Re: [math-fun] Infinite product vector space puzzle
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
I thought about the same question, in a slightly more general setting a few years ago. Specifically, let k be a field, and let Map(N, k) be the k-vector space of all functions from N (the set of natural numbers) to k . The question I considered, was "what is the dimension of Map(N, k) over k ?" I found it very counterintuitive that the answer depends upon the field k ! I did not solve it in all cases, but I can do Dan's case, in which case the dimension is c = 2^Aleph_0 . In general, the dimension is at least max(|k|, c) . I do not know if this is the right answer if k has cardinality strictly between Aleph_0 and c (in which case CH is false). For x in k , consider the function n |--> x^n . These functions are linearly independent over k , so the dimension is at least |k| . On the other hand, the vector space V = Map(N, k) has cardinality at least c . If k is an infinite field, and B is a basis of V , then |V| = |B| |k| = max(|B|, |k|) . This shows that if |k| < c , then |B| >= c . If k is finite, then it's easy to see that |V| = c , and B must be infinite, in which case |V| = |B| , so the dimension is c . (When k = R , the vector space has cardinality c , so a basis can't have cardinality larger than c .) Michael Reid
Nice and general, Michael! My solution just determines dim(R^oo). Here of course oo denotes aleph_0. In what follows we identify aleph_0 with the first countable ordinal, w, and so we denote R^oo by R^w. Lemma: ------ |R^w| = c. Proof: ------ Since |R| = 2^ω = c, it follows that |R^w| = (2^w)^w = 2^(wxw) = 2^w = c. Hence, |R^w| = c is an upper bound for any basis of R^w, i.e., dim(R^w) <= c. Definition: ----------- B = the set of vectors {e_j in R^w | j in Z+} such that the jth coordinate of e_j is 1 and the rest are 0. Let f: Z+ -> Q (the rationals) be any fixed bijection. Definition: ----------- For each real number r, define vector v_r in R^w via v_r satisfies pi_n(v_r) is 1 if f(n) < r, and is 0 otherwise. (I.e., each v_r corresponds to the indicator function of the Dedekind cut for r.) Lemma: ------ No nontrivial linear combination of the v_r's can be the 0 vector. Proof: ------ In any linear combination of the v_r's with nonzero coefficients, the largest r, among the v_r's occurring in it, contains coordinates that are 1, but which are 0 in all the other v_r's occurring in it. Hence the set of all v_r's is linearly independent Since |{v_r | r in R}| = c, we have c <= dim(R^w) <= |R^w| = c, so |R^w| = c. --Dan _____________________________________________ Michael Reid wrote: << I wrote: << Let R^oo denote the real vector space that is the countable direct product of copies of the reals. I.e., all countable-tuples of reals with componentwise addition. Puzzle: What is the dimension of the real vector space R^oo ???
I thought about the same question, in a slightly more general setting a few years ago. Specifically, let k be a field, and let Map(N, k) be the k-vector space of all functions from N (the set of natural numbers) to k . The question I considered, was "what is the dimension of Map(N, k) over k ?" I found it very counterintuitive that the answer depends upon the field k ! I did not solve it in all cases, but I can do Dan's case, in which case the dimension is c = 2^Aleph_0 . In general, the dimension is at least max(|k|, c) . I do not know if this is the right answer if k has cardinality strictly between Aleph_0 and c (in which case CH is false). For x in k , consider the function n |--> x^n . These functions are linearly independent over k , so the dimension is at least |k| . On the other hand, the vector space V = Map(N, k) has cardinality at least c . If k is an infinite field, and B is a basis of V , then |V| = |B| |k| = max(|B|, |k|) . This shows that if |k| < c , then |B| >= c . If k is finite, then it's easy to see that |V| = c , and B must be infinite, in which case |V| = |B| , so the dimension is c . (When k = R , the vector space has cardinality c , so a basis can't have cardinality larger than c .)
Michael Reid's example of functions f(n) = x^n is illuminating. In particular, to support my doubt about whether this definition of "dimension" leads to a sensible results in Hilbert space, I should need to answer affirmatively the following question: Given an arbitrary real sequence X = [x_i] with |x_i| < 1 , does there exist a real sequence A = [a_i] such that for all natural n \sum_i a_i (x_i)^n = 0 over natural i ? Consider for example specialising to x_i = 2^(-i) : we require A such that \sum_i a_i = \sum_i a_i/2^i = \sum_i a_i/4^i = ... = 0. Restricting 0 <= n < m, 0 <= i <= m yields homogeneous simultaneous linear equations; do their solutions A^m converge as m -> oo ? Resulting A^m for m = 1..8, fixing a_0 = 1 : [1,-1] [1, -3, 2] [1, -7, 14, -8] [1, -15, 70, -120, 64] [1, -31, 310, -1240, 1984, -1024] [1, -63, 1302, -11160, 41664, -64512, 32768] [1, -127, 5334, -94488, 755904, -2731008, 4161536, -2097152] [1, -255, 21590, -777240, 12850368, -99486720, 353730560, -534773760, 268435456] Apparently (-1)^m A^m is unimodal at |a_(m-1)| = 2^(m(m-1)/2) - 2^((m-1)(m-2)/2) ; so normalising to \sum (a_i)^2 = 1, as m -> oo all initial coefficients approach zero: in the limit, there is after all no (nonzero, infinite) linear relation! Tradition 1, Lunnon 0. Often happens. WFL On 8/27/12, Michael Reid <reid@gauss.math.ucf.edu> wrote:
Let R^oo denote the real vector space that is the countable direct product of copies of the reals.
I.e., all countable-tuples of reals with componentwise addition.
Puzzle: What is the dimension of the real vector space R^oo ???
I thought about the same question, in a slightly more general setting a few years ago. Specifically, let k be a field, and let Map(N, k) be the k-vector space of all functions from N (the set of natural numbers) to k . The question I considered, was "what is the dimension of Map(N, k) over k ?"
I found it very counterintuitive that the answer depends upon the field k !
I did not solve it in all cases, but I can do Dan's case, in which case the dimension is c = 2^Aleph_0 .
In general, the dimension is at least max(|k|, c) . I do not know if this is the right answer if k has cardinality strictly between Aleph_0 and c (in which case CH is false).
For x in k , consider the function n |--> x^n . These functions are linearly independent over k , so the dimension is at least |k| . On the other hand, the vector space V = Map(N, k) has cardinality at least c . If k is an infinite field, and B is a basis of V , then |V| = |B| |k| = max(|B|, |k|) . This shows that if |k| < c , then |B| >= c . If k is finite, then it's easy to see that |V| = c , and B must be infinite, in which case |V| = |B| , so the dimension is c .
(When k = R , the vector space has cardinality c , so a basis can't have cardinality larger than c .)
Michael Reid
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This is the wrong question. For Hilbert spaces (such as Hilbert space) there is a separate concept of Hilbert basis, whose cardinality is used to define the "Hilbert dimension" of such a space. (See http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension.) But a Hilbert space still has an underlying vector space structure and vector space dimension. At least for the standard Hilbert space H (take its incarnation as square-summable sequences of reals), the "Dedekind cut" method, that works to show that the vector space dimension of the direct product R^oo is c, works to show that the vector space dimension of H is also c. For, instead of the vectors {e_j} used in the proof for R^oo, use the vectors {e_j/j}. Again letting f: Z+ -> Q be a bijection to the rationals, let w_r denote the Hilbert space sum of all the e_j/j such that f(j) < r. (w_r is easily seen to be well-defined in H.) Then mimicking the argument for R^oo shows that the set {w_r | r in R} is a linearly independent set, so once again c = |{w_r | r in R}| <= dim(H) <= |H| = c, so dim(H) = c. Whereas, the Hilbert dimension of H is aleph_0. --Dan On 2012-08-27, at 12:49 PM, Fred lunnon wrote: << . . . . . . whether this definition of "dimension" leads to a sensible results in Hilbert space . . . . . . . . .
So the core paradox seems to reside in the following counterintuitive phenomenon, once (whatever) dimension of a vector space etc. becomes infinite. The cardinal K1 such that every vector is expressible as a linear combination of some K1 points, and the cardinal K2 such that every subset of K2+1 points is linearly dependent, are no longer necessarily equal. Is it obvious that (Hilbert dimension) K1 <= K2 (vector-space dimension) ? Regarding my earlier question concerning representation of a bounded linear operator in (separable) Hilbert space by an denumerably infinite matrix, the answer appears to be affirmative: see eg. http://www.math.psu.edu/yzheng/m597k/m597kLIII5.pdf WFL On 8/27/12, Dan Asimov <dasimov@earthlink.net> wrote:
This is the wrong question. For Hilbert spaces (such as Hilbert space) there is a separate concept of Hilbert basis, whose cardinality is used to define the "Hilbert dimension" of such a space. (See http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension.)
But a Hilbert space still has an underlying vector space structure and vector space dimension.
At least for the standard Hilbert space H (take its incarnation as square-summable sequences of reals), the "Dedekind cut" method, that works to show that the vector space dimension of the direct product R^oo is c, works to show that the vector space dimension of H is also c.
For, instead of the vectors {e_j} used in the proof for R^oo, use the vectors {e_j/j}. Again letting f: Z+ -> Q be a bijection to the rationals, let w_r denote the Hilbert space sum of all the e_j/j such that f(j) < r. (w_r is easily seen to be well-defined in H.) Then mimicking the argument for R^oo shows that the set {w_r | r in R} is a linearly independent set, so once again c = |{w_r | r in R}| <= dim(H) <= |H| = c, so dim(H) = c.
Whereas, the Hilbert dimension of H is aleph_0.
--Dan
On 2012-08-27, at 12:49 PM, Fred lunnon wrote:
<< . . . . . . whether this definition of "dimension" leads to a sensible results in Hilbert space . . . . . . . . .
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On second thoughts, I still don't have it quite nailed down: for a start, it's not clear that "linear dependence" can be assigned any meaning over a non-denumerable set. What I'm trying to do here is get away from the conventional but arbitrary restriction to linear sums over finite sets, which now appears as merely a consequence of the paradox elicited by Dan's original question. WFL On 8/28/12, Fred lunnon <fred.lunnon@gmail.com> wrote:
So the core paradox seems to reside in the following counterintuitive phenomenon, once (whatever) dimension of a vector space etc. becomes infinite. The cardinal K1 such that every vector is expressible as a linear combination of some K1 points, and the cardinal K2 such that every subset of K2+1 points is linearly dependent, are no longer necessarily equal.
Is it obvious that (Hilbert dimension) K1 <= K2 (vector-space dimension) ?
Regarding my earlier question concerning representation of a bounded linear operator in (separable) Hilbert space by an denumerably infinite matrix, the answer appears to be affirmative: see eg. http://www.math.psu.edu/yzheng/m597k/m597kLIII5.pdf
WFL
On 8/27/12, Dan Asimov <dasimov@earthlink.net> wrote:
This is the wrong question. For Hilbert spaces (such as Hilbert space) there is a separate concept of Hilbert basis, whose cardinality is used to define the "Hilbert dimension" of such a space. (See http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension.)
But a Hilbert space still has an underlying vector space structure and vector space dimension.
At least for the standard Hilbert space H (take its incarnation as square-summable sequences of reals), the "Dedekind cut" method, that works to show that the vector space dimension of the direct product R^oo is c, works to show that the vector space dimension of H is also c.
For, instead of the vectors {e_j} used in the proof for R^oo, use the vectors {e_j/j}. Again letting f: Z+ -> Q be a bijection to the rationals, let w_r denote the Hilbert space sum of all the e_j/j such that f(j) < r. (w_r is easily seen to be well-defined in H.) Then mimicking the argument for R^oo shows that the set {w_r | r in R} is a linearly independent set, so once again c = |{w_r | r in R}| <= dim(H) <= |H| = c, so dim(H) = c.
Whereas, the Hilbert dimension of H is aleph_0.
--Dan
On 2012-08-27, at 12:49 PM, Fred lunnon wrote:
<< . . . . . . whether this definition of "dimension" leads to a sensible results in Hilbert space . . . . . . . . .
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Mon, Aug 27, 2012 at 7:06 PM, Fred lunnon <fred.lunnon@gmail.com> wrote:
So the core paradox seems to reside in the following counterintuitive phenomenon, once (whatever) dimension of a vector space etc. becomes infinite. The cardinal K1 such that every vector is expressible as a linear combination of some K1 points, and the cardinal K2 such that every subset of K2+1 points is linearly dependent, are no longer necessarily equal.
This is just because you're stating things in a way that works for finite and not infinite sets. The K1 such that every vector is expressible as a linear combination of some K1 points, and the cardinal K2 such that every set of cardinality > K2 is linearly dependent, are equal. It's just that using K2+1 gives the same result as > K2 when K2 is finite, but not when it's infinite. This is all using vector space, not Hilbert space, definitions, where there is no topology, so no way to add more than finitely many vectors, and linear combination and linear dependence are both defined in terms of finite sums. The theorem also works for Hilbert basis, where you allow countable sums. The cardinal K1 such that every vectorr is expressible as a linear combination of countably many of some K1 points, and the cardinal K2 such that every subset of > K2 points is linearly dependent in the sense that there is a countable set of multiples of elements of the set, not all 0, that sums to 0. If you use finite sums to define a basis, and countable sums to define linear independence, or vice versa, you get different cardinals, but that's to be expected.
Is it obvious that (Hilbert dimension) K1 <= K2 (vector-space dimension) ?
Yes; if every vector is a finite sum of elements of S, it's also a countable sum of elements of S. As far as your later comment goes, I don't know of any useful definition of either basis or linear independence that makes use of sums of more than countably many vectors. I don't know of any notion of summability of uncountable sets that doesn't consider any sum with uncountably nonzero elements to diverge. Andy
participants (4)
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Andy Latto -
Dan Asimov -
Fred lunnon -
Michael Reid