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I'm trying to prove the following statement related to the Zermelo–Fraenkel set theory, which looks rather basic yet I'm still unable to solve it.

Problem: Let $A$ be a nonempty well-ordered set. Prove that for every set $B$ the following two statements are equivalent:

  • There exists an injective function from $A$ to $B$.
  • There exists surjective function from $B$ to $A$.

Any comments, ideas and suggestions would be very much welcome. As usual, thank you in advance!

EDIT: Here is the a modification of the original:

Let $A$ be a nonempty well-ordered set. Prove that for every set $B$ such that there exists an injective function from $A$ to $B$ then there exists surjective function from $B$ to $A$ and vice versa.

e2l3n
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1 Answers1

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Pedro is right. You need $B$ to be well-ordered.

It is possible that there is a set $A$ and an ordinal $\alpha$ such that $A$ can be mapped onto $\alpha$, and $\alpha$ cannot be mapped injectively into $A$. Some examples include:

  1. You can have $A=\Bbb R$ and $\alpha=\omega_1$ (the least uncountable ordinal).
  2. You can have $A\subseteq\Bbb R$ and $\alpha=\omega$.

On the other hand, if $B$ is well-ordered, and not $A$ (or if $A$ is mapped onto $B$, and $B$ is injected into $A$), then you can do the following:

  1. Prove that in general, if $X$ is non-empty, and $f\colon X\to Y$ is injective, then there is a surjective map $g\colon Y\to X$; by dealing with those elements of $Y$ which are in the image of $f$; and separately with those which are not in the image of $f$ (here you use the fact $X$ is non-empty).

  2. Use the fact that $A$ is well-ordered, so if $f\colon A\to B$ is surjective, you can pick canonically $a_b\in A$ such that $f(a_b)=b$ for each $b\in B$; and this guarantees that you defined an injection (why?)

Asaf Karagila
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  • Thank you for the detailed explanation! Just one thing that is still unclear to me - why do you say "$α$ cannot be mapped injectively into $A$" since by hypotehsis the injection is $f:A→α$, not $f:α→A$ ? – e2l3n Feb 18 '15 at 13:29
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    Here $A$ is not well-ordered; $\alpha$ is a well-ordered set. The roles of $A$ and $B$ are reversed here. – Asaf Karagila Feb 18 '15 at 13:32
  • I wonder what if we assume that B is such set that there exists an injection from A to B. Actually this is the easy part to prove the existence of a surjection. I'm more interested in the other way round that is if there exists surjection from B to A then there is an injection from A to B. In particular I wonder if it's possible to find such an injection using the fact that A is well-ordered set. – e2l3n Feb 21 '15 at 17:20
  • No, and these are the two examples I gave. It is possible that there is a surjection from a set $A$ onto $\omega$, but there is no injection from $\omega$ into $A$. – Asaf Karagila Feb 21 '15 at 17:22
  • I understand this, but isn't it also possible to exist a set B for which exists surjection from B to A. If we assume it exists then there might be an injection from A to B. You are saying that the rule doesn't apply for two or more particular cases. – e2l3n Feb 21 '15 at 17:28
  • I *really* have a hard time understanding what you're trying to say. Possible to exist? Sure. Take $B=A$ then there is an injection, surjection, bijection, anything you like. – Asaf Karagila Feb 21 '15 at 17:30
  • Sorry for not being clear enough. Please take a look at the edit on my original post. – e2l3n Feb 21 '15 at 17:34
  • The first part is actually what I proved in the last half of my answer here; the "vice versa" part is simply not provable. I don't know how to make that any clearer than what I have said already, sorry. – Asaf Karagila Feb 21 '15 at 17:36