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Does there exist a non empty perfect subset of $R$ containing only transcendental numbers ? I am looking for an example although even an existential proof will suffice.

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Here's a non-constructive proof. I'm still thinking about a constructive one.

The Cantor set is homeomorphic to its own square (i.e. $C \simeq C \times C$), and from this it follows that $C$ contains infinitely many disjoint copies of itself. If we construct $C$ on an interval with transcendental endpoints, then since the algebraic numbers are countable, then at least one of these copies of $C$ cannot contain any algebraic numbers.

I think a constructive proof using Cantor sets is also possible, but I'm not 100% sure. If I figure something out, I'll update.

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  • $\begingroup$ Sry, but I don't understand your argument, why a copy of $C$ cannot contain any algebraic numbers from the fact, that the algrebaic numbers are countable. $\endgroup$ – Mundron Schmidt Jun 6 '17 at 9:12
  • $\begingroup$ C has uncountable copies of itself, but algebraic numbers are countable. No matter how you spread those algebraic numbers out, there will be copies of C with no algebraic numbers. $\endgroup$ – Alfred Yerger Jun 6 '17 at 15:35

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