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I am having trouble seeing why Cantor set has uncountably many elements.

A cantor set $C$ is closed. So $[0,1] - C = \bigcup\limits_{n=1}^{\infty} I_n$ is open and is countable union of disjoint open intervals. I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them. So between $I_n=(a_n,b_n)$ and $I_{n+1} = (a_{n+1},b_{n+1})$, we must have $a_n < b_n \leq a_{n+1} < b_{n+1}$. If $b_n < a_{n+1}$, then the Cantor set $C$ consists of an interval, which is a contradiction, so $b_n = a_{n+1}$ for all $n$, and thus Cantor set can have at most countably many points.

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    $\begingroup$ "I can further assume that I can order the {In} by their left endpoints since there are only countably many of them." You can list the left endpoints in an order but it needn't (and won't) but an order based on magnitude. Note: the rationals are also countable and can not be so ordered either. $\endgroup$ – fleablood Nov 25 '20 at 5:35
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    $\begingroup$ You confuse countability with order type; given a countable set which has a total order defined on it, it is in general not possible to enumerate it by the natural numbers in a way that preserves the original order (see rationals with their usual order); $\endgroup$ – Conrad Nov 25 '20 at 5:36
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The error in your reasoning is the assumption that a countable set of numbers can be ordered. For example, consider the set of rational numbers, countable, but can't be ordered ('ordering' here means enumerating in a sequence such that $\alpha_1<\alpha_2<\dots$).

A simple way to see that the cantor set is uncountable is to observe that all numbers between $0$ and $1$ with ternary expansion consisting of only $0$ and $2$ are part of cantor set. Since there are uncountably many such sequences, so cantor set is uncountable.

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    $\begingroup$ @BrianM.Scott, I've edited it. Thank you. $\endgroup$ – Martund Nov 25 '20 at 5:40
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I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them.

No. Why would you think you can? Consider for instance the countably many numbers $$ \bigl\{\tfrac1n:\ n\in\mathbb N\bigr\}\cup\bigl\{\tfrac12-\tfrac1n:\ n\in\mathbb N\bigr\}. $$ As long as there is more than one accumulation point, you cannot expect to order them indexed by integers.

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I can further assume that I can order the $\{I_n\}$ by their left endpoints since there are only countably many of them.

By this logic, it should also be possible to enumerate the rational numbers in order. But that's absurd.

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I'm not following your argument well enough to see exactly where it goes wrong... One question you might ask yourself is "does this show that every closed set is countable?" What is special about the cantor set here? I'm not seeing it.

As for why the cantor set is uncountable, consider this:

At each finite level of the cantor set construction, we "throw out" the middle third of each piece. So we have a decision to make at each stage: Do we go left? or do we go right?

E.g., We start in $[0,1]$. Then We have to decide to go into $[0,\frac{1}{3}]$ or into $[\frac{2}{3},1]$. Let's say we go left. Now we have the choice of going into $[0,\frac{1}{9}]$ or $[\frac{2}{9},\frac{1}{3}]$.

You can see that every countable sequence of choices (left or right) gives a unique point of the cantor set. Moreover, every point of the cantor set corresponds to such a sequence of choices. So if we write $0$ for "left" and $1$ for "right, the points of cantor set are in bijection with the infinite strings of $0$s and $1$s.

As a fun aside, the topological structure actually agrees as well! That's why you'll often see people call the cantor set $2^\omega$. In set theoretic language, that basically translates to "infinite sequences of $0$s and $1$s".

Ok, but now there must be uncountably many infinite sequences of $0$s and $1$s by a diagonalization argument. So the cantor set is uncountable too.


I hope this helps ^_^

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