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How can I show that if a set of real numbers is countably infinite, then its complement is infinite but not countably infinite?

Thanks a lot in advance!

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    $\begingroup$ HINT: The union of two countable sets is countable. $\endgroup$ – Brian M. Scott Nov 20 '13 at 20:39
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Let the set you're looking for be $$ \{a_1,a_2,a_3,\ldots\}. $$ Suppose the complement is countably infinite, so that it is $$ \{b_1,b_2,b_3,\ldots\}. $$ Then consider the sequence $$ \{c_1,c_2,c_3,c_4,c_5,c_6\ldots\}=\{a_1,b_1,a_2,b_2,a_3,b_3,\ldots\}. $$ That is countably infinite and contains all real numbers. But that's impossible. So the complement of a countably infinite set of reals relative to the set of all reals cannot be countably infinite.

(Of course, this works only if you've proved the set of all reals is not countably infinite. If that's the point you're stuck on, it's time to ask how to do that. But maybe it's best to ask that question elsewhere than in this forum first and then come here if you run into difficulties.)

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HINT: If $|A|,|B|\leq\aleph_0$ then $|A\cup B|\leq\aleph_0$.

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