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Math Help - Variation on the Cantor Set

  1. #1
    Super Member redsoxfan325's Avatar
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    Variation on the Cantor Set

    We're going to create a set C that is a variation on the Cantor Set. Start with the open interval C_0=(0,1). Remove the segment \left[\frac{1}{3},\frac{2}{3}\right]. Now you have C_1=\left(0,\frac{1}{3}\right)\cup\left(\frac{2}{3  },1\right). Next remove \left[\frac{1}{9},\frac{2}{9}\right] and \left[\frac{7}{9},\frac{8}{9}\right] and you're left with C_2=\left(0,\frac{1}{9}\right)\cup\left(\frac{2}{9  },\frac{1}{3}\right)\cup\left(\frac{2}{3},\frac{7}  {9}\right)\cup\left(\frac{8}{9},1\right). Repeat ad infinitum...

    I suspect this set is empty. Proof:

    Want to Prove: There are no interior points.

    Assume that \exists~ x\in C_n, \exists~ \epsilon>0 such that N(x,\epsilon)\subset C_n.

    For C_0, \epsilon<\frac{1}{2} to create a neighborhood contained in C_0.

    For C_1, \epsilon<\frac{1}{2}\cdot\frac{1}{3}.

    For C_2, \epsilon<\frac{1}{2}\cdot\frac{1}{9}.

    For C_n, \epsilon<\frac{1}{2\cdot3^n}.

    Thus, as n\to\infty, we have that \epsilon<0, which is clearly a contradiction. Thus there are no interior points in this set. However, C is an open set (I'm worried this is false), so all points are interior points, leading us to the conclusion that C is empty.

    Is this correct?
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  2. #2
    Super Member Gamma's Avatar
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    It is not immediate to me that this set would be open. The standard trick to show the cantor set is closed is because it's complement is the union of all the open intervals you removed, which is open, thus the cantor set is closed. Unfortunately, in this case the complement is a union of closed intervals which need not be closed.

    It might actually be easier to just take the regular cantor set and show the only thing in it is exactly the endpoints of the intervals. But in this case you tossed those out too (and 0 and 1 are removed from the beginning).

    But if you really need to prove it from scratch, I suggest thinking about ternary expansions (like the decimal expansion except with only coefficients 0,1,2 ie base 3 instead of 10 so it is like a_k(1/3)^k) of the cantor set. Like when you partition it up you are remove the middle third of the interval each time which corresponds to tossing everything that can be represented with a 1 in that time's digit. Know what I mean? I think you can probably show it this way as well, but it takes some serious elbow grease proving all this stuff.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Gamma View Post
    It is not immediate to me that this set would be open. The standard trick to show the cantor set is closed is because it's complement is the union of all the open intervals you removed, which is open, thus the cantor set is closed. Unfortunately, in this case the complement is a union of closed intervals which need not be closed.

    It might actually be easier to just take the regular cantor set and show the only thing in it is exactly the endpoints of the intervals. But in this case you tossed those out too (and 0 and 1 are removed from the beginning).

    But if you really need to prove it from scratch, I suggest thinking about ternary expansions (like the decimal expansion except with only coefficients 0,1,2 ie base 3 instead of 10 so it is like a_k(1/3)^k) of the cantor set. Like when you partition it up you are remove the middle third of the interval each time which corresponds to tossing everything that can be represented with a 1 in that time's digit. Know what I mean? I think you can probably show it this way as well, but it takes some serious elbow grease proving all this stuff.
    But there are many more points in the original Cantor Set than just the endpoints of the intervals. There are countably many endpoints but there are uncountably many points in the Cantor Set. Interestingly though, there are no isolated points AND no interior points in the Cantor Set.

    If I could somehow prove that this modified Cantor Set is open, the conclusion would follow. But we both agree that \bigcap_{n=1}^{\infty}C_n is not necessarily open.

    Also, I'm just conjecturing that this modified Cantor Set is empty. I don't actually know whether it is or not.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Actually, now I'm beginning to think that it might not be empty. It might consist of all transcendental ternary numbers. For instance, if a ternary decimal is represented as 0.a_1a_2a_3..., then something like 0.220002000000000000000002... where

    a_i=\left\{\begin{array}{lr}2&:i=n!\\0&:i\neq n!\end{array}\right\} for n\in\mathbb{N}

    might be in this modified Cantor set.
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  5. #5
    Super Member Gamma's Avatar
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    Haha, I was just taking a shower and was thinking more about this problem and realized exactly the same things you just pointed out. I think this ternary argument is the way to go. Definitely has to have combinations of 0 and 2, it cant end in an infinite string of 2s either, furthermore, those endpoints of the intervals can't be in there either, so it cant end in a string of 0s either. Outside of that I am really not sure what this thing is going to look like. Like maybe you see something that I am not seeing but I dont see why it couldnt even be something less transcendental looking you know, like why couldnt .2020202020... be in there, like just kinda skipping from left to right intervals each time? I am thinking it is just everything in the Cantor set, but not the endpoints of the intervals which i believe are the things ending in all 2s or all 0s.

    What class is this for, this is a really interesting problem?
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  6. #6
    Super Member redsoxfan325's Avatar
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    It's not really for a class. We were discussing the Cantor set in class and I just started wondering about this. It's not a homework problem or anything.

    I feel like rational ternary numbers wouldn't work, just because they seem too clean. I'm pretty sure there would be some infinite sequence of boundary points that converged to something like 0.202020202... It seems like for a number to be in this set, it would at least have to be irrational. However, I started doing out the Cantor set in ternary and for any finite n, the boundary points of C_n definitely all end in either an infinite string of 0's or 2's.

    This is what C_2 looks like in ternary. (I've added the \bar{0}'s for emphasis.)

    C_2 = (0.\bar{0},0.00\bar{2})\cup(0.02\bar{0},0.0\bar{2}  )\cup\underbrace{(0.2\bar{0},0.20\bar{2})}_{that~i  nterval}\cup(0.22\bar{0},0.\bar{2})

    Your suggested number would lie in "that interval" somewhere. On the next iteration, we'd have a boundary point 0.202\bar{0}; on the iteration after that we'd have a boundary point 0.2020\bar{2}; etc. After infinitely many iterations, I think we'd get 0.202020... as a boundary point. However, just writing this now makes me think that if what I just said is true, then this set will be empty, as even the number I mentioned before would be the convergence of the sequence of boundary points 0.2\bar{0}, 0.22\bar{0}, 0.220\bar{0}, 0.2200\bar{0},...,0.220002000000000000000002\bar{0  },.... I don't know what to think anymore. I keep confusing myself!
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  7. #7
    Super Member redsoxfan325's Avatar
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    Looking at your number 0.202020... in base ten, we have \sum_{n=0}^{\infty}\frac{2}{3^{2n+1}} = \frac{3}{4}.

    Now look at this sequence:

    0.\bar{2}, 0.2\bar{0}, 0.20\bar{2}, 0.202\bar{0}, 0.2020\bar{2}, 0.20202\bar{0},...

    In base 10, (I'm not going to go through the calculations as I did them all on Maple) this sequence looks like this: 1, \frac{2}{3}, \frac{7}{9}, \frac{20}{27}, \frac{61}{81}, \frac{182}{243},...

    This sequence of what are clearly boundary points converges to \frac{3}{4}. Even with only six steps, \frac{182}{243}\approx 0.748971

    I'm definitely starting to lean towards the empty set again...
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  8. #8
    Super Member Gamma's Avatar
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    those are boundary points though right? like the fact that it is the boundary points that are converging to it is the whole point of why I thought that would be a good pick. it means the endpoints of the open intervals are closing in on this point, this point is in the regular cantor set and you know its there because it is like a nested sequence of closed intervals that all contain this point so their infinite intersection contains this point. In this case, I have no idea what you can say about this, it is like at each iteration it is pretty clear that 3/4 is going to be in there, but those boundary points are going to form a cauchy sequence converging to 3/4. I just don't know what that means, lol.

    Fascinating problem though, you will have to keep me updated on what conclusions you come to.

    I think it will not be the empty set simply because like you pointed out there are an uncountable number of points in the cantor set, but all we are losing from that are two endpoints for each of a finite number of intervals over a countable number of interations which is countable. So we should still have an uncountable number of things in this set, but a lot stranger things have happened when dealing with the cantor set than that, lol.
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  9. #9
    Super Member redsoxfan325's Avatar
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    So this is what we know:

    1.) The Cantor set is uncountable.
    2.) The Cantor set has a countable number of boundary points.
    3.) The Cantor set has no interior points or isolated points.
    4.) This set has no boundary points, interior points, or isolated points.
    5.) Any point we can come up with is the limit of a sequence of boundary points in the Cantor set.

    I have no idea what conclusion to draw from this. I'll keep you posted though if I find out anything.
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  10. #10
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    Quote Originally Posted by redsoxfan325 View Post
    We're going to create a set C that is a variation on the Cantor Set. Start with the open interval C_0=(0,1). Remove the segment \left[\frac{1}{3},\frac{2}{3}\right]. Now you have C_1=\left(0,\frac{1}{3}\right)\cup\left(\frac{2}{3  },1\right). Next remove \left[\frac{1}{9},\frac{2}{9}\right] and \left[\frac{7}{9},\frac{8}{9}\right] and you're left with C_2=\left(0,\frac{1}{9}\right)\cup\left(\frac{2}{9  },\frac{1}{3}\right)\cup\left(\frac{2}{3},\frac{7}  {9}\right)\cup\left(\frac{8}{9},1\right). Repeat ad infinitum...

    I suspect this set is empty. Proof:

    Want to Prove: There are no interior points.

    Assume that \exists~ x\in C_n, \exists~ \epsilon>0 such that N(x,\epsilon)\subset C_n.

    For C_0, \epsilon<\frac{1}{2} to create a neighborhood contained in C_0.

    For C_1, \epsilon<\frac{1}{2}\cdot\frac{1}{3}.

    For C_2, \epsilon<\frac{1}{2}\cdot\frac{1}{9}.

    For C_n, \epsilon<\frac{1}{2\cdot3^n}.

    Thus, as n\to\infty, we have that \epsilon<0, which is clearly a contradiction. Thus there are no interior points in this set. However, C is an open set (I'm worried this is false), so all points are interior points, leading us to the conclusion that C is empty.

    Is this correct?
    This is my attempt on your problem.

    The Cantor set relies on the "nested interval theorem". It says
    "If \{[a_n, b_n]_{n=1}^{\infty} is a nested sequence of closed and bounded intervals, then \bigcap_{n=1}^{\infty}[a_n, b_n] is not empty. If, in addition, the diameters of the intervals converge to 0, then \bigcap_{n=1}^{\infty}[a_n, b_n] has precisely one member."


    We shall show that every open interval in your Cantor set, let's say C_\infty, is empty. Let (x,y) be an arbitrary open interval in C_\infty. Since you remove every point that has a ternary expansion using 0, 2(You can check this by induction), both x and y can be represented as ternary expansions using 0 and 2 only.
    Let N be a positive integer for which \frac{2}{3^N} < \epsilon where \epsilon is a positive number. Since x=0.x_1x_2x_3.... has a ternary expansion where each x_n is 0 or 2, we let y=0.y_1y_2y_3... be the real number having the indicated ternary expansion with y_n=x_n for n \neq N and y_N differing from x_N such that y_N is 0 if x_N is 2, and y_N is 2 if x_N is 0. Then, the distance between x and y is within the distance \epsilon,

    |x - y| = \frac{2}{3^N} < \epsilon .

    In other words, as N goes to infinity, the distance between x and y goes to 0.

    Since the union of empty sets is empty, C_\infty is empty.
    Last edited by aliceinwonderland; May 4th 2009 at 03:24 PM. Reason: This proof is not correct. See the Opalg's post 16.
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  11. #11
    Super Member redsoxfan325's Avatar
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    I definitely agree that intuitively, it should be empty. However, very little about the Cantor set is intuitive. My only issue is that this set is the Cantor set minus the boundary points. There are countably many boundary points and uncountably many points in the Cantor set, so removing the boundary points should still leave an uncountable set. However, it seems as if any number we try is the limit of a sequence of boundary points, and thus would be a boundary point in C_{\infty} and therefore not in the set.

    I've emailed my professor, TA, and a fifth-year analysis grad student, so hopefully one of them should be able to put this to rest.
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  12. #12
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    The Cantor set has a Lebesgue measure 0, but is not empty (as far as I can remember )
    But intuitively, a thing that has a Lebesgue measure 0 may be empty... lol
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  13. #13
    Super Member redsoxfan325's Avatar
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    Yes, the Cantor set has measure zero, and is uncountable. Our problem here is that we're removing all the boundary points of the Cantor set. That means that the resulting set has no boundary points, no interior points, and no isolated points. So what kind of points, if any, does it have???
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  14. #14
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    Quote Originally Posted by redsoxfan325 View Post
    Yes, the Cantor set has measure zero, and is uncountable. Our problem here is that we're removing all the boundary points of the Cantor set. That means that the resulting set has no boundary points, no interior points, and no isolated points. So what kind of points, if any, does it have???
    Every point in the Cantor set is a boundary point, but in this construction it is not true that all the boundary points have been removed. The points that are removed are the boundary points of the intervals that define the complement of the Cantor set. As pointed out above, there are only countably many such points, whereas the Cantor set contains uncountably many points. So in this modification of the usual construction, "most" of the Cantor set is unaffected.
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  15. #15
    Super Member redsoxfan325's Avatar
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    So would 0.2020202020... = \frac{3}{4} be in this set?
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