Proof that the set of irrational numbers is uncountable?

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brulotfao · Accepted Answer

Given that the reals are uncountable (which can be shown via Cantor diagonalization) and the rationals are countable, the irrationals are the reals with the rationals removed, which is uncountable. (Or, since the reals are the union of the rationals and the irrationals, if the irrationals were countable, the reals would be the union of two countable sets and would have to be countable, so the irrationals must be uncountable.)

Madison Costa · Accepted Answer

Assume for contradiction that the irrational numbers are countable. Now let       q    1    ,      q    2    ,  … be an enumeration of the rationals, and let       r    1    ,      r    2    ,  … be an enumeration of the irrationals. Now set       F    i    =      R    ∖  {      q    i    ,      r    i    }. Then the sets       F    i   are open and dense in the usual topology on       R  , and so by the Baire Category Theorem,       ⋂          i      =      1        ∞        F    i   is dense in       R  . However,       ⋂          i      =      1        ∞        F    i    =  ∅ which is not dense, and hence the irrationals mustn&#039;t have been countable.

Proof that the set of irrational numbers is uncountable?

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