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Online I found the following statement that for $X$ an infinite set, if there is a bijection from $X$ to $\mathbb{N}$ this is equivalent to there is an injection from $X$ to $\mathbb{N}$. How would I prove this?

Attempt: The fact that if there is a bijection it is also an injection is true by definition of bijection. Beyond that, I don't know how to demonstrate that an injection is equivalent to a bijection for an infinite set $X$ to $\mathbb{N}$.

Asaf Karagila
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frog1944
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  • "If there is a bijection from $X$ to $\Bbb N$ then there is an injection from $X$ to $\Bbb N$"... the bijection whose existence is guaranteed in the hypothesis is by definition of what a bijection is also an injection as well. "how to demonstrate an injection is equivalent to a bijection" That is false. The converse of the statement is false for general infinite sets. Just because an injection exists between infinite sets $X$ and $Y$ does not imply a bijection exists, nor does it imply that the injection talked about is a bijection itself either. – JMoravitz Mar 15 '18 at 05:42
  • That being said, $\Bbb N$ is something of a special case as it can be thought of as "the smallest" infinite set in a way. There are still able to be injections from $X$ to $\Bbb N$ which are not bijections (e.g. the function $f~:~\Bbb N\to\Bbb N$ given by $f(n)=2n$ is not surjective but is injective). – JMoravitz Mar 15 '18 at 05:46
  • Would you be able to check my solution @JMoravitz ? – Yunus Syed Mar 15 '18 at 05:49
  • So is fact 6 incorrect? https://gowers.wordpress.com/2011/11/28/a-short-post-on-countability-and-uncountability/ – frog1944 Mar 15 '18 at 09:19

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We start off by observing that $X$ can not be uncountable as $card(X) \le card(\mathbb{N})$ due to the existence of an injection $f: X \to \mathbb{N}$. However, since $X$ is infinite, it is countable. Hence, $X$ is in bijection with $\mathbb{N}$.

Yunus Syed
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