I was reading this (problem no 18, second method). There, I can understand the proof:

The main idea in attached link is: take a sequence of all the rational numbers $\{x_n\}$, then pick an open interval $U_1$ centered at $x_1$, next find out the first element of sequence that is not in $U_1$, around this, find open nbd $U_2$ such that $\overline U_1\cap \overline U_2$ is empty. Constructing this way, you get $U_n$'s. We observe that $$E=(\bigcup_1^\infty U_n)^c$$ is nonempty perfect set with no rational.

But I find this very unintuitive. Any $x\in E$ is irrational, and it seems like all the points in $E$ are endpoints of $U_n$(although that violates uncountability of perfect set). And any open neighborhood around $x$ will contain infinitely many rationals. How in that case it is possible that $\overline U_n$'s are mutually disjoint?

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    $\begingroup$ I would try reviewing the proof of why the Cantor middle-third set is not entirely composed of endpoints, and particularly, why $\frac{3}{4}$ is not an endpoint, but does lie in the set. $\endgroup$ – Theo Bendit Aug 18 '18 at 11:20
  • $\begingroup$ @TheoBendit, thank you very much for your comment. i have thought in that direction for last couple of hours, and have seen here (in last para of that section) that why even if $1/4$ is not endpoint, it is in cantor set. But, then I read Rudin's book's theorem that "If $K_n$ is sequence of compact sets such that $K_{n+1}\subset K_n$ and if diameter of $K_n$ tends to $0$, then $\cap K_n$ consists of exactly one point." In that situation how can $1/4$ or $3/4$ are elements of Cantor set? I will be obliged with explanation. $\endgroup$ – Silent Aug 18 '18 at 13:57
  • $\begingroup$ Here I think that the theorem I mentioned is applied to each leftmost interval, for example, in reference to this picture. @TheoBendit $\endgroup$ – Silent Aug 18 '18 at 14:01

Since you've read up about the Cantor set, let's examine it first. The number $\frac{1}{4}$ lies in the Cantor set. The proof of this typically involves using ternary expansions. In particular, the ternary expansion of $\frac{1}{4}$ is $$\frac{1}{4} = 0.\overline{02}$$ Elements of the Cantor set have a ternary expansion that consist of $0$ or $2$. Why is this? Well, essentially the process of removing middle thirds of the interval removes numbers whose ternary expansion(s) contain a $1$. Removing the first middle third $\left(\frac{1}{3}, \frac{2}{3}\right)$ removes all numbers beginning with $0.1\ldots$ (with the exception of $0.1\overline{2} = \frac{2}{3}$, but this can also be expressed as $0.2$). Similarly, removing the middle third from $\left[0, \frac{1}{3}\right]$ will remove numbers of the form $0.01\ldots$, except $0.01\overline{2} = 0.02$, etc.

The fact is, if $\frac{1}{4}$ were not an element of the Cantor set, it would have to be in one of these middle third sets. It would have to have a $1$ somewhere in one of its ternary expansions. Because it doesn't have a terminating ternary expansion, its expansion is unique. The fact that it repeats $0$ and $2$, tells us that it will be in the left part after removing the first middle third, then on the right after removing another middle third, then on the left, then on the right, etc. Notably, it's never in the middle, so it's never removed!

Cantor's intersection theorem can't be applied to the Cantor set as a whole, as the diameter of the sets we are intersecting don't limit to $0$ (in fact, $0$ and $1$ are never removed, so the diameter is constantly $|1 - 0| = 1$). This doesn't mean that the set is empty, of course, just that the theorem fails to apply.

The point of the theorem is more to utilise completeness. You don't ever use it to show a particular point happens to lie in an intersection; you use it to prove the existence of a point with desirable properties, and one that you typically can't explicitly construct.

But, that said, we can actually still use it. Take $K_1 = [0, 1]$. Let $K_2$ be the first third of $K_1$, and $K_3$ be the third third of $K_2$, then $K_4$ be the first third of $K_3$, and $K_5$ be the third third of $K_4$, etc. Then we have $\operatorname{diam} K_n = \frac{1}{3^n} \to 0$. By Cantor's intersection theorem, there must be a unique point in the intersection of the $K_n$s. We just happen to know what it is: $\frac{1}{4}$.

More generally, we can form other points in the Cantor set by choosing a sequence of nested "thirds". Always, the diameter will approach $0$, so always there will be a unique point of intersection. If your choice of first third or third third ends up repeating indefinitely, you've chosen an endpoint! If not, then the point you've chosen is not an endpoint, but still a member of the Cantor set.

Essentially, we have a binary tree of possibilities, where each vertex represents a middle third removed, and branching to the left means "first third" and branching to the right means "third third". Then each point in the Cantor set can be uniquely represented by an infinite chain from the root vertex, i.e. starting with the root vertex, and working your way down the tree indefinitely.

The set $E$ constructed in the solutions follows much the same idea. You won't be removing the middle third, per se, but you will be removing an open interval from $\mathbb{R}$, thus cutting it into two disjoint closed intervals. The construction doesn't explicitly remove an interval from the left portion and the right portion, but that's what happens eventually. Then, from the four portions remaining, there will eventually be intervals removed from each of them, etc.

Basically, the construction is essentially the same as the Cantor set, but it starts with $\mathbb{R}$ instead of $[0, 1]$, and we aren't so regular about which open interval we remove (i.e. it's not always just the "middle third" that we remove).

But this still produces a binary tree of possibilities. We can still pick one open interval to be our root, and everything less than this interval is our left choice and everything greater is our right choice. From our left half, we can pick on another removed open interval, and everything from the left of that is our left branch, and everything to the right is our right branch, etc.

Then, again, we form elements of this set from infinite chains beginning at the root vertex. If our sequence of choices of left and right eventually becomes "left, left, left, etc" or "right, right, right, etc", then we have an endpoint, but if they never turn into this pattern, then the resulting element of $E$ is not an endpoint.

  • $\begingroup$ Thank you very much for your extraordinary explanation. So, what I got from your cantor set explanation is, since 1/4 in ternary expansion is $0.0202..$, it is rational number, and if instead, something like $0.02002000200002..$ where number of zeros before occurrence of next 2 is increasing gives irrational, since it has non-repeating decimals. Here in ternary expansion 0's and 2's have special significance of choosing branch at each stage, and cantor intersection theorem guarantees that a point exists after all the intersection is carried out. $\endgroup$ – Silent Aug 19 '18 at 8:33
  • $\begingroup$ Just finished second half! This answer is so satisfying! I can upvote only once, but I think this deserves more upvotes. Thank you very much for allotting your precious time here. $\endgroup$ – Silent Aug 19 '18 at 8:40
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    $\begingroup$ @Silent Thank you for your kind words! Yes, Cantor's intersection theorem will guarantee the existence of these (often otherwise difficult to represent) irrational points in the Cantor set. More generally, this is another way one can use completeness to show the existence and uniqueness of real numbers given a $n$-ary epansion. $\endgroup$ – Theo Bendit Aug 19 '18 at 12:45
  • $\begingroup$ @TheoBendit Does this construction depend on AC? $\endgroup$ – YuiTo Cheng Jan 20 at 16:03
  • $\begingroup$ @YuiToCheng This particular construction is obviously very "choice-y", and indeed if I were to make this more formal, I'd probably use Zorn's lemma. However, other methods exist to make a perfect subset of the irrationals, such as the other two methods outlined in the pdf in the question. I haven't thoroughly checked, but neither of them seem to depend on AC. $\endgroup$ – Theo Bendit Jan 21 at 2:12

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