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This question already has an answer here:

Sorry if this problem is repeated.

Is there a nonempty perfect set in $\mathbb{R}^1$ which contains no rational number?

Proof sketch: This set must be uncountable because any nonempty perfect set in $\mathbb{R}^k$ is uncountable.

We know that Cantor set (C) is perfect set on a line. If we take $E=C+\sqrt{2}=\{x+\sqrt{2}: x\in C\}$. Will this set be an answer to above problem?

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marked as duplicate by Henno Brandsma general-topology May 11 '18 at 13:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See example 5 here. $\endgroup$ – David Mitra Jul 12 '15 at 15:56
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    $\begingroup$ See this also. $\endgroup$ – David Mitra Jul 12 '15 at 15:58
  • $\begingroup$ I read this. I concluded that my example is also satisfies. Am I true? $\endgroup$ – ZFR Jul 12 '15 at 16:21
  • $\begingroup$ Some translate would work See this. I'm not sure if $\sqrt 2$ does it... $\endgroup$ – David Mitra Jul 12 '15 at 16:37
  • $\begingroup$ I understood you! Because Cantor set contains irrational numbers. Quite possible that $x+r\in \mathbb{Q}$ where $x\in C$ and $r\in \mathbb{Q}^c$. Am I right? $\endgroup$ – ZFR Jul 12 '15 at 16:49
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The irrationals are homeomorphic to the space $\Bbb N^{\Bbb N}$ with the product topology; you’ll find two proofs here. The Cantor set is well-known to be homeomorphic to $\{0,1\}^{\Bbb N}$ with the product topology, where $\{0,1\}$ has the discrete topology. Thus, if $h:\Bbb N^{\Bbb N}\to\Bbb R\setminus\Bbb Q$ is a homeomorphism, $h\left[\{0,1\}^{\Bbb N}\right]$ is a Cantor set contained in the irrationals.

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I didn't look at the links posted in the comments. I may repeat an argument there in the below.

Suppose $E$ is the set of all $x\in [0,1]$ such that $x$ has a decimal expansion of the form

$$\tag 1 .\text {_} 0 \text {_ _} 0 \text {_ _ _} 0 \text {_ _ _ _} 0 \dots ,$$

where the entries in the blanks _ are either $1$ or $2.$ Such expansions are unique; we don't run into the "string of $9$'s" business. Therefore $E$ has the same cardinality as the set of binary sequences, hence is uncountable.

Because the expansions in $(1)$ are nonrepeating, $E$ contains no rational. If a sequence in $E$ converges, the corresponding decimal expansions converge in each slot (we don't run into the $.4999\bar 9 = .5000\bar 0$ problem). Hence $E$ is closed.

Finally $E$ is perfect: Let $x\in E.$ Suppose $x$ has infinitely many $1$'s in its expansion, say in the slots $n_1,n_2, \dots.$ Then $x+1/10^{n_k} \to x,$ and each $x+1/10^{n_k} \in E.$ If $x$ had $2$'s in the slots $n_1,n_2, \dots,$ we would instead look at $x-1/10^{n_k} \to x.$ So $E$ is perfect as desired.

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