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Let $G$ be the Cantor set. It is well known that:

  1. $G$ is perfect and hence closed.
  2. $G$ has the cardinality of the continuum.
  3. $G$ has measure zero.
  4. For any set $S \subset \mathbb{R}$ (I will not keep writing that we are in $\mathbb{R}$) we have $S \text{ is closed} \Leftrightarrow S^c \text{ is open}$.
  5. Any open set $O$ can be written as a --- in fact unique --- countable union of disjoint open intervals.

$G^c$ can thus be written as a countable union of disjoint open intervals. We now imagine this union as being superposed on the real line graphically as follows:

R: <<<----(....)---(..)--(.)---------(...)--->>>

where the (...) represents the open disjoint intervals (of differing size) composing $G^c$, and the --- represents the remaining non-covered real numbers (that is those in $G$). Now we can cover $\mathbb{R}$ in its entirety by "collecting" the --- into disjoint closed intervals. Any one of these closed intervals might of course consist of only a single element. We now obtain:

R: <<<[--](....)[-](..)[](.)[-------](...)[-]>>>

Take the union of these disjoint closed intervals. This must be $(G^c)^c = G$. Now it is not hard to imagine a mapping from the (...)'s to the [...]'s. Just take the next (...) in line for each [...] (and do some trivial fixing at the ends). Therefore we have written $G$ as a countable union of disjoint closed sets. However, $G$ has measure zero and therefore cannot contain any closed sets other than the single element type. Hence $G$ is countable. Contradiction.

Where do I go wrong?

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    $\begingroup$ One note to add to the existing answer: Only the points having a ternary expression ending in either all $2$'s or all $0$'s after a point can be the endpoint of an open interval composing the complement. All the other points only manage to be on the boundary because there are sequences of intervals outside of the set whose endpoints tend there, despite no single interval being witness to this. $\endgroup$ – Milo Brandt Sep 16 '16 at 1:42
  • $\begingroup$ @MiloBrandt Thanks for your comment, but I do not quite follow? $\endgroup$ – Jori Sep 16 '16 at 2:49
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    $\begingroup$ To understand Milo's point, consider the collection of open intervals $\{(\frac 1{n+1}, \frac 1 n) \mid n\in \Bbb Z, n \ne 0, n \ne -1\}$. If you remove the sets from $(-1,1)$, this complement also is a perfect set. Note, though, that $0$ is in the complement, but is not a boundary point of any of these intervals. For the Cantor set, this happens uncountably often. $\endgroup$ – Paul Sinclair Sep 16 '16 at 13:58
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You're imagining that the open intervals of $G^c$ are ordered discretely, like the integers, so you have alternating open intervals in $G^c$ and closed intervals in $G$. But actually, the open intervals of $G^c$ are densely ordered, and order-isomorphic to the rationals. As a result, there is no "next (...) in line for each [...]" like you claim there is. There are uncountably many closed intervals (actually, all of them are just single points) in between these open intervals, much like how there are uncountably many irrational numbers in between the rational numbers. There is no "next rational number" after each irrational number that you can use to get a bijection between rationals and irrationals, and the same thing is happening here.

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    $\begingroup$ I knew there was something fishy with that line... I just could not put my finger on it precisely. So basically the problem is that --- by the proof of (5) in fact --- we cannot "zoom in" enough in order to get the discrete picture I proposed (we cannot pick two (...)'s with no (...) in between). Hence the bijection is bogus. Very interesting. I'm just a beginner in analysis and in my opinion this shows some really deep and exciting stuff about the real number line. For some reason I found the problem a lot harder to spot than in your analogy. $\endgroup$ – Jori Sep 15 '16 at 22:13
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    $\begingroup$ I only now truly realize that even though something might be countable, it might not be possible to count them in an arbitrary order. This must mean that we cannot make infinite changes to functions and still consider them generally valid functions. Otherwise I could just take your favorite f : N -> Q and just interchange every pair that is not ordered right. This even though a single interchange is obviously allowed. In simple terms: what is order-isomorphism? $\endgroup$ – Jori Sep 15 '16 at 22:19
  • $\begingroup$ I find this is some sense very counter intuitive. You would expect that we can make infinite arbitrary changes to functions if every step was valid in the process and interpret them as such. Are there other examples of this phenomena that are somewhat more clearly absurd? Anyhow, many many thanks for your answer! +1 $\endgroup$ – Jori Sep 15 '16 at 22:22
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    $\begingroup$ The problem with "interchanging every pair" is that a single element might end up changing spots infinitely many times, and then there is no way to define where it ends up "in the limit". For an example that is more obviously absurd, here is a "proof" that there is a natural number $n\in\mathbb{N}$ such that $n>m$ for all $m\in\mathbb{N}$. The proof is to start with one natural number $n$, and then one by one modify it so that it is larger than each $m\in\mathbb{N}$. After doing this infinitely many times, you've found a natural number $n$ that is greater than every $m$! $\endgroup$ – Eric Wofsey Sep 15 '16 at 22:34
  • $\begingroup$ Thank-you that was very helpful. $\endgroup$ – Jori Sep 15 '16 at 23:11

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