A set $A$ is set to be infinite if it is not finite, i.e. if there exists no $n\in\mathbb{N}$ such that $|A|=n$, meaning there exists a bijection $A\leftrightarrow\{1,\dotsc,n\}$. How do I prove that for any such set there necessarily exists an injective function $f:\mathbb{N}\to A$, meaning there are no infinite cardinalities under countability?
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I believe that if you find a subset of $A$ that has a bijection with $\mathbb{N}$, that will do. – Mike Pierce Dec 07 '14 at 22:39
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There are probably many other threads on the subject, and I encourage other people to suggest other duplicates. – Asaf Karagila Dec 07 '14 at 22:53
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Suppose $|A|<|\mathbb{N}|$. Then there is an injection $A\to \mathbb{N}$ onto a proper subset that has smaller cardinality than $\mathbb{N}$. However, all infinite subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$, so it follows that $A$ is finite.
Matt Samuel
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But you assume that either $|A|<\mathbb{N}$ or $\mathbb{N}<|A|$. Don't you use the axiom of choice here? – Peter Franek Dec 07 '14 at 23:45