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I know that the Cantor Set is uncountable (this is a well-known result), so I know that there must be something wrong with the following method for counting its elements, but I'm not sure where the flaw is.

I'll assume that the Cantor set in question is a subset of the interval $[0,1] \subseteq \mathbb{R}$ obtained through iterative elimination of middle thirds.

The point with coordinate $0$ is numbered $1$, and the point with coordinate $1$ is numbered $2$. After the first iteration, the middle third of the unit interval is deleted, and we find two more points that must be in the Cantor set. These are the points with coordinates $1/3$ and $2/3$. Number them $3$ and $4$, respectively.

After the next iteration, there are four new endpoints of line segments that must now be counted. Number the points with coordinates $1/9$, $2/9$, $7/9$ and $8/9$ as $5$, $6$, $7$, and $8$, as depicted below.

numbering of endpoints for the first few iterations

Every point in the unit interval must either not be in the Cantor set, or be the endpoint of a line segment at some finite iteration of the deletion process. But every such endpoint is numbered by the process above, so every point in the Cantor set should correspond to a non-negative integer, and vice versa.

I think the problem is that there exist points that are in the Cantor set that are not endpoints of a line segment at some iteration, and so would not be counted. However, I can't think of an example of such a point ...

I thought about this while musing over the following questions:

  • Is it possible to draw an uncountably infinite number of non-intersecting circles in the plane? (clearly yes, e.g. concentric circles with radii in $[0,1]$)

  • Is it possible to draw an uncountably infinite number of non-intersecting figure eights in the plane? (I think the answer is no, but I am not sure why)

While thinking about the second of these, I considered a figure eight with two figure eights inside of it, one in each half, and two inside each of the halves of those, and so on:

nested figure eights

These figure eights are easily countable. Number the biggest one $1$, the second biggest ones $2$ and $3$, and so on. This is closely similar to how the points in the Cantor set are "counted" in the first part of this question.

I'm dealing with two sets. One is the Cantor set; the other is the infinite set of nested figure eights, the first few of which are in the diagram above. The latter is clearly countable, while the former is ostensibly not. What is the difference between the two?

Is there possibly some way to construct a set of figure eights with centers at points in the Cantor set so that these figure eights don't intersect? Would this be a way to draw an uncountable infinity of figure eights in the plane? If not, is there another way?

richarddedekind
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Zubin Mukerjee
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    There are more points on the cantor set than just the endpoints. There is also the limit of the endpoints like $1/4$. On the topic of the figure eights the points on the cantor set are all accumulation points so I'm pretty sure your figure eights would intersect(being infinitely close). – ruler501 Mar 29 '14 at 04:45
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    The figure-8 question is discussed at http://math.stackexchange.com/questions/78018/countable-or-uncountable-set-8-signs – Gerry Myerson Mar 29 '14 at 05:50

2 Answers2

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I think the problem is that there exist points that are in the Cantor set that are not endpoints of a line segment at some iteration, and so would not be counted. However, I can't think of an example of such a point…

Here is one example. Start with the whole interval, $[0,1]$. In the first step, go to the left sub-interval, $[0/3,1/3]$. In the second step, go to the right interval, $[2/9,3/9]$. Continue this process, alternating between left and right child, an infinite number of times. Choosing any element from the interval at each step, you obtain a sequence of numbers which converges. And since $\mathbb R$ contains the limits of all convergent series, that sequence corresponds to a real number. On the other hand, that real number can not be an endpoint at any level of the iteration, because at every finite level, the corresponding sequence element will be a whole interval, and two steps later both endpoints of that interval will have been cut away. Therefore you have a number which is in the Cantor set but not an endpoint. Changing the pattern of how you alternate between left and right children will result in other examples as well.

If you want to be more explicit, the limit of the series described above satisfies the condition $x=\frac19x+\frac29$ so you can conclude that $x=\frac14$. So in fact this is the number that ruler501 mentioned in a comment to your question, although I didn't realize this when I wrote my answer.

Regarding the second part of your question, about the figure-eight, you should probably follow the post mentioned by Gerry Myerson, Countable or uncountable set 8 signs. I must confess that dealing with the first part of your question, I myself would have imagined this to be possible, similar to the Cantor set construction but using areas at each step which thin out to curves in the limit. Although the mentioned post has me convinced, I still don't see exactly where such an approach would fail.

Community
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MvG
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Your problem is that you compare the nested eight figures to the Cantor set, but you should be comparing them to the intervals that we intersect instead.

Note that the Cantor set is the intersection of countably many intervals, and the result is indeed an uncountable set, but this set contains no interval. Similarly, the eight figures are countable, but if you take the intersection of those you will be left with uncountably many points in the plane.

Asaf Karagila
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    Wow! Upvote and downvote within five seconds! (If only there was a way for me to know what's wrong with the answer...) – Asaf Karagila Mar 29 '14 at 05:14
  • Thanks for the answer! I've never thought of the Cantor set as an intersection of intervals, but that seems interesting.

    I'm not sure what you mean by the intersection of the figure eights leaving an uncountable number of points in the plane?

     – Zubin Mukerjee Mar 29 '14 at 05:34
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    Zubin, recall that a point is in the Cantor set if and only if it is in all the "levels" of the constructions of the Cantor set (i.e. at each step, when we remove a third of the previous step, we didn't remove the point). This is exactly as saying that the Cantor set is the intersection of the unions of closed intervals. Similarly, the figure eight shapes enclose a region of the plane, and if we consider "steps" as the first figure, then the union of the regions inside and so on; the intersection of these "steps" gives us an uncountable subset of the plane. – Asaf Karagila Mar 29 '14 at 05:38
  • (I am fairly sleep deprived at this moment, so I apologize if I'm not making too much sense. I'll try my best, and if nothing else then I'll edit my answer later on.) – Asaf Karagila Mar 29 '14 at 05:39