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Show that there exists a bijection from a set that is countable and infinite into natural numbers.

I know that this question is a bit dumb but I can't prove it explicitly. I mean I can't reconcile the following two statements:

  1. Set $A$ is countably infinite, meaning there exists a bijection from $A$ to $N$.
  2. Set $A$ is countable, meaning there exists an injection from $A$ to $N$ and $A$ is infinite.

It looks intuitive but I can't find a clear bijection given (2).

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closed as off-topic by Andrés E. Caicedo, Daniel W. Farlow, Leucippus, John B, Tim Raczkowski Feb 26 '16 at 1:19

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  • $\begingroup$ This is directly by defintion $\endgroup$ – Micael Jul 12 '15 at 3:18
  • $\begingroup$ @bof sorry typo $\endgroup$ – user10024395 Jul 12 '15 at 3:31
  • $\begingroup$ @Micael the definition of countable infinite is 1. Therefore, I don't think it is directly by definition. $\endgroup$ – user10024395 Jul 12 '15 at 3:33
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I think you're asking the following:

Suppose $A$ is infinite, and has an injection into $\mathbb{N}$. Show that there is a bijection between $A$ and $\mathbb{N}$.

Here's how to show this:

Let $f: A\rightarrow \mathbb{N}$ be injective, and consider the image of $f$ - this image can be written as $im(f)=\{a_0<a_1<a_2< . . .\}$. Now, the image of $f$ need not be $\mathbb{N}$ itself - for instance, maybe $f$ maps $A$ onto the even numbers. However, this is easily fixed by providing a bijection $b$ between $im(f)$ and $\mathbb{N}$: $a_0\mapsto 0, a_1\mapsto 1, a_2\mapsto 2, . . . $

Then $b\circ f$ is a bijection from $A$ to $\mathbb{N}$.


EDIT: In detail, here's why any infinite subset of $\mathbb{N}$ is in bijection with all of $\mathbb{N}$: fix $A\subseteq\mathbb{N}$ infinite, and consider the following map $$F: A\rightarrow\mathbb{N}: a\mapsto\vert\{n\in A: n\le a\}\vert.$$ So, for instance, if $A=\{1, 3, 7, . . .\}$, then we would have $F(1)=1$, $F(3)=2$, $F(7)=3$. Let $B=\{F(a): a\in A\}$, and note that $F$ provides a bijection between $A$ and $B$; it is enough to show that $B=\mathbb{N}$.

To see this, note that $B$ is downwards closed: given any $n\in B$ and $m<n$, we have $m\in B$. Now $\mathbb{N}$ has the property that its only infinite downwards-closed subsets is the whole of $\mathbb{N}$. This is proved by induction: since $B$ is infinite, we know that for each $m\in B$ there is a $n\in B$ with $n>m$; since $B$ is downwards closed, this means we have $$m\in B\implies m+1\in B.$$ But this means that $B=\mathbb{N}$. (If you phrase induction as "the only inductive subset of $\mathbb{N}$ is $\mathbb{N}$," this is immediate; if you phrase induction as "any property $P$ which holds of 0, such that $P(n)\implies P(n+1)$, holds of all natural numbers," then this follows by considering the property "is in $B$.")

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  • $\begingroup$ why are you sure that there won't be duplicate in the set im(f)? I still don't quite understand. From what I see from your construction, I can turn any injection into N into a bijection. Is that true? $\endgroup$ – user10024395 Jul 12 '15 at 5:40
  • $\begingroup$ Since I only have injection into N, I can't be sure that im(f) is the entire N, such that I can have a bijection between im(f) and N, right? What happens if |im(f)| < |N|? Or can there be nothing that is infinite yet smaller than N? If so, how can one prove it? $\endgroup$ – user10024395 Jul 12 '15 at 5:48
  • $\begingroup$ @user136266 see my edit. $\endgroup$ – Noah Schweber Jul 12 '15 at 6:37
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Your set is countable and infinite. As your definition says countable infinite means there exist an injection from $A$ to $\mathbb N$. Let write $A$ as $\{a_1,a_2,a_3,...\}$. Suppose under the injective map $a_1\rightarrow n_1$, $a_2\rightarrow n_2,\dots$. Now to get bijection you simply map $n_1$ to $1$ $n_2$ to $2$, $\dots$. Thus we get a bijection from $A$ to $\mathbb N$.

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The Schroder--Bernstein theorem constructs a bijection between two sets $A$ and $B$ if there exists injections $$ f: A \to B $$ and $$ g: B \to A. $$ I think this answers your question.

(see my book Proof Patterns for further discussion and the proof of the theorem.)

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  • $\begingroup$ While technically true, this answer is far more complicated than necessary: there's no need to invoke SB to show that infinite sets injecting into $\mathbb{N}$ have bijections with $\mathbb{N}$. In particular, the proof of SB does not have a short easy proof, whereas the question the OP asks does. $\endgroup$ – Noah Schweber Jul 12 '15 at 3:44
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    $\begingroup$ whether or not it is necessary for this problem, it is something that the OP ought to know, and it certainly answers his question. $\endgroup$ – Mark Joshi Jul 12 '15 at 4:51

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