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Old November 28th, 2008, 12:09 PM
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Default Countability

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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Old November 28th, 2008, 12:49 PM
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Originally Posted by jas_viru View Post
Q:
Let A,B≠0 and let f:A→B be an onto mapping. Then if A is countable then prove B is countable.
This is one of the most important theorems in theory of cardinality of sets: 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 g(B)\subset A,\; g(B)\text { is countable }

That means that B \text{ is countable }.
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