# Why isn't the Cantor Set contradictory?

So you start with a 1-dimensional stick, remove the middle third of it, leaving 2 pieces. From each of these 2 pieces, remove the middle third. Etc. Whatever is left at the end of infinitely many stages, call "the Cantor Dust".

The short version of my problem: Here are two things I read about the Cantor Dust: a. It does not consist of isolated points. b. It also does not contain any segments of nonzero length. (a) and (b) seem to me incompatible with each other. I don’t see what the third alternative could be.

Longer version: Here is what seems to me the case:

1. At the end of the process, there are aleph-null pieces of the stick left. This must be so, since the cuts only occurred at locations with rational coordinates (calling one endpoint of the stick “0” and the other end “1”). So there are aleph-null cuts, so there must be only aleph-null pieces.
2. Each piece has a size of zero. This must be so, since the measure of the stuff removed is 1, so the total measure of the Dust is 0.
3. If a piece has a size of zero, it’s a point. Imagine overlaying one of the pieces of Cantor Dust on top of a geometric point. How far would it stick out? Zero distance. Therefore, the piece would coincide with the point, and therefore the piece itself is just a point.
4. Therefore, the Cantor set consists of aleph-null points. (From 1-3.)
5. But I also read that the Cantor set contains uncountably many points.

What has gone wrong here? In the sources I looked at, no one addresses the "long version” argument above. None even address the "short version", which surprises me.

Amendment: I intend "pieces" to be the largest connected parts that are left (so a piece might be a single point). I intend the term to capture the sense in which, after the first stage, we say "there are two pieces"; after the second stage "there are four pieces", etc.

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As an example to see that 1. is wrong think of "cutting" the real line at every rational point. –  martini Jul 10 '14 at 17:39
The fact that (a) and (b) seem to be contradictory is part of the point of constructing the Cantor set. –  Jack M Jul 10 '14 at 17:39
Note that viewing $\mathbb{Q}$ as a subset of $\mathbb{R}$, we have that $\mathbb{Q}$ does not consist of isolated points, but contains no non-trivial intervals. –  goblin Jul 10 '14 at 17:47
Five answers so far, and I'm still the only one who's up-voted the question. As often happens, that gets neglected. –  Michael Hardy Jul 10 '14 at 18:33
My posted answer below has evolved as I have further digested the posted question. The question quite correctly says that the number of endpoints of intervals remaining in the Cantor set after each deletion of middle thirds is $\aleph_0$. Moreover, the argument given in the question to establish that fact is valid. Apparently it is assumed that the Cantor set is built up of pieces bounded by those endpoints. I hope my posted answer shows why that is wrong. –  Michael Hardy Jul 10 '14 at 19:01

"a. It does not consist of isolated points. b. It also does not contain any segments of nonzero length. (a) and (b) seem to me incompatible with each other."

Consider the set of all rational numbers.

• It has no isolated points since every open interval about a rational number contains other rational numbers.
• It includes no intervals of positive length since every open interval about a rational number contains some irrational numbers. (That last statement is a bit more work to prove than may superficially appear, although it may naively seem obvious.)

You correctly observe that the number of endpoints of remaining "pieces" is $\aleph_0$. It cannot exceed that because every endpoint is rational and there are only countably many rational numbers.

Then you incorrectly infer from that that the Cantor set is just a union of "pieces" each of which is bounded by some of those endpoints. This would make sense if the Cantor set were simply the union of intervals whose endpoints were those pieces.

Think of the number $1/4$: it is a member of the Cantor set. Which "piece" would it belong to? It is a member of the lowest third of the interval from $0$ to $1$, so it does not get deleted at the first step. It is a member of the highest third of that, so it does not get deleted at the second step. It is a member of the lowest third of that, then of the highest third of that, and so on, alternating. The fact that it alternates like that should become clear as soon as you see that $1/4$ is located just $1/4$ of the way from $1/3$ down to $0$, i.e. $1/4$ of the way from the top of the lowest third to the bottom. If the Cantor set consisted of "pieces" whose endpoints were those of deleted middle thirds, then the question of which such "piece" the number $1/4$ belongs to would make sense.

No interval of positive length is a subset of the Cantor set since some parts of every interval will get deleted as the process of deleting middle thirds progresses.

If you have $\aleph_0$ pieces each of size $0$, then they don't fill up the whole interval. Measure is "countably additive", i.e. if the measure of each piece $A$ is $m(A)$ and they don't intersect each other (except possibly in sets whose measure is $0$) then $$m\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty m(A_n).$$ If the union on the left is the whole interval, then the sum of measures on the right is $1$.

You cannot express the interval from $0$ to $1$ as a union of $\aleph_0$ subsets of equal measure.

When I first saw the Cantor set, I thought it would contain only the endpoints of the deleted middle thirds. But the fact $1/4$ is a member of the Cantor set is a counterexample. So is $3/10$. But both of those have repeating patterns; it is the presence of non-repeating patters that makes it possible for the cardinality to exceed $\aleph_0$.

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One way to see just how many points are in the Cantor set is to realize that every point with ternary expansion containing only "$0$"s and "$2$"s survives. The first deletion removes all the one with "$1$" in the $3^{-1}$ place, the second deletions remove all those with with "$1$" in the $3^{-2}$ place, and so on. This is also how to "immediately see" a bijection between binary expansions of everything in the unit interval and ternary expansions of points in the Cantor set. –  Eric Towers Jul 11 '14 at 3:44
Thanks, Mike. I was tripped up by a couple of thoughts: (a) the thought that each 'piece' was created by some particular pair of 'cuts'. But actually, each remaining piece is a single point, and there's no particular pair of cuts that created it. (b) The thought that for each piece, the stuff "immediately to the left" or "immediately to the right" of it must be an interval that was removed. But actually, there's no uniform interval that's "the interval immediately to the left/right", because every interval contains a mix of removed bits and remaining bits. –  Owl Jul 11 '14 at 17:18
I still find it bizarre that the number of remaining pieces is infinitely greater than the number of removed pieces... –  Owl Jul 11 '14 at 17:20
Lots of things like that are bizarre. There are only countably many rational numbers, but uncountably many gaps between them. (A gap is a pair of sets $A,B$ of rational numbers such that every member of $A$ is less than every member of $B$ and $A\cup B$ contains all rational numbers and $A$ has no maximum and $B$ has no minimum.) –  Michael Hardy Jul 11 '14 at 17:28
@Owl: Because you think about "pieces". Remove the rational numbers, you removed countably many "pieces" and the remaining is a totally disconnected set, so each "piece" is a singleton again. Therefore the pieces remaining are larger than the pieces removed. –  Asaf Karagila Jul 11 '14 at 17:29

Yes, the Cantor set is counterintuitive, or rather Cantorintuitive... okay, that's not that funny.

There are points in the Cantor set which are not interval endpoints. For example points which are limit points of the interval endpoints. Since the Cantor set is closed, it contains all its limit points, and remember that a countable set can have an uncountable number of limit points.

For example, $\frac14$ is written in trenary base as $0.\overline{02}$. Therefore it is an element of the Cantor set, but it is quite easy to see that $\frac14$ is not an interval endpoint.

Let me also add that the third point is reversed. Single points have measure zero, but not every measure zero set is a point. For example a finite set is the union of finitely many points and therefore its measure is the sum of finitely many $0$'s, which is $0$.

If you also accept the fact that Borel and Lebesgue measures are $\sigma$-additive, this extends to every countable set. For example the rational numbers, the algebraic numbers, and so on.

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A related remark: measure and topological "smallness" are very much incompatible with one another. In particular, the union of Cantor sets of measure $1-1/n$ is a set of full measure which is topologically "small" (i.e. first category/meager). Dually, the complement of this union has measure zero but is "big" (i.e. second category/nonmeager). –  Ian Jul 10 '14 at 17:40
"Cantorintuitive" -- funny. –  user4894 Jul 10 '14 at 20:00
To clarify: I intended the "pieces" to be the largest connected parts that are left of the line. So (3) was supposed to mean: if a largest-connected-part of a 1-dimensional object has measure 0, then it's a point. –  Owl Jul 11 '14 at 17:24
@Owl: Well, in that case it is true. Connected + measure zero implies a singleton. I hope this will serve as a lesson for you for being more precise... –  Asaf Karagila Jul 11 '14 at 17:27

3 is false. There are plenty of sets of measure zero which are infinite. Take the rationals, for instance. We can cover them by intervals of size $\epsilon 2^{-n}$ around the $n$th rational number. Then we know that the total measure of the entire set of rationals is:

$$\mu(\Bbb Q)\le\sum_{n=1}^\infty \epsilon 2^{-n}=\epsilon$$

hence the rationals have measure zero, but the set of rationals is certainly not a single point.

Incidentally: the rationals contain no intervals either and have no isolated points and have measure $0$.

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What this answer says is true, but in the context of the question I find this misleading. It's true that "measure zero" does not imply "one point" (this is trivial: just consider two points). But the OP is asking about the "pieces" of the set; and each "piece" (as thought by the OP) is a connected (convex) set. For a connected set it's true that "measure zero" implies one point. The error in the reasoning is not here. –  leonbloy Jul 10 '14 at 23:09
No, the "pieces" aren't intervals, so the error is still there. Really, the op should use better mathematical language. –  Adam Hughes Jul 10 '14 at 23:10
Thanks, Leonbloy, that's right. Sorry I wasn't clear. I intended "pieces" to include single points, if that's the largest connected part. I think the error is in (1), not in (3) (per Mike Hardy's answer). –  Owl Jul 11 '14 at 17:29
@Owl both 1 and 3 are flawed, I opted just to address 3 because derailing any part is enough to derail an entire argument. 3's assumption that size zero implies a point is crucial to the rest of the misunderstanding. –  Adam Hughes Jul 11 '14 at 20:35
But 3 doesn't say that just any set with measure 0 is a point (and I never would have said that). –  Owl Jul 12 '14 at 16:53

There is a problem already in speaking of "at the end of the process". If we have a sequence of rational numbers $3, 3.1, 3.14, 3.141, 3.1415, 3,14159, \ldots$, how can there be the irrational number $\pi$ "at the end" of the sequence if all terms in the sequence are rational? Well, that's just it: $\pi$ is the limit of the sequence, but not in the sequence. When switching from a process to its limit, weird things can happen. Sometimes it is even problematic what the "limit" should be (consider a lamp switched off at times $1-\frac1{2n}$ and on at times $1-\frac1{2n-1}$; what is the limit state of the lamp at $t=1$?). For number sequences we get a notion of limit via the metric on the reals; for the Cantor dust, the limit is simply the intersection of the individual steps; but still we must acknowledge that many properties cannot be transferred too "naively" from the terms of the sequence to the limit.

From a different point of view: You know that between two rational numbers there is always at least one irrational number; and between two irrational numbers, there is always at least one rational number. So the situation is symmetric - and yet there are $\aleph_0$ many rationals and $\mathfrak c$ many irrationals.

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Your last paragraph took a moment: until you realize there's actually an infinite number between in both cases, and still countable or not, respectively. –  Mark Hurd Jul 11 '14 at 14:53
Thanks, Scott. You're right. One of the things boggling my mind is that within the series, the number of pieces left is always 1 + the number of pieces removed. But at the end, the # of pieces left is infinitely greater than the # removed. –  Owl Jul 11 '14 at 17:32

Perhaps the physical metaphor you are using is the problem. You certainly have discrete pieces at the $n$th step of the cutting, but that doesn't mean there are discrete pieces when you are finished with infinite removals.

These sentence is meaningless: "If a piece has a size of zero, it’s a point. Imagine overlaying one of the pieces of Cantor Dust on top of a geometric point. How far would it stick out? Zero distance. Therefore, the piece would coincide with the point, and therefore the piece itself is just a point."

What does that mean?

How does (1)-(3) let you conclude that the points are countable?

The notion of "piece" is a what is confusing you. What do you mean? What you get is a set of points, which is what you start with. It's true that, at the $n$th step, you can represent that set as a union of $2^n$ closed intervals (pieces,) but the intuition that what you are left with after repeating an infinite set of removals is "pieces" rather than "points" is a intuition, not rigor.

They are like the rational numbers in this way, but they are unlike the natural numbers in that they are uncountable. That is what makes the Cantor set interesting.

Here's a useful question. What does it mean for a set $X$ to be measure zero? For that, you need the concept of Lebesgue measure, but it basically means that if $\epsilon>0$ then there is a possibly infinite set of (non-trivial) intervals which cover $X$, such that the sum of the lengths of the intervals is $<\epsilon$. Countable sets can be easily covered this way - if the set is $\{x_1,x_2,\dots\}$ then draw an interval of length $e/2^i$ around $x_i$ to get such a covering.

Now, the Cantor set is the intersection of sets $C_0\supset C_1\supset C_2\supset C_3\dots$. Where $C_i$ is the state after the removing the intervals at the $i$th step. And $C_i$ is already the union of $2^i$ intervals of length $3^{-i}$. So we have that $C=\cap C_i$ also has measure zero, since each each $C_i$ covers $C$.

The opposite notion - that "positive measure" is some summation of "positive parts" - is wrong, too. You can take the set of irrational numbers in $(0,1)$, and that set has measure $1$. There are no "parts" of this set - it contains no intervals. That is one reason we use the term "measure" rather than "length." "Length" gives the wrong impression.

(W also use "measure" because it generalizes in a lot of ways, of course.)

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Thanks, Thomas. I intended "piece" to include single points, if that's the largest connected part that's left. I think my mistake was in (1), inferring from "there are aleph-null cuts" to "there are aleph-null pieces". –  Owl Jul 11 '14 at 17:36
Yeah, if you just enumerate the rational numbers between $[0,1]$ and simply remove each from $[0,1]$ one at a time in that order, you'd also have finitely many "pieces" at each step, but end up with the set of irrationals in $[0,1]$, which is uncountable, and yet there would be no "pieces" other than the individual points. –  Thomas Andrews Jul 11 '14 at 17:46