Suppose $A$ is infinite. Also suppose $\exists g:A \rightarrow \mathbb {N}$ that is 1-1. This is all the info I have. I need to show the result in the title because then I would have some function $h$ that is a bijection from $A $ to $\mathbb{N} $ and that would mean that $A $ is countable. But I just do not see how to obtain that function $f $ in title...
Showing that $\exists f: \mathbb {N} \rightarrow A$ that is injection
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real-analysis
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0As long as $A$ contains at least one element, there is a function $f:\Bbb N\to A$. If you want something to do with injections, surjections and bijections, then you should say exactly what you're after, and exactly what you're given. – 2017-02-20
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0I assume you meant to say that $g$ is injective and that you'd like $f$ to be injective as well? If so, then note that [any infinite set has a countably infinite subset](http://math.stackexchange.com/questions/1126636/every-infinite-set-has-an-infinite-countable-subset) – 2017-02-20
1 Answers
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$|A| \leq \mathbb N$, by the existence of one injection ($g:A \to \mathbb N$). On the other hand, $|A|$ is infinite, so we know that $|\mathbb N| \leq |A|$. Hence there exists a bijection between them, and consider its inverse for the other injection.