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Continuum hypothesis and fractals

  1. Feb 25, 2016 #1
    Wikipedia: "The hypothesis states that the set of real numbers has minimal possible cardinality which is greater than the cardinality of the set of integers" ; i.e., let cardinality of integers = ℵ0 and cardinality of reals = ℵ; then there is no ℵ such that ℵ0 < ℵ < ℵ . But what about fractals? In his book The Fractal Geometry of Nature, Mandelbrot gives a formula for the fractal/Hausdorff dimension of Cantor's set - it's DF = log 2/log 3 ≈ 0.6309 - as opposed to the topological dimension DT = 1; clearly, DF < DT, and, if you see one of the pictures in the book, it's kinda intuitive that "there's less stuff" in Cantor's set than in, say, the [0; 1] interval - whose cardinality is also ℵ -, so whatever DF is, it seems to be more descriptive of "size" than DT (actually, Mandelbrot mentions there are technicalities regarding these definitions of dimension, which I'm completely ignorant of). Now, I'll naively assume that the cardinality of ℝ2 is (ℵ)2; for ℝ2, I believe DF = DT = 2, but, if I (also naivelly) take the exponent to be (numerically equal to) the fractal rather than the topological dimension, I'd have Cantor's set cardinality ℵC = (ℵ)0.6309 < (ℵ)1 . Sooo, if CH is true, what is wrong in my intuition? The technicalities previously mentioned? Or, say, can you prove that there's a bijection between a fractal and ℝDT? (I don't think there's one, self-similarity and "ruggedness" kinda exclude continuity in my POV). (P.S.: I'm prejudiced; regardless of ZFC or any other scheme, I think CH is false, even tho' I ain't no mathematician, and that may be an issue here; just so you know ;P )
     
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  3. Feb 25, 2016 #2

    jedishrfu

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    Can you provide the wikipedia reference you mentioned?
     
  4. Feb 25, 2016 #3
  5. Feb 25, 2016 #4
    Uh, sorry, I think I glossed over: when I said that for Cantor's dust DT = 1, I actually meant that that's the initiator of the fractal's - i.e., [0; 1] - DT; for the dust, DT = 0, as it is basically a bunch of points. The extrapolation made assumes that, given (DF)fractal < (DT)initiator , the cardinalities of these two sets oughtta follow the same trend.
     
  6. Feb 25, 2016 #5

    rubi

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    I don't know about fractals, but you can check that ##|C|=|\mathbb R|## by noting that ##|\mathbb R| \leq 2^{\aleph_0}## (for instance because of the Dedekind cut construction) and since every element of ##C## can be written as ##\sum \frac{a_n}{3^n}## with ##a_n\in\{0,2\}##, we have ##|C|=2^{\aleph_0}##. Since ##C\subseteq\mathbb R##, we have ##|C|=|\mathbb R|##.
     
  7. Feb 25, 2016 #6

    mathman

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    In general dimensionality (as long as it is finite) and cardinality have no relationship.
     
  8. Feb 25, 2016 #7
    Thanks for the rep; though I think I get the essential message here, it still feels odd to state that the Cantor set C is same cardinality than ℝ , not by your construction but rather concept: while, in your notation, |[0; 1]| = |ℝ|, if we "subtract" from [0; 1] all reals but, say, those of the form y = 0.xx...(finite # x's)...x000... , we'd have |S = {y}| = |ℤ|; if CH is correct, then there is no "smooth" way to go between these two extremes, which is uncomfortable. You can access any real in [0; 1] by pratically any number of appropriate limiting procedures, but the ones that lead to C are more restricted by the construction of the set itself, so it might be wrong but at least feels sensible enough that |ℤ| = |S| < |C| < |ℝ| by furthering the definition of |set| ; for instance, it sure feels sensible that S ⊆ ℝ and |S| = |ℝ| ; but, is it enough? Does one always imply the other, no matter the structure, topology, continuity (or lack of), etc., of any arbitrary S? Maybe I'm only talkin' BS here, but try addressing this more down-to-earth thing: find a one-to-one mapping between C and ℝ. Maybe it's even trivial, but I can't quite see it.
     
    Last edited: Feb 25, 2016
  9. Feb 25, 2016 #8
    Hmm... where could I read about that, is it a general result proven by someone or something? In any case, like I said in the previous rep., I don't see an impossibility in making dimensionality and cardinality have to do with each other - one simply generalizes the definition of the latter, like one generalizes pretty much anything in math according to one's own whims/inclinations, though exactly how, it seems, would be matter for speculation and beyond me.

    On a side note: if dimension has nothing to do with cardinality, then |ℝ2| = |ℝ|2 is either wrong or accidental; (basic question) how would I figure |ℝ2| without using |ℝ| ? I naively suppose |ℝD| = |ℝ|D , D the topological dimension, is what one invokes, that's why I drew an analogy.
     
    Last edited: Feb 25, 2016
  10. Feb 25, 2016 #9

    rubi

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    Some things in math feel odd, but they are true nevertheless.

    Well, my proof already establishes that one exists, but here we go: Every real number ##x\in(0,1)## can be written as ##x=\sum \frac{a_n}{2^n}## with ##a_n\in\{0,1\}## (binary expansion). Some numbers have 2 such expansions, for example ##(0.100000\ldots)_2 = (0.011111\ldots)_2##. In that case, always choose the non-periodic one. Define ##f:(0,1)\rightarrow C##, ##f(\sum \frac{a_n}{2^n}) = \sum \frac{2a_n}{3^n}##. This is a bijection between ##(0,1)## and ##C##. Then ##f(\frac{1}{\pi}\mathrm{arctan}(x)+\frac{1}{2})## provides a bijection between ##\mathbb R## and ##C##.

    Edit: Actually this is only an injection ##\mathbb R\rightarrow C##, so ##C## might even have more elements than ##\mathbb R##. You get a bijection by defining another injection ##g:C\rightarrow\mathbb R##, ##g(x)=x## and applying Cantor-Bernstein-Schröder. There is an explicit construction given in the "Book of proof".
     
    Last edited: Feb 25, 2016
  11. Feb 25, 2016 #10

    FactChecker

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    The mapping between [0,1] and [0,1]x[0,1] where .d1d2d3d4d5d6.... <=> (.d1d3d5d7d9d11...., .d2d4d6d8d10d12....) shows that cardinality does not increase when going from 1 dimension to 2. So |ℝ2| = |ℝ|. And clearly you can do something similar for any finite number of dimensions.
     
  12. Feb 26, 2016 #11
    Fair enough; it also further shows me how uncountable sets are counterintuitive: say you had n evenly spaced integers in the real line, call it V; clearly, |V X V| = |V|2 , yet this kind of reasoning fails for |ℝ X ℝ| . I suspect this kinda thing is more important than most physicists make it be: for instance, there are non-perturbative calculations in QCD done over lattices, rather than the whole 4D spacetime. Stuff has been derived in this regime and compared to real life, but I can't but feel uneasy about that because you're throwing away a lot of stuff in the process (maybe one can justify it is irrelevant stuff, but the whole business looks arbitrary). No matter, I digress.
     
  13. Feb 26, 2016 #12
    OK, the first sentence is a weird statement (as it sounds eerily Banach-Tarskiesque); I'd expect the possibility of |ℝ| ≥ |C|, but not that |ℝ| ≤ |C| ! Also, your construction in base 2 was a bit strange to me; suppose we expand all reals in I = [0; 1] in, say, decimals - that includes the elements of C , as well - ; since these expansions are unique (right?), a starting point to the mapping I → C would be simply the identity Id: I → I , which is clearly a bijection; but, since we explicitly ditch some numbers (like 0.5) in the construction of C , restricting the image of Id to C would not give a bijection unless you also restrict the domain, which would be saying explicitly that |I| > |C| ; but I don't know, even if you can't reason this way one could possibly cook up a map M that is a bijection - which you did.
    Anyway, I'll check out the explicit construction, provided you give me more details of the ref. you cited.
     
  14. Feb 26, 2016 #13
    It also fails for countably infnite sets - or do you think that there are more rational numbers (pairs of integers) than there are integers?
     
  15. Feb 26, 2016 #14
    The usual fractal dimension — Hausdorff dimension — of a fractal set in Euclidean space Rn has little to do with its cardinality (which is what the continuum Hypothesis is about). Any subset of Rn with positive Hausdorff dimension must have its cardinality greater than aleph0. (Note: I've edited this last sentence to correct an earlier mistake.)

    One very nice way to think of the Cantor set is that it is topologically equivalent to the countable direct product of a 2-element set with itself:

    K ≈ {0, 1}aleph0.​

    This provides a mapping from the Cantor set onto the real numbers between 0 and 1, by mapping a point (ε1, ε2, ε3, ...) of the Cantor set as above to the binary number having the same sequence of bits. It's easy to show that by ignoring binary expansions ending with all 1's (i.e., of form .???.......1111111111...) — which form a merely countable set — this can be made into a bijection. Which shows that

    card(K) = 2aleph0.​

    This can also be seen more directly from the rules about taking powers of cardinalities.

    Maybe the easiest way to see this, however, is that K as above is immediately in bijective correspondence with the set of all subsets of the positive integers. (If εj = 1, we include j in the subset; otherwise not.) The cardinality of the set of all subsets of any set X (the power set of X) is by definition 2card(X). And we know from Cantor's theorem that every power set has greater cardinality than the original set, so in particular:

    2aleph0 > aleph0.
     
    Last edited: Feb 26, 2016
  16. Feb 26, 2016 #15

    FactChecker

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    This may be surprising, but once you realize how it works, it can become fairly intuitive. Even countably infinite sets have the property. If you have a 2-dimensional grid of NxN, you can clearly count them all in order: (0,0), (1,0),(1,1),(0,1), (2,0),(2,1),(2,2),(0,2),(1,2), (3,0),(3,1),(3,2),(3,3),(0,3,),(1,3),(2,3), .....
    So |NxN| = |N|. You can do something similar for any finite number of dimensions.
     
  17. Feb 26, 2016 #16
    I think that's what I said.
     
  18. Feb 29, 2016 #17
    I'm Reading Manfred Schroeder's "Notes from an Infinite Paradise" (so slowly) and really struggling to get my head around Cantor Dust and continuing fractions - to me there is an intuitive connection between this puzzle and all the when-does-the-wave-not-wave-collapse or not debate as well as the tension between discrete quantum space-time geometry and observer independent continuum.

    Not sure if that is sensible or not... which is why I'm struggling. But your question definitely captures my imagination

    So the Cantor Space/Set...
    Is Totally Disconnected:

    • It can only be represented as the union of disjoint sets, a discrete space.
    Is Nowhere Dense:
    • It's closure has empty interior, all points are boundary points. The reals on the other hand are not nowhere dense so differential calculus is legal.
    Is totally Perfect:
    • It is a closed set with no isolated points. So how does nowhere dense work if no points are isolated? And if there are no isolated points why isn't differentiation of a Cantor Space legal (or is it)?
    Is Compact:
    • It Contains all its limit points. How is that possible if it is discrete? Doesn't discrete mean that some point at the shrinking limit of difference between some other two points is - undefined? In other words where are the points at the limit of between the disjoint points that are totally disconnected? They can't be in the set can they?
    Is a Metric Space:
    • Distances of all its points are defined. This one really breaks my head. How do you tell two self-similar subsets or points in Cantor Space apart? What's the distance between them if they are self-similar? Euclidean distance would have to be zero right? Is the index of iteration somehow included in the notion of measure? Schroeder does the example of shift of decimal representation of ternary numbers to build the "Devil's Staircase". His point I thought is that the representation literally repeats. But If you are counting the number of shifts then it's not really a repetition per se' is it? It's just a similar-looking event n shifts away from the other? Also, what kind of metric space is not differentiable!?
    ?:) !

    Million questions. Sorry.
    It is a really bizarre combination of perfect-compact-metric-cardinality and infinitely-recursive-non-linear-diffusion(or evolution) - which just sounds so much like all the Quantum Gravity puzzles.

    Also, I blame Lee Smolin, Per Bak, and Steven Strogatz for my combination of confusion and enthusiasm.
    http://arxiv.org/pdf/1506.02938v1.pdf See "Real Ensemble Hypothesis" in this one.

    Also I took all the definitions and explanations of the properties of Cantor Set/Space from Wiki (after reading Schroeder and others though). That said I very likely botched them in simple ways.
     
  19. Feb 29, 2016 #18

    WWGD

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    Compactness: By Zinq's description as the countable product of the compact set {0,1} with itself (though products of any cardinality will be compact, by Tychonoff's theorem) and

    Metrizable: Countable product of metrizable spaces is metrizable, though an uncountable product is not -- 1st countability is lost.

    I don't mean to be glib about this; you can see the general proofs and adapt them to the case
    of the Cantor set.
    Sorry, I don't have time to look at the others now.
     
    Last edited: Feb 29, 2016
  20. Mar 1, 2016 #19

    Samy_A

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    I'll address this one.
    The Cantor set is a subset of ##\mathbb R##. The distance between two points is just the usual distance between two real numbers.
     
  21. Mar 1, 2016 #20
    "It can only be represented as the union of disjoint sets, a discrete space."

    These describe two separate qualities that a space can have. Actually, every nonempty set is the union of disjoint sets.

    But the Cantor set is not a "discrete" space, which means that each point has a neighborhood with no other points in it.

    Rather, the Cantor set is totally disconnected, meaning that any two points have disjoint neighborhoods. This is true also of the set of rational numbers, but they aren't a discrete set, either.
     
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