bonanova 85 Posted October 16, 2012 Report Share Posted October 16, 2012 Consider the unit cube x, y, z = [0,1]. Does it contain more points than its edge x = [0,1], y=z=0? Edit. Let's state it simply: Does a cube contain more points than are contained in one of its edges? Quote Link to post Share on other sites

0 Solution bushindo 14 Posted October 17, 2012 Solution Report Share Posted October 17, 2012 My take on this There are the 'same' number of points on the cube as there are on the edge. That is because we can construct a 1-to-1 mapping between any point on the cube and the corresponding point on the edge. Let (x,y,z) be the 3-dimensional coordinate of any point in the cube (surface included) where x, y, z = [0,1]. In decimal form, let's write x as x = x_{0}.x_{1}x_{2}x_{3}x_{4}x_{5}... where x_{0} = 0 or 1 is the number before the decimal point, and x_{i} = 0, 1, ..., or 9 for i > 0. (Of course, if x_{0} = 1, then x_{i} must all be equal to 0) Similarly, we can write y and z in the same decimal form, y = y_{0}.y_{1}y_{2}y_{3}y_{4}y_{5}... z = z_{0}.z_{1}z_{2}z_{3}z_{4}z_{5}... Now, we can establish a 1-to-1 relationship between any point (x,y,z) on the cube to a point e on line segment [0, .111 ]. To do that, we can simply interleave digits and construct e from (x,y,z) as follows e = .x_{0}y_{0}z_{0}x_{1}y_{1}z_{1}x_{2}y_{2}z_{2}... It is easy to see that we can reverse the process and recover a 3-D coordinate in the cube (x,y,z) given a point e. Now that we have a strict 1-to-1 matching between points on a cube to the segment [0, .111], we can do a strict 1-to-1 matching between any point e in [0, .111] and any point s in [0, 1] using a linear function s = e * (1/.111 ) 1 Quote Link to post Share on other sites

0 Rob_Gandy 2 Posted October 16, 2012 Report Share Posted October 16, 2012 yes. Even though both are infinite I would say the number of points in the cube is larger. Similar to how I would say the number of integers is larger than the number of natural numbers. Quote Link to post Share on other sites

0 mmiguel 1 Posted October 17, 2012 Report Share Posted October 17, 2012 Consider the unit cube x, y, z = [0,1]. Does it contain more points than its edge x = [0,1], y=z=0? Edit. Let's state it simply: Does a cube contain more points than are contained in one of its edges? yes, though both are uncountably infinite sets of points, an edge of a cube is a subset of the cube itself, so the cardinality of the edge is less than or equal to the cardinality of the cube. it is not equal, because there exist points which belong to the cube, but do not belong to the edge. therefore, the only remaining option is that the cardinality of the edge is less than the cardinality of the cube. Quote Link to post Share on other sites

0 EventHorizon 16 Posted October 17, 2012 Report Share Posted October 17, 2012 First lets look at the degree of infinity of the number of points in each. The number of points in the edge has the same degree of infinity as the same as the number of real numbers, which is 2^{Aleph}_{0}. If you look at where that 2 came from, it really is just any finite number greater than 1. You can see this easily by noting that n*Aleph_{0}=Aleph_{0}, where n is any positive finite number. If you assume the number of points on an edge was x, then the number of points in the cube would be x^{3}. Substituting 2^{Aleph}_{0} for x gives, (2^{Aleph}_{0})^{3} = 2^{3*Aleph}_{0} = 2^{Aleph}_{0}. So the degree of infinity of the number of points in the edge are the same as that of the number of points in the cube. The cube contains 12 edges, so it certainly contains more points... right? That's even ignoring the internal points and points on the faces of the cube! The degree of infinity is the same, but there are surely points more in the cube because it contains the edge and more. (infinite copies of the edge even! You can let y and z be any constant value in [0,1] to get another copy) So the answer is yes if only because every point in the edge is in the cube and not vice versa. (e.g., a set contains more elements than any strict subset of itself) Quote Link to post Share on other sites

0 bonanova 85 Posted October 17, 2012 Author Report Share Posted October 17, 2012 mmiguel, would you say there are different cardinal numbers for one edge, 12 edges, faces, solids? BTW just to clarify, I mean by cube to denote a solid, as per EH's reasoning. Any other views? Quote Link to post Share on other sites

0 vinay.singh84 0 Posted October 17, 2012 Report Share Posted October 17, 2012 To solve this, one needs cardinal numbers. The edge is an uncountable set of points with specific cardinality of Aleph_{1}. This can be shown by creating a simple bijection y=tan(pi*(x-0.5)) of (0,1) to R. Similarly, each plan contains an uncountable number of edges and the cube and an uncountable number of planes. Thus, the cardinality of the set of points in the cube == Aleph_{1}*Aleph_{1}*Aleph_{1} == Aleph_{1} So the cube contains the same number of points as any of its edges. Quote Link to post Share on other sites

0 bonanova 85 Posted October 17, 2012 Author Report Share Posted October 17, 2012 And has the title become germane? Quote Link to post Share on other sites

0 mmiguel 1 Posted October 18, 2012 Report Share Posted October 18, 2012 (edited) mmiguel, would you say there are different cardinal numbers for one edge, 12 edges, faces, solids? BTW just to clarify, I mean by cube to denote a solid, as per EH's reasoning. Any other views? i was interpreting the OP to mean solid. i did a little research, and it seems that by the standard mathematical definition of cardinality, it seems that just because an infinite set A is a superset of an infinite set B, this does not mean that |A| > |B|. the concept that the superset operator translates into a "greater-than" inequality when going from sets to the cardinalities of those sets is true for finite sets, as is the bijective mapping view of cardinality that mathematicians have chosen to stick to when generalizing from finite sets to infinite sets. under this generalization, the two concepts are no longer equivalent. this is something i may have read a long time ago, but did not recall when providing my answer. http://en.wikipedia....iki/Cardinality "Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets and proper supersets of an infinite setS that have the same size as S, although S contains elements that do not belong to its subsets, and the supersets of S contain elements that are not included in it." so what this means is that a solid cube has the same cardinality as an edge of the cube because a bijective mapping can be formed between the two sets. i wonder, is there a standard definition for the infinite-set generalization of the subset/superset interpretation of cardinality from the finite case? perhaps such a definition was deemed to be not useful... thinking over this, it is all so very strange: the cube contains all the points that the edge contains, and then some.... yet each point in the cube can be paired with a point in the edge, with no points in the edge being left unpaired... if these are both true, than this must be a contradiction (mustn't it?). if one is false, then why is it false? i think it must be a contradiction if both are true because suppose we set about pairing points in the cube to the edge. let's say we restrict ourselves to first pair all the points in the edge of the cube before pairing any other points of the cube. since the cube contains the edge, we could pair each point of the edge from the cube set perspective, to the same point in the edge set perspective. each point in the edge gets paired with itself. thus there are no longer any points in the edge that are unpaired. but there are points not on the edge, which still are part of the cube which are not paired to any points within the edge, and cannot be, since all the points in the edge set are already paired (with themselves from the cube set perspective). hence no bijective mapping can be formed if restrict ourselves to pairing the points within the edge first in the cube set perspective.... perhaps someone can enlighten me? Edited October 18, 2012 by mmiguel Quote Link to post Share on other sites

0 mmiguel 1 Posted October 18, 2012 Report Share Posted October 18, 2012 (edited) sorry, thought a little more, wanted to write a little more is being able to form a 1-1 mapping really equivalent to having the same number of things? obviously yes for the finite case in the infinite case, it seems to me that this definition of cardinality is more like: starting from the cube set or the edge set, there are always enough available point resources in the other set to support a mapping no matter how many points you ask for a partner for (i.e. there exists an injective mapping). this doesn't to me seem to mean that when you have paired all the points in your starting set, that there are no longer any point resources in the other set left (and are therefore unpartnered) i.e. does not imply that the mapping is surjective. to prove surjectivity, for example in bushindo's case, you invert the algorithm and show that it is injective when starting from the other set as well. the crux of the counter-intuitiveness is that one real number, can contain the information of three real numbers, as bushindo showed. what if there is an infinite amount of information? (by information i mean the ordered set of decimal digits making up a number) ----> what if x=y=z=pi or some other transcendental number? i guess e would be another transcendental number, but everything would still work. but this almost seems like it's cheating. logically what we are doing is mapping one number to three numbers using a clever encoding scheme. any real number can be thought of as an infinite ordered set of natural numbers (the digits of the number). by saying we can construct the number e, from numbers x,y, and z, we are implicitly assuming that e has the capacity to store x, y, and z. this assumption is logically equivalent to the conclusion that a single edge has the capacity to provide partner points to every point in a cube. doesn't that make assuming that e can be constructed from x,y, and z, to prove that edge and cube cardinalities are the same, a circular fallacy? if we are not allowed to assume that e has the capacity to completely store x, y, and z, but rather the capacity to store only one of them, then I think we would conclude the cardinality of the cube is greater than the edge. so how do we prove that e can be constructed for any real x,y,z? perhaps the argument is, I can try to construct it and I know what I would do at every step, so therefore it is constructable? but if x,y, or z is transcendental, then such a construction would never end.... well at least it couldn't be described in terms of a simple repetitive instruction loop (e.g. the rest of the digits are all zero, or the rest of the digits are 142857 repeating). i suppose it's always an infinite construction even in the simplest of cases. perhaps the concept that a single real number can store the information of an arbitrary number of other real numbers is an axiom rather than something proven from simpler axioms. if that is the case, then the original question is essentially asking whether we choose to adopt this axiom or not. anyone care to offer some help? Edited October 18, 2012 by mmiguel Quote Link to post Share on other sites

0 mmiguel 1 Posted October 18, 2012 Report Share Posted October 18, 2012 (edited) choice It is an axiom - the axiom of choice (well... maybe it is)...! Here is my attempt at connecting the two ideas. http://en.wikipedia....Axiom_of_choice Here is a variant: "Given any set X of pairwise disjoint non-empty sets, there exists at least one set C that contains exactly one element in common with each of the sets in X." Treating a triplet of digits at positions (k,k+1,k+2) for k = floor(n/3)+1 (where the 10^0 decimal is at position 0) in bushindo's e as the C above and X as the set containing the nth digits of x,y, and z, the axiom of choice guarantees bushindo's construction exists. I'm not sure how to prove that if the axiom of choice is not taken, that bushindo's construction cannot certainly be constructed... so maybe i am wrong here. i am assuming ordered sets (tuples), so that is slightly different than the statement of the axioms of choice i have. but if i am right, then the title of this thread is actually relevant http://en.wikipedia....ki/Ordered_pair "In 1921 Kazimierz Kuratowski offered the now-accepted definition^{[4]} of the ordered pair (a, b): " let x be the set of ordered pairs (v,n) where n is an integer indicating the position of a digit, and v is the digit itself. an ordered triplet (a,b,c) can be defined as (a,(b,c)) = {{a},{a,(b,c)}} = {{a},{a,{{b},{b,c}}}} so given n, to construct the triplet C_n = (x_n,y_n,z_n) = {{x_n},{x_n,{{y_n},{y_n,z_n}}}}, we must be able to select: {{x_n},{x_n,n}} from x {{y_n},{y_n,n}} from y {{z_n},{z_n,n}} from z i.e. elements from x,y,z where the second value in the ordered pair equals n. by the way we have defined things, such an element must always exist for x,y, and z and any n, so i don't see why we wouldn't be able to select it. the only time i have seen (in my limited experience) where the axiom of choice actually plays a useful role is when there is some sort of indistinguishability between multiple things, and the axiom of choice kicks in to say, even though these things are indistinguishable, you can still pick one. by the way we defined this, i don't see any indistinguishability. we could even guarantee that x,y, and z distinguishable, by making them ordered pairs themselves and putting some identifier (e.g. 1 for x, 2 for y, 3 for z) as the first element of the ordered pair, and then previously described countably infinite set of natural numbers as the second element of the pair. so maybe the axiom of choice doesn't apply here.... on the other hand... the AC is equivalent to the well-ordering theorem in first order logic, and perhaps we somehow implicitly assumed that in our construction of these real numbers. i think i've probably reached the limit in what i know about this topic (as well as the limit of the time i have to spend on this tonight), so hopefully someone else can step in and show the light. Edited October 18, 2012 by mmiguel Quote Link to post Share on other sites

0 EventHorizon 16 Posted October 18, 2012 Report Share Posted October 18, 2012 Let Z be the set of all integers and Q be the rationals. We know that |Z| = |Q| and that |Q|-|Z| is undefined since |Q| and |Z| are both infinity. But isn't it true that Q contains more than just integers? Since Q is a superset of Z and the extra elements of the set Q are still numbers, can't it be said Q contains more numbers than the numbers contained in the set Z... even though the cardinality of the sets are equal? In other words the number of elements in each is the same, and yet one set contains more elements in addition to those contained in the other. Let C be the set of points in the cube. Let E be the set of points in the edge. We know |C| = |E|. I've shown they're the same degree of infinity and others have given one-to-one mappings. So we know the number of points in each is the same. But does that necessarily mean one doesn't contain more points than are contained in the other? We know |C| - |E| is undefined since both |C| and |E| are infinity. However, |C-E| is infinity. So if we wait to abstract away the actual sets, we can have an answer that isn't just undefined. So I guess what I'm saying is that if the question was "is the number of points in the cube greater than the number of points in the edge", the answer would be no since both are infinity. But if the question isn't outright asking about the cardinalities and comparing their values, it could be different. If it is only asking about the points not contained in the edge, then the answer is yes. We have the sets and know their difference. So we know what points are contained in the cube in addition to those contained in the specified edge. So just like I would say Q contains more than just the numbers contained in Z (even though they both contain the same number of numbers), the cube contains more points in addition to those contained in the edge. So, yes, we all agree the cardinality is the same. The question now is whether the original question is asking "|C|>|E|" or "|C-E|>0?" "Does the cube contain a greater quantity of points than the quantity contained in one of its edges?" - Answer: No. "Does the cube contain more points than the quantity of points contained in one of its edges?" - Answer: No. "Does the cube contain more points than just those points contained in one of its edges?" - Answer: Yes. "Does the cube contain extra points in addition to those contained in one of its edges?" - Answer: Yes. "Does a cube contain more points than those contained in one of its edges?" - I'd still answer yes. Actual question: "Does a cube contain more points than are contained in one of its edges?" It does sound slightly vague to me as to which it is asking, but I'm still leaning towards the latter interpretation (ie, answer is yes). I'm sure if I ask bonanova which interpretation he meant... he'll say "I can't or I won't say." Quote Link to post Share on other sites

0 bonanova 85 Posted October 19, 2012 Author Report Share Posted October 19, 2012 Let Z be the set of all integers and Q be the rationals. We know that |Z| = |Q| and that |Q|-|Z| is undefined since |Q| and |Z| are both infinity. But isn't it true that Q contains more than just integers? Since Q is a superset of Z and the extra elements of the set Q are still numbers, can't it be said Q contains more numbers than the numbers contained in the set Z... even though the cardinality of the sets are equal? In other words the number of elements in each is the same, and yet one set contains more elements in addition to those contained in the other. Let C be the set of points in the cube. Let E be the set of points in the edge. We know |C| = |E|. I've shown they're the same degree of infinity and others have given one-to-one mappings. So we know the number of points in each is the same. But does that necessarily mean one doesn't contain more points than are contained in the other? We know |C| - |E| is undefined since both |C| and |E| are infinity. However, |C-E| is infinity. So if we wait to abstract away the actual sets, we can have an answer that isn't just undefined. So I guess what I'm saying is that if the question was "is the number of points in the cube greater than the number of points in the edge", the answer would be no since both are infinity. But if the question isn't outright asking about the cardinalities and comparing their values, it could be different. If it is only asking about the points not contained in the edge, then the answer is yes. We have the sets and know their difference. So we know what points are contained in the cube in addition to those contained in the specified edge. So just like I would say Q contains more than just the numbers contained in Z (even though they both contain the same number of numbers), the cube contains more points in addition to those contained in the edge. So, yes, we all agree the cardinality is the same. The question now is whether the original question is asking "|C|>|E|" or "|C-E|>0?" "Does the cube contain a greater quantity of points than the quantity contained in one of its edges?" - Answer: No. "Does the cube contain more points than the quantity of points contained in one of its edges?" - Answer: No. "Does the cube contain more points than just those points contained in one of its edges?" - Answer: Yes. "Does the cube contain extra points in addition to those contained in one of its edges?" - Answer: Yes. "Does a cube contain more points than those contained in one of its edges?" - I'd still answer yes. Actual question: "Does a cube contain more points than are contained in one of its edges?" It does sound slightly vague to me as to which it is asking, but I'm still leaning towards the latter interpretation (ie, answer is yes). I'm sure if I ask bonanova which interpretation he meant... he'll say "I can't or I won't say." If extra, or better, other, were used in OP, the answer would be Yes. But with more, as in OP, the answer is No. But it's best to say the points in the cube and its edge have the same cardinality. That gets us off the hook a little, because cardinality is a well behaved algebraic number only for finite sets. For infinite sets, Aleph0, Aleph1, Aleph2, are symbols that are not like algebraic symbols. Specifically, they are invariant to addition, multiplication, squaring, cubing ... Aleph1 == Aleph1 + 1 == Aleph1 x Aleph1 == ... And so forth. That's why asking questions about the Alephs that include words like "more than" lead to counterintuitive results. But 2^{Aleph1} == Aleph2. That is, the set of all subsets [power set] of an infinite set jumps up to the next higher cardinality. The proof that the elements of any set can not be paired 1-1 with the elements of its power set is both simple and elegant. Cantor [can't or] originally thought Aleph1 x Aleph1 would lead to Aleph2. That is, the points in a plane would have next higher cardinality than the points in a line. And a cube would have the next higher cardinality than the plane. And so on. But when he tried to show this, he instead proved all three cardinalities were the same. Much to his surprise, and now to ours. So the point of this puzzle was to pause and wonder at the result that adding and multiplying, squaring or cubing do not affect the cardinality of infinite sets. And that result can be shown, as Bushindo did, by finding a bijection between the cube and an edge. Quote Link to post Share on other sites

0 mmiguel 1 Posted October 19, 2012 Report Share Posted October 19, 2012 If extra, or better, other, were used in OP, the answer would be Yes. But with more, as in OP, the answer is No. But it's best to say the points in the cube and its edge have the same cardinality. That gets us off the hook a little, because cardinality is a well behaved algebraic number only for finite sets. For infinite sets, Aleph0, Aleph1, Aleph2, are symbols that are not like algebraic symbols. Specifically, they are invariant to addition, multiplication, squaring, cubing ... Aleph1 == Aleph1 + 1 == Aleph1 x Aleph1 == ... And so forth. That's why asking questions about the Alephs that include words like "more than" lead to counterintuitive results. But 2^{Aleph1} == Aleph2. That is, the set of all subsets [power set] of an infinite set jumps up to the next higher cardinality. The proof that the elements of any set can not be paired 1-1 with the elements of its power set is both simple and elegant. Cantor [can't or] originally thought Aleph1 x Aleph1 would lead to Aleph2. That is, the points in a plane would have next higher cardinality than the points in a line. And a cube would have the next higher cardinality than the plane. And so on. But when he tried to show this, he instead proved all three cardinalities were the same. Much to his surprise, and now to ours. So the point of this puzzle was to pause and wonder at the result that adding and multiplying, squaring or cubing do not affect the cardinality of infinite sets. And that result can be shown, as Bushindo did, by finding a bijection between the cube and an edge. using bushindo's method, we can represent n real numbers (a vector) using a single real number for any finite n. each element of such a vector is also a real number, and can therefore represent another vector of real numbers. following along these lines, any real tensor can be uniquely mapped to a single real number. a single real number, therefore has the same level of complexity as a 100 by 100 by 100 tensor of arbitrary real numbers (i.e. they can be used interchangeably to communicate the same thing). practically any set of data one could ever want to use or interact with, regardless of how large, could be encoded as a single numerical position on a number line (with higher precision required for more info). forget about achilles passing the tortoise... what about when he passes the point on the racetrack whose real position contains all of the data in his genetic makeup under an encoding scheme like the one bushindo showed? that is far more interesting! Quote Link to post Share on other sites

0 EventHorizon 16 Posted October 20, 2012 Report Share Posted October 20, 2012 If extra, or better, other, were used in OP, the answer would be Yes. But with more, as in OP, the answer is No. But it's best to say the points in the cube and its edge have the same cardinality. That gets us off the hook a little, because cardinality is a well behaved algebraic number only for finite sets. For infinite sets, Aleph0, Aleph1, Aleph2, are symbols that are not like algebraic symbols. Specifically, they are invariant to addition, multiplication, squaring, cubing ... Aleph1 == Aleph1 + 1 == Aleph1 x Aleph1 == ... And so forth. That's why asking questions about the Alephs that include words like "more than" lead to counterintuitive results. But 2^{Aleph1} == Aleph2. That is, the set of all subsets [power set] of an infinite set jumps up to the next higher cardinality. The proof that the elements of any set can not be paired 1-1 with the elements of its power set is both simple and elegant. Cantor [can't or] originally thought Aleph1 x Aleph1 would lead to Aleph2. That is, the points in a plane would have next higher cardinality than the points in a line. And a cube would have the next higher cardinality than the plane. And so on. But when he tried to show this, he instead proved all three cardinalities were the same. Much to his surprise, and now to ours. So the point of this puzzle was to pause and wonder at the result that adding and multiplying, squaring or cubing do not affect the cardinality of infinite sets. And that result can be shown, as Bushindo did, by finding a bijection between the cube and an edge. One point I omitted to state outright (but meant to include) was that English (and spoken language in general) is quite subjective. For instance, looking up "more" in the dictionary gives one definition as being "additional or further." Extra's definition sounds just like that ("beyond or more than what is usual, expected, or necessary; additional"). So I still think the question vague (if only because it is in English), and it seems I was mistaken when I thought it intentionally vague. Also, even though it is commonplace nowadays and I'm positive I've done it before, I intentionally didn't call the cardinality of the set of reals Aleph_{1} due to the generalized continuum hypothesis being unproven (unprovable with common set theory, and some have even said it is not well defined so it has no truth value). I did point out that (2^{Aleph}_{0})^{3} = 2^{(Aleph}_{0}^{*3)} = 2^{Aleph}_{0}. And hinted that, where n is a positive number, n^{Aleph}_{0} = (2^{log}_{2}^{ n})^{Aleph}_{0} = 2^{(Aleph}_{0}^{*}^{log}_{2}^{ n}^{)} = 2^{Aleph}_{0}. Since Aleph_{n} = Aleph_{n} * Aleph_{n} for all non-negative integer n, the same could be done for any Aleph_{n}. This means, assuming GCH, that Aleph_{n} raised to any non-zero finite number is Aleph_{n} and any positive finite number raised to Aleph_{n} is Aleph_{n+1}. What is Aleph_{n}^{Aleph}_{n}? Since x^{x} grows similarly to x!, is Aleph_{n}! the same as Aleph_{n}^{Aleph}_{n}? And is any of this math useful/valid/defined or does working with infinities mean you cannot use it so bijections are absolutely necessary? Quote Link to post Share on other sites

0 bonanova 85 Posted October 20, 2012 Author Report Share Posted October 20, 2012 One point I omitted to state outright (but meant to include) was that English (and spoken language in general) is quite subjective. For instance, looking up "more" in the dictionary gives one definition as being "additional or further." Extra's definition sounds just like that ("beyond or more than what is usual, expected, or necessary; additional"). So I still think the question vague (if only because it is in English), and it seems I was mistaken when I thought it intentionally vague. Also, even though it is commonplace nowadays and I'm positive I've done it before, I intentionally didn't call the cardinality of the set of reals Aleph_{1} due to the generalized continuum hypothesis being unproven (unprovable with common set theory, and some have even said it is not well defined so it has no truth value). I did point out that (2^{Aleph}_{0})^{3} = 2^{(Aleph}_{0}^{*3)} = 2^{Aleph}_{0}. And hinted that, where n is a positive number, n^{Aleph}_{0} = (2^{log}_{2}^{ n})^{Aleph}_{0} = 2^{(Aleph}_{0}^{*}^{log}_{2}^{ n}^{)} = 2^{Aleph}_{0}. Since Aleph_{n} = Aleph_{n} * Aleph_{n} for all non-negative integer n, the same could be done for any Aleph_{n}. This means, assuming GCH, that Aleph_{n} raised to any non-zero finite number is Aleph_{n} and any positive finite number raised to Aleph_{n} is Aleph_{n+1}. What is Aleph_{n}^{Aleph}_{n}? Since x^{x} grows similarly to x!, is Aleph_{n}! the same as Aleph_{n}^{Aleph}_{n}? And is any of this math useful/valid/defined or does working with infinities mean you cannot use it so bijections are absolutely necessary? Well we've finished the discussion of the puzzle so I won't spoiler this. You made several points and raised new questions. I'll try to reply to the extent that I know. Inarguably the cube contains points that its edge does not. For finite sets that gives a Yes answer. For infinite sets, and these sets are uncountable, we express size by cardinality. The edge has the cardinality of the reals, so the question really asks the cardinality of the cube. Bushindo's bijection expresses Cantor's finding that geometric figures of all dimensions, arbitrarily large ones, have the cardinality of the reals. So the cube has other points, but not more points. The power set is the set of all subsets. Is there a bijection between the elements of a set and the elements of its power set? The answer is No, and so the cardinality of the power set is greater than the cardinality of the set. For finite sets, it can be seen by inspection. The set {1,2,3} has 2^{3} = 8 subsets, including the empty set. No bijection exists. For infinite sets, the proof is almost as simple. It's a proof by contradiction. Suppose the elements of an infinite set have been paired with the elements of its power set. We then color the set's elements blue that are paired with a subset that contains the element, and red if they are not a member of their paired subset. Now consider the element of the set that is paired with the subset of all the red colored elements. Can it be red? No, because red elements are paired with subsets that do not contain them. Can it be blue? No, because blue elements are paired with subsets that do contain them. So the subset of all red elements cannot be paired, and this contradict the premise of a bijection. So power sets have increasing cardinalities. We know the integers are countably infinite, and the reals are not; and we have an understanding of both these sets. Do we have an understanding of a set that has cardinality Aleph2? One such set is the set of all functions of real numbers. Or of all curves in a plane. No one has suggested a set of cardinality Aleph3 other than the power set of the functions of a real number. What is Aleph_{n}^{Aleph}_{n}? I would say it lacks any recognizable similarity with a power set or a real number raised to the power Alpha_{n} because the infinite cardinals are not algebraic. So it's a symbol raised to its own power. What value or significance we could thus assign to it I do not know. Quote Link to post Share on other sites

0 EventHorizon 16 Posted October 20, 2012 Report Share Posted October 20, 2012 What things can I include in this post more than are contained in my previous posts? Notice I used more in the sense of other, extra, additional, etc. We agree, except I'm harping on the various definitions of more. Just wait until I start arguing over what the definition of is is... I've seen those examples of sets of various degrees of infinity, and have had to come up with various bijections including them. But it doesn't say whether bijections are always necessary or if math can be used instead. Perhaps in these cases the symbolic math is just shorthand for the bijections already given. But that seems to be the way with math in general... operators are defined by previously defined operators and/or algorithms including other previously defined operators. Aleph_{0}! = Aleph_{0} * (Aleph_{0}-1) * ... * 1. Since Aleph_{0} = Aleph_{0} * Aleph_{0}, the limit seems to be Aleph_{0}. But if you replace those all with 2's... it is equal to Aleph_{1}. Aleph_{0}^{Aleph}_{0} = x ... you'd expect it to be >= Aleph_{1} since 2^{Aleph}_{0}=Aleph_{1} (assuming GCH). log(Aleph_{0}^{Aleph}_{0}) = log(x) ...ignoring the question of what that means. Aleph_{0} * log(Aleph_{0}) = log(x) ...log seems to subtract 1 from the aleph number... what is Aleph_{-1}? Aleph_{0} = log(x) e^{Aleph}_{0} = x Aleph_{1} = x = Aleph_{0}^{Aleph}_{0} Since 2^{Aleph}_{0} <= Aleph_{0}! <= Aleph_{0}^{Aleph}_{0} and 2^{Aleph}_{0} = Aleph_{0}^{Aleph}_{0}, 2^{Aleph}_{0} = Aleph_{0}! = Aleph_{0}^{Aleph}_{0}. SUM_{i=1}^{Aleph}_{0} 0 = 0 ...additive identity SUM_{i=1}^{Aleph}_{0} 1 = Aleph_{0} ...could use anything greater than the additive identity. SUM_{i=1}^{Aleph}_{0} i = Aleph_{0} SUM_{i=1}^{Aleph}_{0} Aleph_{0} = Aleph_{0} ....since Aleph_{0} = Aleph_{0} * Aleph_{0} PROD_{i=1}^{Aleph}_{0} 1 = 1 ...multiplicative identity PROD_{i=1}^{Aleph}_{0} 2 = Aleph_{1} ...could use anything greater than the multiplicative identity. PROD_{i=1}^{Aleph}_{0} i = Aleph_{0}! = Aleph_{1} PROD_{i=1}^{Aleph}_{0} Aleph_{0} = Aleph_{0}^{Aleph}_{0} = Aleph_{1} Makes perfect non-sense to me .... Actually, if you have multiplication representing a cross product of sets (ie, all possible tuples... and the obvious/common/correct interpretation when applied to sets), then Aleph_{0}^{Aleph}_{0} is the set of infinite length tuples whose elements are any integer (wlog, i'll say positive integers... easier for the bijection). (Edit: could also interpret it as the set of all multisets of a countable set. Instead of included or not, it is how many are included.) If the integers that are the tuple elements were bounded, it would obviously be the same cardinality as the reals (represented in the base number of the bound). A bijection between the reals and this set would be the following. Let the reals be given in binary. Two 0's in a row is a sentry marking the end of the current segment. For a 1, list a '1'. For a 0, list a '01'. Each segment represents an integer in the infinite length tuple. All of the set of infinite tuples are covered and the reals between 0 and 1 are used to map onto (so '0.' whatever is needed for the matching infinite length tuple). The tuple is [1,2,3,4,5,6,7,...] The real is 0.(1)00(101)00(11)00(10101)00(1011)00(1101)00(111)00... with the integers from the tuple marked in parenthesis. So, assuming GCH, Aleph_{0}^{Aleph}_{0} = Aleph_{1} after all Quote Link to post Share on other sites

0 mmiguel 1 Posted October 21, 2012 Report Share Posted October 21, 2012 (edited) So the cube has other points, but not more points. I don't think we can say the the same about the edge with respect to the cube: The edge does not have other points than those contained in the cube. Do you agree? The subset operator would still show asymetry here. The less-than operator would not show asymmetry between the cardinalities of the two sets. Our intuition tells us, if A is a strict subset of B, then |A| < |B|. Our conclusion from this discussion is that this transformation, from a statement of sets, to a statement of cardinalities, does not hold when sets are infinite. Is there some other commonly accepted mathematical construct, e.g. denoted for set A by [A], that evaluates to a number, such that if A is a strict subset of B, then [A] < even for infinite sets? Edited October 21, 2012 by mmiguel Quote Link to post Share on other sites

0 mmiguel 1 Posted October 21, 2012 Report Share Posted October 21, 2012 I don't think we can say the the same about the edge with respect to the cube: The edge does not have other points than those contained in the cube. Do you agree? The subset operator would still show asymetry here. The less-than operator would not show asymmetry between the cardinalities of the two sets. Our intuition tells us, if A is a strict subset of B, then |A| < |B|. Our conclusion from this discussion is that this transformation, from a statement of sets, to a statement of cardinalities, does not hold when sets are infinite. Is there some other commonly accepted mathematical construct, e.g. denoted for set A by [A], that evaluates to a number, such that if A is a strict subset of B, then [A] < even for infinite sets? One more requirement for [A]: for finite sets, [A] = |A| Quote Link to post Share on other sites

0 mmiguel 1 Posted October 21, 2012 Report Share Posted October 21, 2012 One more requirement for [A]: for finite sets, [A] = |A| One more requirement: If A is the real interval [0,1], and B = [0,0.5) U (0.5,1] U {2} Then [A] = B is the same as A, except that 0.5 is removed, and 2 is added. B is no longer a subset of A, but from an intuitive perspective should have the same "quantity of points". Quote Link to post Share on other sites

0 EventHorizon 16 Posted October 22, 2012 Report Share Posted October 22, 2012 Is there some other commonly accepted mathematical construct, e.g. denoted for set A by [A], that evaluates to a number, such that if A is a strict subset of B, then [A] < even for infinite sets? One more requirement for [A]: for finite sets, [A] = |A| One more requirement: If A is the real interval [0,1], and B = [0,0.5) U (0.5,1] U {2} Then [A] = B is the same as A, except that 0.5 is removed, and 2 is added. B is no longer a subset of A, but from an intuitive perspective should have the same "quantity of points". For your example A and B, there's always set difference (eg, |A-B| = |B-A| ). For continuous segments of reals and n-tuples of reals there's length, area, volume, hyper-volume, and further n-spaces. But you'd still need to represent the missing lower order segments within. An option would be to have a tree structure whose child nodes are the missing or added segments of lower order. But this could make comparisons anywhere from difficult to meaningless. This would still have problems representing things that aren't extremely well-behaved and contrived. An example would be [0,1] - Q, where Q is the set of all rationals if Q was a set that wasn't a named and known set. It seems there are possibilities where you could have one set's cardinality less than another in the limit as some finite intersecting set grows. This would still be meaningless with sufficiently complex sets, and would be highly dependent on the intersecting set and the way it grows (as different choices could give drastically different results). This would also have problems due to the intersecting set being finite when both sets are uncountable degrees of infinity (may just compare a tiny infinitesimal portion). I'm fairly positive any such operator would only be useful with the most well-behaved and contrived of sets, and be easily 'fooled' and/or misleading with the rest... (sort of like = with cardinalities of infinite sets) A big part of the problem is that you can map (eg, bijection) all reals to a small section of the reals. You can't assign a different number as the "quantity" of the points within. So the "quantity" should be the same after abstracting away the sets. Such is the nature of uncountablility. Set operations seem to me like the only "good" way to compare them. Quote Link to post Share on other sites

## Question

## bonanova 85

Consider the unit cube x, y, z = [0,1].

Does it contain more points than its edge x = [0,1], y=z=0?

Edit. Let's state it simply:

Does a cube contain more points than are contained in one of its edges?

## Link to post

## Share on other sites

## 20 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.