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Suppose $A$ and $B$ are two sets with $B\subset A.$ Let $f:A\to B$ be injective. Then show that $\exists$ a bijection $h:A\to B.$

The proof given was :

Let $$X=(A-B)\cup f(A-B)\cup f^2(A-B)\cup \cdots=\bigcup_{n=0}^{\infty}f^n(A- B);f^0(A-B)=A-B.$$

Then clearly if $x\in X,x\in f^n(A-B)$ for some $n.$ Then $f(x)\in f^{n+1}(A-B)\implies f(x)\in X.$

Thus, $f(X)\subset X.$ We define, $$h:A\to B$$ by

$$h(x)=f(x)\text{, if }x\in X,$$ $$x\text{ otherwise. }$$

To show that $h$ is one-one, let $x,y\in A$ and $h(x)=h(y).$

If $x,y\in X$ then $f(x)=f(y)\implies x=y$

If $x,y\in A-X$ then, $x=y,$ by definition of $h.$

If $x\in X$ and $y\in A-X$ then $$h(x)=h(y)\implies f(x)=y\implies y\in f(X)\subset X,$$ a contradiction.

So, the last case can't occur. Thus, $h$ is one-one.

To show that $h$ is onto, let $y\in B.$

If $y\in X$ then $y\in f^n(A-B)$ for some $n\geq 1\implies y=f^n(x),$ for some $x\in A-B$ $$\implies y=f(x')$$ where $x'=f^{n-1}(x)\in f(A-B)\subset X.$ $$\implies y=h(x').$$

If $y\notin X$ thhen, $h(y)=y.$ Thus, $h$ is onto. This completes the proof.

However, I find this proof much convoluted, which was not at all necessary. I tried devising an alternative proff, which is quite straight-foerward in it's working and might even be considered simpler.

Here it is :

Notation: If S is a set, then by $|S|$ I mean, Cardinality of the set $S$

Given,

$B$ is a subset of $A$ and $f:A\to B$ is an injective mapping.

Since, $B$ is a subset of $A,$ we have, $|B|\leq |A|.$

Also, $f$ is an injective mapping from $A$ to $B$ implies, $|A|\leq |B|.$

Hence, $|A|=|B|$ is the only possible conclusion. Again, as $f$ is an injection from $A$ to $B$ (and with the (now) established fact that, $|A|=|B|,$ ) $f$ is a bijection. Thus, $f=h$ and $h:A\to B$ is a bijection.

I hope that my alternative approach is a correct one. Is there any way, I can improve the proof? I am looking, for some specific suggestions that might make my proof look more readable and understandable. Any suggestions/remarks against ways of improving it, will be highly appreciated. Also, please do point out if any requisite changes should be made or not. Lastly, if there are any errors in my proof, please correct me.

Thomas Finley
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    Pick $A=\mathbb{N}=B$ and $f:\mathbb{N}\rightarrow \mathbb{N}, f(n)=42 n$. The map $f$ is injective, but surely not bijective ($41$ is not in the image). Also, how do you know $\vert A\vert \leq \vert B\vert$ and $\vert B\vert \leq \vert A\vert$ implies $\vert A\vert =\vert B\vert$? This is pretty much the statement you wish to prove. – Severin Schraven May 22 '23 at 07:49
  • You might find this interesting https://math.stackexchange.com/questions/659964/when-do-surjections-split-in-zf-two-surjections-imply-bijection – Severin Schraven May 22 '23 at 07:56
  • The result is an immediate corollary of the Cantor-Schroeder-Bernstein theorem (a.k.a Cantor-Bernstein, a.k.a. Schroeder-Bernstein): If $f:A\to B$ is injective and $g:B\to A$ is injective, then there exists a bijection $h:A\to B$. Here, we can let $g=id_B$. I have seen a long complicated proof of the C-S-B theorem,and a short simple proof, e.g. in Introduction To Topology And Modern Analysis, by Simmons. – DanielWainfleet May 22 '23 at 08:26
  • How do you conclude: "Hence, |A|=|B| is the only possible conclusion."? If you have the Cantor-Bernstein theorem, this follows directly (and the proof I have is about a page long). And in the absence of anything similar, it's not apparent. – Porky May 22 '23 at 08:52
  • @Porky If $a\leq b$ and $b\leq a$ then $a=b$ ? This is obvious. Am I missing something? – Thomas Finley May 22 '23 at 08:54
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    If by $|A| \le |B|$ you mean there is an injection from $A$ into $B$, (Or $A$ is dominated by $B$) which it looks like you do, then it's not obvious. What do you mean by $|A| \le |B|$? – Porky May 22 '23 at 09:00
  • @Porky Well, I mean cardinality of $A$ is less than or equal to cardinality of $B$ . – Thomas Finley May 22 '23 at 09:05
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    How do you compare cardinalities, i.e. what means "cardinality of $A$ is less than the cardinality of $B$"? Porky gave you the standard definition of the order relation on cardinal numbers, what is yours? – Severin Schraven May 22 '23 at 09:11
  • Yes, to say the cardinality of $A$ is less than or equal the cardinality of $B$ means the same thing as stating that there is an injection from $A$ into $B$. – Porky May 22 '23 at 09:16
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    @SeverinSchraven I think I understand my mistake now, from the point where you mention the word "cardinal numbers". I am considering $|A|$ to be a real number , just like we say, "number of elements in $A$ " but my proof is valid only if, each of the sets are finite . If we consider the sets to be infinite, then my proof dont have any meaning. Is my understanding of this situation correct?( I ask Porky the same thing as well.) – Thomas Finley May 22 '23 at 09:17
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    Indeed, if you are only considering finite sets, then your statements become correct. If you throw infinite sets into the mix, things become more complicated. It would be nice if you could post an extended version of your comment as an answer. I think it is both a valuable question and answer for people starting off with set theory. – Severin Schraven May 22 '23 at 09:18
  • @SeverinSchraven Ah, That's where the problem is! Thank you! The mistake I was making was that, I was not taking infinite sets into consideration. Can you suggest some reference, where I can learn about cardinal numbers of infinite sets? – Thomas Finley May 22 '23 at 09:20
  • I am really sorry, I do not know much set theory and cannot point you to any reasonable source. But there are a lot of people here that would know. – Severin Schraven May 22 '23 at 09:25
  • @SeverinSchraven No problem! I have added an answer as you suggested. I hope this answer fits good in here. Please feel free to correct me if something is incorrect in the answer I posted. Thank you! – Thomas Finley May 22 '23 at 09:35
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    I like your answer :) I must admit that I would not have managed to guess that this was the issue. Porky saved the day. – Severin Schraven May 22 '23 at 09:40
  • @SeverinSchraven Ha ha! Yes you are so right! – Thomas Finley May 22 '23 at 09:42

1 Answers1

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The above alternative proof is justified/valid only under the assumption, that we are working with finite sets.

To elaborate, the proof might be true in general, but it's not a complete one. It is incomplete because, while dealing with cardinalities, of $A$ and $B$, i.e $|A|$ and $|B|$ we have used the following implication :

If $|A|\leq |B|$ and $|B|\leq |A|$ , then $|A|=|B|.$

This is obvious, if $|A|,|B|$ are real numbers. More appropriate to say, the above fact, makes sense only if $A,B$ are finite sets.

The problem comes, when we add infinite sets to the picture. If atleast either of the sets $A$ or $B$ are infinite, we can't treat $|A|$ and $|B|$ as real numbers. They are then, treated as so called "Cardinal Numbers ". In case of Cardinal numbers, the properties applicable for real numbers might not be even valid. One must prove them, in order to establish their validity/correctness.

In short: The proof in OP is correct if all the sets considered in the question, are finite sets. If either of $A$ or $B$ is infinite, then the proof is invalid and erroneous.

Thomas Finley
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