the finite-dimensional case, but, as already alluded to, the infinite-dimensional case has qualitatively different properties.
Also, it does not really give me an intuitive idea of how the limit would look like...

Finally, what is the definition of "limit" used here ?

Last fiddled with by m_f_h on 2008-03-08 at 20:22

2008-03-09, 01:40   #7
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

10100000012 Posts

Quote:
 Originally Posted by m_f_h That's complicated... and maybe a bit misleading, since after any finite number of steps, you are always still in the finite-dimensional case, but, as already alluded to, the infinite-dimensional case has qualitatively different properties. Also, it does not really give me an intuitive idea of how the limit would look like... Finally, what is the definition of "limit" used here ?
Consider $S^n$ as defined above (i.e. iteratively) and let $S^\infty = \bigcup S^n$

2008-03-13, 21:18   #8
m_f_h

Feb 2007

24·33 Posts

Quote:
 Originally Posted by Orgasmic Troll Consider $S^n$ as defined above (i.e. iteratively) and let $S^\infty = \bigcup S^n$
This notation does not make sense. You know that things like "the set of all sets" lead to logical inconsistency (if it is a set, it cannot be included in itself). You have to specify in which "universe" the objects you are considering live in.

Also, even assuming that we have given a rigorous meaning to this notation, it would imply that any element of S°° would be the element of some S^n, and reciprocally. This also seems not obvious / correct to me. (Why should a 1-sphere also be a oo-sphere ? It is not a 2-sphere, though.)

2008-03-14, 06:03   #9
Orgasmic Troll
Cranksta Rap Ayatollah

Jul 2003

641 Posts

Quote:
 Originally Posted by m_f_h This notation does not make sense. You know that things like "the set of all sets" lead to logical inconsistency (if it is a set, it cannot be included in itself). You have to specify in which "universe" the objects you are considering live in. Also, even assuming that we have given a rigorous meaning to this notation, it would imply that any element of S°° would be the element of some S^n, and reciprocally. This also seems not obvious / correct to me. (Why should a 1-sphere also be a oo-sphere ? It is not a 2-sphere, though.)
There is no "set of all sets" notation here. It's a union of sets over a countable index.

Sure, every element of $S^\infty$ is an element of (an infinite number of) $S^n$. Every element of the reals is also contained in an infinite number of intervals of the form [-a,a].

I don't know where you're getting the notion that a 1-sphere is a $\infty$-sphere. pi is a real number, but that doesn't make it all of the reals.

 2008-03-14, 14:07 #10 wblipp     "William" May 2003 New Haven 2,371 Posts The sequence xn = 2-n/2 has the sum of squares = 1 but is not in any of the Sn - although it IS in the completion of the Union.
 2008-03-29, 17:25 #11 Kevin     Aug 2002 Ann Arbor, MI 6618 Posts You have two main choices: One is just to take the union of all the S^n's (basically infinite sequences that are zero for all but finitely many terms, where the sum of the squares of the non-zero terms is 1), or in a sense taking the completion of this, and allowing all infinite sequences where the sum of the squares of the entries add up to 1 (allowing things like 6/pi^2*(1,1/2,1/3,1/4,...) . If you know topology, it's basically the difference between choosing the product topology and the box topology (http://mathworld.wolfram.com/ProductTopology.html). I have no idea which way is the standard way to define S^inf, and without knowing the question we can't really judge which one makes more sense in this situation.

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