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I tried to prove this using statement using the difference of sets $\mathbb{R}-\mathbb{Q}$ and the fact that $\mathbb{R}$ is not countable and $\mathbb{Q}$ is countable.

In general, is it possible to say that "Let $A$, $B$ and $C$ be sets such that $B$ is not countable and $C$ is countable. If $A=B-C$, then A is countable." ?

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    $\begingroup$ An uncountable set missing a countable subset is still uncountable. So no, $A$ is not enumerable. $\endgroup$
    – QC_QAOA
    Commented Apr 24, 2022 at 21:12
  • $\begingroup$ I think you may have an issue with language and negatives - what you have written does not make much sense, but you seem to have a better idea of the concepts than you have expressed in your question (assuming English is not your natural language). $\endgroup$
    – Mark Bennet
    Commented Apr 24, 2022 at 21:25
  • $\begingroup$ There are even subsets of the irrational numbers that are still uncountable, including every interval with a positive length , the transcendental numbers and the uncomputable numbers (being a subset of the transcendental numbers) $\endgroup$
    – Peter
    Commented Apr 25, 2022 at 10:41

1 Answer 1

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The set of irrational numbers is no enumarable. If $\mathbb{R}$ - $\mathbb{Q}$ was enumarable, $\mathbb{R}$ would be union two countable sets which is countable.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 24, 2022 at 21:22

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