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This question is about a blog post by Terrence Tao showing that the real line $\mathbb{R}$ cannot be expressed as a disjoint union of countably many closed intervals. There are many proofs for this result, but I'm interested in one particular method. We suppose that $\mathbb{R} = \bigsqcup_{n=1}^\infty [a_n,b_n]$, and let $E := \bigsqcup_{n=1}^\infty \{a_n, b_n\}$ be the set of endpoints of the intervals. From here Tao shows that $E$ is closed and perfect, and thus uncountable by a weaker form of the Baire Category Theorem. This is a contradiction.

At the end of the post, Tao mentions that we could alternatively show that $E$ has the topological structure of a Cantor set to conclude that $E$ is uncountable. I would like to formalize this by constructing a homeomorphism $f : C \to E$, where $C \subseteq [0,1]$ is the middle-thirds Cantor set; but I'm not sure how to do this. $E$ certainly "looks" like $C$ from an intuitive standpoint, but has some key differences. Most notably, $C$ has a largest and a smallest element, whereas $E$ does not. Thus we can't hope to make $f$ order-preserving, which was my original plan of attack. Maybe $E$ isn't actually homeomorphic to $C$, but rather to $C \setminus \{0,1\}$? This feels plausible, but again I'm not sure. How should I proceed?

Edit: I realize now that $E$ and $C$ can't possibly be homeomorphic, since $C$ is compact while $E$ is not. So maybe the homeomorphism is in fact between $C \setminus \{0,1\}$ and $E$?

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You are correct, it should be the Cantor set without endpoints.

The way we construct it is quite similar: define $A_0=\mathbb{R}$ and $A_n=A_{n-1}\backslash (a_n,b_n)$. Note that intervals are open. Then define $C=\bigcap_{i=1}^\infty A_n$. So, if all $[a_n,b_n]$ cover $\mathbb{R}$ then we have to have $C=E$.

This $C$ space is a variant of Cantor set, except non-compact and with non-equal pieces at each step of the construction. It deifnitely is uncountable by the same argument why the Cantor set is. And it goes as follows:

Given a function $f:\mathbb{N}\to\{-1,1\}$ we can associate it with a point in $C$. First define a sequence $I_n$ of subsets as follows: $I_0=\mathbb{R}$, while $I_n$ is the intersection of $I_{n-1}$ with either $(-\infty,a_m)$ if $f(n)=-1$ or with $(b_{m'},\infty)$ if $f(n)=1$. Those $m,m'$ are the smallest $m,m'$ making the intersection nonempty (if you draw how those intervals may distribute you will know why it is so). So in other words the $f$ sequence gives us a binary walk in the Cantor set construction: $-1$ tells us to go left, while $1$ to go right at the next $[a_m,b_m]$ interval we see. Then define a sequence $x_n$ as an arbitrary point of $I_n$. As long as $f$ changes sign at least once it will give us a bounded sequence, and in fact convergent. Then the limit has to belong to $C$, and it is a matter of calculation that the association $f\mapsto\lim x_n$ has to be injective. Of course we need to get rid of the two problematic constant functions $f\equiv 1$ and $f\equiv -1$ (which in the normal Cantor set would yield endpoints).

And so since there are uncountable many functions $\mathbb{N}\to\{-1,1\}$ we must have that $C$ is uncountable. The association $f\mapsto\lim x_n$ is also surjective. And it is likely that it is continuous (with the infinite product of discrete spaces as topology for $\{-1,1\}^\mathbb{N}$), which leads to the conclusion that $E$ has to be homeomorphic to the standard Cantor set minus two endpoints. Although honestly I did not check continuity, and can't guarantee that.

That's the idea with a sketch anyway.

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