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Give an example of a perfect set in $\mathbb R$ that does not contain any of the rationals.

I found out a proof here using continued fractions. I have trying to understand the proof.

In continued fraction, I studied (from Burton's Elementary Number Theory) that if $\frac{p_n}{q_n}$ is the $n$th convergent i.e., if $\frac{p_n}{q_n} =[a_0;, a_1,a_2,\cdots,a_n]$ convergent to the irrational number $x$ then $$\Big|\frac{p_n}{q_n}-x\Big|<\frac{1}{q_{n+1} q_n}$$

But in the proof (in the given link), the author has written $\Big|x_n-x\Big|<\frac{1}{q_{n-1} q_n}$.

How to construct $x_n$? what is the relation between $x_n$ and $[a_0;, a_1,a_2,\cdots,a_n]$ in this context?

I did not understand it. Please help me.

user1234
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  • Can you recall what a perfect set is? – Giuseppe Negro Jul 25 '22 at 18:15
  • @GiuseppeNegro $E$ is said to be perfect if $E = E^\prime$. Here $E^\prime$ is the set of all limit points of $E$ – user1234 Jul 25 '22 at 18:18
  • Another definition: $E$ is closed and every point of $E$ is a limit point of $E$. – user1234 Jul 25 '22 at 18:19
  • @user1234 Note that the empty set is a perfect subset of $\mathbb R$ and contains no rational numbers. Presemably the OP wants a nonempty perfect set. – bof Jul 25 '22 at 20:50
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    If $C$ is the Cantor set and $T={t\in\mathbb R:(C+t)\cap\mathbb Q\ne\varnothing}$ then the set $T=\bigcup_{r\in\mathbb Q}(r-C)$ is meager and has Lebesgue measure zero. If $t\in\mathbb R\setminus T$ then $C+t$ is a nonempty perfect set containing no rational numbers. – bof Jul 25 '22 at 22:41

1 Answers1

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If $x_n = p_n/q_n$ then $$ \left| x_n - x \right| < \frac{1}{q_n q_{n+1}} $$

Robert Israel
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