Why isn't the Cantor Set contradictory?

Solution 1:

"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 patterns that makes it possible for the cardinality to exceed $\aleph_0$.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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 $0$.

Solution 5:

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.)