Q:

Let A,B≠0 and let f:A→B be an onto mapping. Then if A is countable then prove B is countable.

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- Nov 28th 2008, 10:09 AMjas_viruCountability
Q:

Let A,B≠0 and let f:A→B be an onto mapping. Then if A is countable then prove B is countable. - Nov 28th 2008, 10:49 AMPlato
This is one of the most important theorems in theory of cardinality of sets: $\displaystyle f: A \mapsto B \text{ is onto if and only if there is a one-to-one } g: B\mapsto A$.

Now any subset of a countable set is countable. Because $\displaystyle g(B)\subset A,\; g(B)\text { is countable }$

That means that $\displaystyle B \text{ is countable }$.