Asked 8 years ago. Active 7 years, 11 months ago. Viewed 4k times. My Question: Why do we say that the Cantor set has no length? JimmyJackson JimmyJackson 1, 2 2 gold badges 18 18 silver badges 33 33 bronze badges. Add a comment. Active Oldest Votes. DonAntonio DonAntonio k 17 17 gold badges silver badges bronze badges. But, let me read some more about the density of a set. I think this has me going in the right direction thanks!
My answer was only intended to clear up a particular point of confusion in the original post. Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown. Featured on Meta. Now live: A fully responsive profile. Related Hot Network Questions. Question feed. Taking the limit as goes to infinity, we find that.
This means the Cantor set has measure 0 in the real line. The second question, which concerns the number of points in the Cantor set, is a bit trickier. There are a few ways to tackle this. Intuitively, the endpoints and of are in the Cantor set.
So are the endpoints of , etc. Since there are disjoint segments in , it follows there are endpoints in. This implies the number of points in is greater than for every. It follows that has an infinite number of points. In fact, the cardinality of the number of points in Cantor set is uncountably infinite.
One proof is topological as in Munkres, p. An argument earlier showed that is compact. Since every point in the Cantor set is a limit point i. There is a theorem saying that if a nonempty compact Hausdorff space has no isolated points, it is uncountable. Since is a Hausdorff space in fact, it is a metric space , the hypotheses of the theorem are satisfied, and the result follows.
A second proof that the Cantor set is uncountable uses a property of binary and ternary expansions. Write every real number in in ternary expansion. We must be careful, as real numbers do not have unique binary, ternary, or even decimal representations for instance, 0.
Then the Cantor set is precisely those numbers that do not have a 1 in at least one of their ternary expansions. It is clear that the function takes points in the Cantor set to real numbers in.
As shown above, it is not one-to-one, but it is onto. Since is a surjection onto , which is uncountable, it must be that the Cantor set is uncountable.
The Cantor function, also known as the Cantor Staircase, is a bizarre function that is continuous and has a derivative of 0 at every point where it is differentiable. In fact, it is differentiable at every point other than on the Cantor set, which is a set of measure zero. It is not obvious that this function is continuous.
Since I used earlier, I will use here. As it turns out, will not be too different from. First, let map all of to 0 and all of to 1. Then it remains to map onto. For , expand in its ternary expansion. In fact, it is not too hard to see that the function is constant on any removed middle-third interval, i. Let and. We could have chosen and to be the endpoints of any middle-third interval and the argument holds, the only difference being there a finite number of digits in front of the leading 1.
Constant parts of a function are continuous, so it remains to show that is continuous on the Cantor set. Now we may use the old episilon-delta formulation of continuity in calculus. Given a point , we wish to show that for any , there exists such that implies. Given , let be such that. Then suffices. To see this, suppose that , then there must be some binary digit where they first differ, let be the index of this digit.
Now suppose and are separated by no more than , then their ternary expansions are the same up to the first digits. Then and in binary are the same up to the first digits, and thus they are separate by no more than.
Note that the Cantor function hits every value between 0 and 1, i. Thus it manages to rise from 0 to 1 despite being constant almost everywhere. For more information about the Cantor set or the Cantor function , check out the following:.
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