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Is the cantor set diffeomorphic to the fat cantor set?

  1. Oct 17, 2011 #1
    Ok I originally posted this question in the hw section but it doesn't seem like anyone knows the answer there. Hopefully someone can help me out here!

    1. The problem statement, all variables and given/known data

    Ok so the first part of the problem was constructing fat cantor sets, cantor sets with some positive lesbegue measure. The second part involved proving that any fat cantor set is homeomorphic to the regular cantor set. The third part asks whether there is a diffeomorphism from [0,1] to itself such that a fat cantor set is mapped to the ternary cantor set

    3. The attempt at a solution

    I was able to get the first two parts. For constructing the homeomorphism it seemed simple when you look at the formation of the cantor set in an algorithmic way. At each stage lines are removed from the middle and the number of intervals left grows like 2^n (eventually the intervals shrink to points of course).

    What I did was for each stage was define a homeomorphism from the cantor set to the fat cantor set. This is obviously possible since we're just mapping lines segments to line segments. This defines a sequence of functions and as n->∞ we'll have a homeomorphism from a fat cantor set to the regular one

    For the diffeomorphism it appears more difficult. On the one hand it seems simple because both spaces are totally disconnected and are uncountably infinite so we could just map points to points and have diffeomorphisms fitting the missing intervals in between them correctly. On the other hand, the ternary cantor set shrinks much quicker than the fat cantor sets and it would appear that as n->∞ the map won't be diffeomorphic when we're mapping a point from one cantor set to another. The mapping of the intervals around these points seems to be too violent to be a diffeomorphism (derivative wouldn't be defined).

    Thanks for any help!
  2. jcsd
  3. Oct 17, 2011 #2


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    Here's a couple of ideas:

    i)Any compact Hausdorff perfect topological space with countable
    base which is totally disconnected is homeomorphic to the Cantor set C

    ii)When you do a calorically-enhanced , aka, thinness-disadvantaged Cantor set,
    you end with terms that can be expressed as being in some number base in ways
    that are helpful. The standard Cantor set can be seen as the set of numbers in base
    3 that have no 1's in the expansion. Try to express other Cantor sets as elements in
    base-n which do not have a certain element of the base-n in their expansion. This may
    help give you a homeo. between the two.
  4. Oct 17, 2011 #3
    Thanks for the idea. I knew about the theorem about perfect compact topological spaces. I didn't think of expanding into different bases. I think that would be useful if I was say, throwing out the middle fifth or sixth, which I originally did but theres ways to make fat cantor sets without considering proportions. I could always define some converging geometric sum instead.

    Actually, I dont know if I mentioned this, but the fat cantor sets have to have positive measure. If i threw out the middle fourth or fifth or whatever I could work in a different base to prove a homeomorphism, but I think the set will still have measure 0.

    Honestly I'm more confused about the idea about the diffeomorphism. The idea of defining homeomorphisms at each stage of the algorithm seems to be the best idea (since the number of intervals grows at the same rate, as 2^n for both). I feel the spaces should be diffeomorphic by tweaking the map of [0,1] but I'm not sure.
  5. Oct 18, 2011 #4
    You shouldn't be looking for a diffeomorphism, you should be looking for a proof that no such function exists. Remember the following theorem from calculus -- if U and V are open sets in [itex]\mathbb{R}[/itex] and g is a diffeomorphism from U to V, then:

    [tex]\int_{V} f(x)\ dx = \int_{U} f(g(x)) |g'(x)|\ dx[/tex]

    Now, apply this theorem where f is the indicator function of the cantor set.
  6. Oct 18, 2011 #5


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    I think this is true only for isometries, and for general g then we use detg, where g is the Jacobian of the transformation. Take, e.g., the unit interval V=[0,1] and take U=[0,2] , and here U is homeomorphic to V,f(x)=x, and one homeo would be g(x)=2x , so that g'(x)=2 . Letting f(x)=x , then:

    [tex]\int_{[0,1]}x\ dx [/tex]=1/2 , but [tex]\int_{[0,2]}4x\ dx [/tex]=8. g(x)=1/(1-x) gives
    between (0,1) and the whole real line gives you an even more dramatic counter.
    Last edited: Oct 18, 2011
  7. Oct 18, 2011 #6


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    I'm not sure of where the problem is, Uzuki, but the indicator function is discontinuous. Tho I agree with you in the issue of the change-of-variables result. But I think what this result says is that the image is scaled by the determinant of the Jacobian of the map. Maybe I shouldn't be posting this late when my mind is not so clear; I hope I am not missing something obvious here.

    If one were to disprove the fact that both sets are diffeomorphic, one would need to show that there is no diffeomorphism g that does not satisfy the change-of-variable relation described above.
    Last edited: Oct 18, 2011
  8. Oct 18, 2011 #7
    Bacle, you wrote the diffeomorphism backwards. g is a diffeomorphism from U to V. So if U=[0, 2] and V = [0, 1], then the linear diffeomorphism would be g(x) = x/2, so that:
    [tex]\int_{[0, 1]} x\ dx=\frac{1}{2} = \int_{[0, 2]} \frac{x}{2} \cdot \left| \frac{1}{2} \right| \ dx[/tex]

    And yes, in the general case you use the determinant of the Jacobian, but since we are in the one-variable case, the Jacobian (and therefore its determinant) is g'(x).
  9. Oct 18, 2011 #8


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    But how can one tell if the diffeo that increases the measure by a factor k does not also have g'(x)=1/k?

    Yes, I wrote things backward; my bad for posting at 4 a.m,, sorry.
  10. Oct 18, 2011 #9
    If a linear diffeomorphism increases the measure by a factor of k, then the magnitude of the Jacobian will be k. But in the case of U=[0, 2] and V=[0, 1], the measure is being decreased by a factor of 2, not increased. So the Jacobian is 1/2.
  11. Oct 19, 2011 #10


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    My general point, tho, is that one cannot conclude that there is no diffeomorphism between the two cantor sets without showing that there is no diffeo f that will satisfy the conditions of the change-of-variable formula.
  12. Oct 19, 2011 #11
    Hmm... I thought it was obvious why there was no diffeomorphism, but I guess not. Okay, suppose that g:[0, 1] → [0, 1] is a diffeomorphism that maps the fat cantor set to the cantor set. Let f(x) be the indicator function of the cantor set. Then since the cantor set has measure zero we have:

    [tex]\int_{[0, 1]} f(x)\ dx = 0[/tex]

    But on the other hand, by hypothesis g maps the fat cantor set to the cantor set, and so f(g(x)) is the indicator function of the fat cantor set, which has positive measure. Also |g'(x)|>0 everywhere, since g is a diffeomorphism. Therefore f(g(x))|g'(x)| is a nonnegative function which is strictly positive on the fat cantor set, which has positive measure. Therefore:

    [tex]\int_{[0, 1]} f(g(x)) |g'(x)|\ dx > 0[/tex]

    And this inequality is strict. Therefore the two integrals cannot be equal, which is a contradiction.
  13. Oct 19, 2011 #12


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    I don't mean to nitpick, but don't we need f --the indicator function--to be continuous on [0,1]? AFAIK it is part of the statement of the change-of-variable formula?


    And I think the indicator function of the Cantor set is nowhere continuous on [0,1] , since a neighborhood of any point of the set contains points not in the set, right?

    The main reason I have trouble with the result is that the two Cantor sets are homeomorphic,
    and they are both 0-dimensional manifolds, and, for n=/4 , any two homeomorphic manifolds are also diffeomorphic.
    Last edited: Oct 19, 2011
  14. Oct 19, 2011 #13
    Actually the change of variables formula only requires g to be continuous -- f can be any integrable function whatsoever, although deriving this result requires the use of measure theory and Lebesgue integrals, so you frequently see the more conservative result where f is also continuous quoted for the sake of people who have only studied Riemann integrals.

    This statement isn't true - a 0-manifold is a discrete space. The cantor space does not have the discrete topology.
  15. Oct 19, 2011 #14


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    You're right, I missed this obvious fact. Since we were talking about diffeomorphisms, I assumed the spaces in question were manifolds.

    I had done Lebsegue theory, but had never seen this format of CV.

    Hmm... if neither is a manifold, how can we then talk about the spaces being diffeomorphic? Isn't a diffeomorphism a map defined between manifolds?
    Last edited: Oct 19, 2011
  16. Oct 19, 2011 #15
    Remember the precise statement of the problem: "The third part asks whether there is a diffeomorphism from [0,1] to itself such that a fat cantor set is mapped to the ternary cantor set" (emphasis added).
  17. Oct 19, 2011 #16


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    Never mind; sorry for my confusion. When there was a mention of diffeomorphism, I wrongly assumed the two Cantor sets were both manifolds, until Uzuki correctly pointed out that they are not. But there is a definition of diffeomorphism that can be done between non-manifolds. My bad.
  18. Oct 19, 2011 #17


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    Yes, my bad; I worked under assumptions/conditions that existed only in my mind, sorry for the general confusion.
  19. Oct 19, 2011 #18
    Don't worry, even the best of us get confused sometimes. :smile:
  20. Oct 19, 2011 #19
    Uzuki: thanks for the help! I dont have any background in measure theory (this for a class in topology with a focus on differential topology and the textbook is Milnor's so he assumes quite a bit from the reader). Measure 0 must be a diffeomorpic invariant, which is a neat result.

    The issue with manifolds also confused me, I wasn't sure if the Cantor set could be considered a 0 dimensional manifold. It is homeomorphic to an uncountable product of {0,1} if I'm not mistaken, so it would seem the discrete topology would work.

    I dont think this is true. I know theres a unique differential structure for n<4 and for some cases for n>4 but the n dimensional cube and the n dimensional sphere are homeomorphic but not diffeomorphic.
  21. Oct 19, 2011 #20


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    The n-cube can be smoothed into being diffeomorphic to the sphere, and, clearly, viceversa.
    For n=2, inscribe the circle in a square, and draw a line from the center of the circle into the place where it meets the square. The map taking a point in the circle to a point in the square is a diffeomorphism.

    And the topology of the Cantor set does not agree with the discrete topology; there are subsets of the Cantor set that are closed , but not open, and viceversa. The product topology of the discrete topology does not give you a discrete space. There is also the fact that the Cantor set is uncountable and compact (it is closed--as the complement of open sets that are removed--and bounded); an uncountable discrete space is not compact, e.g., a cover by singletons has no finite --not even countable--subcover. There is also the fact that the Cantor
    set has no isolated points.

    Re homeo., but not diffeomorphic, you're right I misspoke. I should have said for n<4
  22. Oct 19, 2011 #21


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    I wonder if Haudorff dimension could also be useful here, tying this with another thread here recently, on fractals. I wonder if Hausdorff dimension is a diffeomorphic invariant.
  23. Oct 19, 2011 #22
    You're right, Bacle. Hausdorff dimension is a diffeomorphic invariant. Here's a proof:

    Suppose that U, V are open subsets of [itex]\mathbb{R}^n[/itex] and that f:U→V is a diffeomorphism. Suppose that A is a bounded subset of U such that [itex]\overline{A} \subseteq U[/itex] and [itex]H^k(A)=0[/itex], where [itex]H^k[/itex] is the k-dimensional Hausdorff measure. Now, let [itex]r=d(A, U^c)[/itex]. Then the set of points [itex]K=\{ x: d(A, x) ≤ \frac{r}{2} \}[/itex] is a compact subset of U, so function [itex]\Vert D_x f \Vert_{op}[/itex] has a maximum M on K. Suppose that B is a subset of A having diameter at most r/2. I claim that diam(f(B)) ≤ M*diam(B). For let x and y be any two points in B, and let [itex]\gamma : [0, 1] \rightarrow \mathbb{R}^n[/itex] be the straight line path between them. Note that every point in the image of γ is at a distance of at most diam(B)≤r/2 from A, and thus contained in K. So we have:

    [tex]\begin{align*} \Vert f(y)-f(x) \Vert &= \left\Vert \int_{0}^{1} \frac{d}{dt}(f(\gamma (t))) \ dt \right\Vert \\ &\leq \int_{0}^{1} \left\Vert \frac{d}{dt}(f(\gamma (t))) \right\Vert \ dt \\ &= \int_{0}^{1} \Vert (D_{\gamma(t)}f)(\gamma'(t)) \Vert \ dt \\ &= \int_{0}^{1} \Vert (D_{\gamma(t)}f)(y-x) \Vert \ dt \\ &\leq \int_{0}^{1} \Vert (D_{\gamma(t)}f)\Vert_{op} \Vert y-x \Vert \ dt \\ &\leq \int_{0}^{1} M\operatorname{diam}(B) \ dt \\ & = M\operatorname{diam}(B) \end{align*}[/tex]

    So it follows that diam(f(B))≤M diam(B), as claimed. Now, since A has measure zero, we have for any ε>0 and any δ>0 that A can be covered by countably many sets {B_m} with diam(B_m)<δ and [itex]\sum_{m=1}^{\infty} \operatorname{diam}(B_m)^k < \varepsilon[/itex] WLOG we may assume that the B_m are all subsets of A. Then as long as 0<δ<r/2, the previous result gives:

    [tex]H_{\delta}^{k}(f(A)) \leq \sum_{m=1}^{\infty} \operatorname{diam}(f(B_m))^k \leq \sum_{m=1}^{\infty} M^k \operatorname{diam}(B_m)^k = M^k \varepsilon[/tex]

    This implies that:

    [tex]H^{k}(f(A)) = \lim_{\delta \rightarrow 0^{+}} H_{\delta}^{k}(f(A)) \leq M^k \varepsilon[/tex]

    And since M^k is constant, taking the limit as [itex]\varepsilon \rightarrow 0[/itex] shows that [itex]H^{k}(f(A)) = 0[/itex]. This shows that whenever A is a bounded set of k-dimensional Hausdorff measure 0 whose closure is contained in U, the measure of f(A) is also 0. For a general set A of k-dimensional Hausdorff measure zero, we can write [itex]A_m = \{ x \in A : d(x, 0) < m \mbox{ and } d(x, U^c) > \frac{1}{m} \}[/itex], so that each A_m is bounded and has closure contained in U. Then since [itex]A=\bigcup_{m=1}^{\infty} A_m[/itex], we have [itex]f(A)=\bigcup_{m=1}^{\infty} f(A_m)[/itex], so f(A) is the countable union of sets of measure zero and thus has measure zero.

    We have shown that if A has k-dimensional Hausdorff measure zero then so does f(A). This tells us that the Hausdorff dimension of A is at least the Hausdorff dimension of f(A). Applying the same logic to the inverse of f tells us the Hausdorff dimension of f(A) is the same as the dimension of A. Q.E.D.
  24. Oct 19, 2011 #23


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    Nice one, Citan, good job!.

    I had been doing algebraic topology for a while, and I was getting tired of all the somewhat-hand-wavy arguments with diagrams. It is nice to see an actual hard-analysis proof without sheafs or lifts or similar arguments, where you can "see under the hood".
    Last edited: Oct 19, 2011
  25. Oct 20, 2011 #24


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    Another approach to solving the OP question:

    I made things easier for myself this time, using a sly trick: I read the actual question!

    A diffeomorphism f:[0,1]-->[0,1] is Lipgarbagez continuous, being C , and,

    by compactness of [0,1], all of its derivatives are bounded. Then Lipgarbagez continuity implies

    absolute continuity, and absolutely continuous functions take sets of measure zero to sets

    of measure zero.
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