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Is the Cardinality of irrational numbers in the interval $(0,1)$ equal to Cardinality of real numbers in the interval $(0,1)$ ? Or, Is the Cardinality of all irrational numbers, equal to Cardinality of all real numbers ?

In other words, is this correct?

$$card (\text{real numbers})=card (\text{irrational numbers})=2^{\aleph_0}$$

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    Hint: The difference between the real numbers and the irrational numbers are the rational numbers, of which there are cardinality $\aleph_0$. – Michael Burr Jul 26 '19 at 18:17
  • Yes, that is correct. The cardinality of the rationals is $\aleph_0$, and a finite union of sets of this size cannot produce cardinality of the reals. – abiessu Jul 26 '19 at 18:18
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    yes, your last line is correct, The cardinality of the union of two infinite sets equals the larger of the two sets (at least with the axiom of choice, and maybe even without it, I don't remember). – Ned Jul 26 '19 at 18:19
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    Incidentally, rather than asking many questions about basic cardinal arithmetic here, you might want to find a good text (not that there's anything wrong with that, it's just an inefficient way to learn). This old thread has some useful sources. – Noah Schweber Jul 26 '19 at 19:15
  • @NoahSchweber Thank you very much. –  Jul 26 '19 at 20:51

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