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Let $A$ be an infinite subset of $\Bbb N$. Then there exists a bijection from $A$ to $\Bbb N$.

My attempt:

We define $f:A \to \Bbb N$ by $f(a)=|\{a'\in A\mid a'<a\}|$.

  1. $f$ is injective

For $a_1,a_2\in A$ and $a_1<a_2$, then $\{a'\in A\mid a'<a_1\} \subsetneq \{a'\in A\mid a'<a_2\}$, then $|\{a'\in A\mid a'<a_1\}| < |\{a'\in A\mid a'<a_2\}|$. Thus $f(a_1)<f(a_2)$. It follows that $f$ is injective.

  1. $f$ is surjective

Assume the contrary that $f$ is not surjective. Let $k=\min \{n \in \Bbb N \mid n \notin \operatorname{ran}f\}$. It's clear that $0 \in \operatorname{ran}f$ since $|f(\min A)|=|\{a'\in A\mid a'<\min A\}|=|\emptyset|=0$. It follows that $0<k$. Let $k=p+1$, then $p = f(b)=|\{a'\in A\mid a'<b\}|$ for some $b \in A$. Let $c=\min \{a' \in A \mid b<a'\}$, then $\{a'\in A\mid b \le a'<c\}=\{b\}$. Next we prove $f(c)=k$.

We have $\{a'\in A\mid a'<c\} =\{a'\in A\mid a'<b\} \bigcup \{a'\in A\mid b \le a'<c\}=\{a'\in A\mid a'<b\} \bigcup \{b\}$, thus $|\{a'\in A\mid a'<c\}|=|\{a'\in A\mid a'<b\}| + |\{b\}|=p+1$. Thus $f(c)=p+1=k$, and consequently $k \in \operatorname{ran}f$. This contradicts the fact that $k=\min \{n \in \Bbb N \mid n \notin \operatorname{ran}f\}$. Hence $f$ is surjective.

To sum up, $f$ is bijective.

Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

Update: As @saz suggested in his answer, I did not explicitly use the fact that $A$ is infinite in my proof.

Let $c=\min \{a' \in A \mid b<a'\}$...

To show that such $c$ does exists, we must prove that $\{a' \in A \mid b<a'\}\neq \emptyset$. This claim is justified by the fact that $A$ is infinite.

Akira
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    Define $$a_n:=\min{x\in A:\forall m<n: x\neq a_m}$$ where $a_0:=\min(A)$. Consider the mapping $f:\Bbb N\to X:n\mapsto a_n$. – Crosby Sep 10 '18 at 20:06
  • @Crosby I got your approach. I did utilize your approach in in another proof at https://math.stackexchange.com/questions/2909659/let-a-be-a-subset-of-bbb-n-without-a-greatest-element-then-there-exists-a. My main concern is whether my proof contains any error. Could you please have a check on it? – Akira Sep 11 '18 at 01:29
  • It honestly takes more effort than most people have to go through a long, complicated argument and check for correctness, when there's a much simpler argument available (like what Crosby suggests here). The point is that your $f$ is the collapse of $A'$ onto its order type. You can get surjectivity by observing that when you iterate least-element-selection $n+1$ times, the final $a$ you get has $f(a)=n.$ (So your $f$ is precisely the inverse of the bijection Crosby recommends you recursively construct.) – spaceisdarkgreen Sep 12 '18 at 04:28
  • Hi @spaceisdarkgreen, As I'm really aware of that, I tried to write the proof as clear as possible. Thus people don't take much time to figure out what's going on :) – Akira Sep 12 '18 at 04:35

2 Answers2

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As far as I can see, your reasoning is fine. Some remarks:

Remark 1: From your proof it is not really clear where you use that $A$ is infinite so I would stress this fact a bit more at the place(s) where you need it.

Remark 2:

Thus $f(a_1)<f(a_2)$. It follows that $f$ is injective

Perhaps you could expand this a bit and write

Thus $f(a_1)<f(a_2)$. This shows that $f$ is strictly increasing and hence injective.

This is probably a matter of taste, but in my opinion it often helps to name things. Many reader are aware of certain implications ( e.g. "strictly monotone $\implies$ injective" or "differentiable $\implies$ continuous or...) and it helps them to point them to the implication which you are using. Here it is quite obvious, but once you are dealing with more abstract/complicated objects, this becomes more important.

Remark 3:

Assume that $f$ is not surjective

This is certainly also a matter of taste, but I would find it more straight-forward to prove the assertion directly by induction. Essentially this is already what you are doing, but you are doing it (kind of artifically, in my opinion) by contradiction.

Claim: $k \in \text{ran} \, f$ for all $k \in \mathbb{N}_0$.

$k=0$: ... that's the first part of the first paragraph on surjectivity.

$k \to k+1$: If $k \in \text{ran} \, f$, then $k=f(b)$ for some $b \in A$ and you can follow exactly your reasoning to construct $c \in A$ such that $f(c)=k+1$.

saz
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  • Regarding your Remark 1, I think I'm sloppy where I Let $c=\min {a' \in A \mid b<a'}$ ... without justifying the existence of such $c$. To prove that such $c$ does exists, we must prove that ${a' \in A \mid b<a'}\neq \emptyset$. This claim is justified by the fact that $A$ is infinite. Did you find other places where I must appeal to the fact that $A$ is infinite but I did not? – Akira Sep 12 '18 at 14:48
  • Regarding your Remark 2, I will try to be as specific as possible from now on. Yes, to name things makes me truly understand what I wrote and helps other people figure out what's going on more easily. That's huge benefit. – Akira Sep 12 '18 at 14:53
  • Regarding your Remark 3, Yes, it's just a matter of taste. Your direct proof is more straightforward and clear. I will try to make use of this direct approach in other proofs. – Akira Sep 12 '18 at 14:55
  • I'm very very thankful for your answer. I must say it's really useful and helps me understand subtle points such as I did not mention the fact that $A$ is infinite where it should be mentioned. I have been waiting for your answer for several days ^^ – Akira Sep 12 '18 at 14:58
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    @LeAnhDung You are welcome. I think the well-definedness of $c$ is indeed the only place where you need that $A$ is infinite. – saz Sep 12 '18 at 15:05
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The proof of injectivity is correct. The fact that $A$ is infinite is not needed for this.

No contradiction is necessary for surjectivity.

First, $0=f(\min A)$.

Suppose $k=f(a)$; we want to find $a'\in A$ with $f(a')=k+1$. This will prove that the image of $f$ is $\mathbb{N}$.

Since $A$ is infinite, the set $B=\{b\in A:b>a\}$ is not empty. Let $a'=\min B$.

Then $\{x\in A:x<a'\}=\{x\in A:x<a\}\cup\{a\}$. One inclusion is clear. Let $y\in\{x\in A:x<a'\}$. If $y<a$, then $y\in\{x\in A:x<a\}$; otherwise $y\ge a$, so $a\le y<a'$ and we conclude $y=a$ by minimality of $a'$.

Since $\{x\in A:x<a\}\cap\{a\}=\emptyset$, we have $$ |\{x\in A:x<a'\}|=|\{x\in A:x<a\}\cup\{a\}|=k+1 $$

A different approach uses Cantor-Schröder-Bernstein and the following theorem (which depends on countable choice).

If $A$ is infinite, there exists an injective map $f\colon\mathbb{N}\to A$. Therefore $|\mathbb{N}|\le|A|$.

Here $A$ need not be a subset of $\mathbb{N}$.

Since $A\subseteq\mathbb{N}$, we have $|A|\le|\mathbb{N}|$.

By Cantor-Schröder-Bernstein, $|A|=|\mathbb{N}|$.

egreg
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