For a set A to be proved uncountable it seems to be that a bijection onto infinite binary strings (B) is sufficient, but I would also expect a surjection from A to B to be sufficient since that would imply that some subset of A can be bijected onto B. Am I correct in this assumption, or is it possible for a superset of an uncountable set to be countable?
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You are right that this is enough. – Wojowu Apr 14 '17 at 10:35
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Yes, this shows uncountability. More precisely, it shows $|A|\ge2^{\aleph_0}$. Re your last question: Look at it the other way round: Every subset of a countable set is countable (because every subset of $\Bbb N$ is) – Hagen von Eitzen Apr 14 '17 at 10:37
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Related but not quite what you want: Schröder–Bernstein theorem
More relevant, this previous question: Is there a Cantor-Schroder-Bernstein statement about surjective maps?
So, if you can get injections both ways between two sets or surjections both ways between two sets then they have the same cardinality.
If you care about the use of the Axiom of Choice then read them carefully.
badjohn
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