Does there exist a subspace of the Cantor set homeomorphic to the rationals? I am looking for a full explanation of the following two questions, as well as a rather hand-wavy intuitive argument (if one exists).

Let $E$ be the set of endpoints in the Cantor set, $C$. Is it true that $E\cong\mathbb{Q}$?
If not, does there exist some $X\subset C$ such that $X\cong\mathbb{Q}$?

This paper mentions that the first question is true, then gives a short proof of the statement that a topological space $X$ can be embedded into $C$ if and only if $X$ is $T_0$, second-countable, and zero-dimensional, which, while certainly answering the second question, doesn't resolve (for me) why the image of such an embedding could be $E$.
 A: Yes, the family of all endpoints of intervals removed in the construction of the Cantor set is homeomorphic to the rationals. A quick way to see this is the following topological characterization of the rationals which can be found in 


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*Sierpiński, W., Sur une propriété topologique des ensembles dénombrables denses en soi., Fundamenta Mathematicae 1, 11-16 (1920). ZBL47.0175.03.



Theorem. Any countable metric space without isolated points is homeomorphic to $\mathbb{Q}$.

Calling this set of endpoints $Q$, it is clear that $Q$ is countable (there are only countably many intervals removed) and metric (as a subspace of $\mathbb R$). That $Q$ has no isolated points can be seen by noting that at stage $n$ in the construction of $C$ there intervals remaining all have length $3^{-n}$ (taking the zero-th step to be the closed unit interval $[0,1]$ of length $1 = 3^{-0}$). So if $x \in Q$ and $\varepsilon > 0$ is given, taking $n$ is large enough that $3^{-n} < \varepsilon$, then in the $n$th stage some interval of length $3^{-n}$ must have been removed from the left (or right) side of $x$ with its right (or left) endpoint within $\varepsilon$ of $x$.
