Let A be a countable set, and B is another set. Assume further that there exists an onto function f:A->B. Is B necessarily countable? Provide a full justification for your answer.
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How the proof proceeds depends on how you define "countable". A nonempty set
is countable if either:
1) there exists a one-to-one function
2) there exists an onto function
It turns out these are equivalent, so go with the second.
Since is countable there exists an onto function , and by hypothesis there is an onto function . The composition is onto, so that B
B is countable.