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Are there any simple proofs that obtain this result? I haven't been able to find one online.

John Allingera
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  • Depends what you mean by simple. Once the idea of enumerability is understood and once the non-numerability of $\Bbb R$ is achieved, the fact you ask about follows quite readily. – Andrea Mori Feb 27 '14 at 20:17
  • The (real or complex) algebraics are countable. – André Nicolas Feb 27 '14 at 20:17
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    I think the "horror" about a century ago of discovering that the cardinality of the transcendentals is greater than that of the algebraic numbers was comparable to the "Pythagorean horror" of the irrational numbers. – colormegone Feb 27 '14 at 20:21

1 Answers1

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The word "simple" is very tricky. The usual proofs goes like this:

  1. Show that the union, and product, of two countable sets is countable.
  2. Show that $\Bbb Q$ is countable.
  3. Show that $\Bbb Q[x]$ is countable.
  4. Show that The algebraic numbers are countable.
  5. Show that the real numbers are uncountable.
  6. Show that the set of transcendental real numbers is uncountable.

The steps are not very difficult if you have seen them before. But if you haven't seen any set theoretic arguments before hand, then it might not be so simple.

Asaf Karagila
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  • @Henning: I didn't want to give out every single step of the proof, you're right, but this falls into the proof of (4). :-) – Asaf Karagila Feb 27 '14 at 20:21
  • On further thought it was a bad idea, because if you split it up that way you end up needing choice. – hmakholm left over Monica Feb 27 '14 at 20:29
  • @Henning: No, you don't. Countable union of finite sets of real numbers (or sets equipotent with the real numbers, or generally of sets which can be linearly ordered) are always countable. If anything the crux of avoiding the axiom of choice is in proving that $\Bbb Q[x]$ is countable. – Asaf Karagila Feb 27 '14 at 21:27
  • x @Asaf: Yes, but my proposal was to prove as a lemma "countable union of finite sets is countable", full stop -- without the assumption that the sets in question are sets of reals. – hmakholm left over Monica Feb 27 '14 at 21:37
  • Oh, in that case you're absolutely right. – Asaf Karagila Feb 27 '14 at 21:37