When there is a one-to-one function from a set $A$ to a set $B$ , we say that $A$ is embeddable in $B$ , and write $A \le B$ . Thus $\le$ is a (class) binary relation on the class $V$ of all sets. This relation is clearly reflexive and transitive. If $A\le B$ and $B\le A$ , then, by Schröder-Bernstein theorem, $A$ is bijective to $B$ , $A\sim B$ . However, clearly $A\ne B$ in general. Therefore $\le$ fails to be a partial order. However, since $A\sim B$ iff they have the same cardinality, $|A|=|B|$ , and since cardinals are by definition sets, the class of all cardinals becomes a partially ordered set with partial order $\le$ . We record this result as a theorem:

With the addition of the axiom of choice, one can show that $\le$ is a linear order on the cardinals. In fact, the statement ``$\le$ is a linear order on the cardinals'' is equivalent to the axiom of choice.

Proof. Restating the second statement, we have that for any two sets $A,B$ , there is an injection from one to the other. The plan is to use Zorn's lemma to prove the second statement, and use the second statement to prove the well-ordering principle (WOP).

Zorn implies Statement 2:

Suppose there are no injections from $A$ to $B$ . We need to find an injection from $B$ to $A$ . We may assume that $B\ne \varnothing$ , for otherwise $\varnothing$ is an injection from $B$ to $A$ . Let $P$ be the collection of all partial injective functions from $B$ to $A$ . $P$ , as a collection of relations between $B$ and $A$ , is a set. $P\ne \varnothing$ , since any function from a singleton subset of $B$ into $A$ is in $P$ . Order $P$ by set inclusion, so $P$ becomes a poset. Suppose $F$ is a chain of partial functions in $P$ , let us look at $f:=\bigcup F$ . Suppose $(a,b),(a,c)\in f$ . Then $(a,b)\in p$ and $(a,c)\in q$ for some $p,q\in F$ . Since $F$ is a chain, one is a subset of the other, so say, $p\subseteq q$ . Then $(a,b)\in q$ , and since $q$ is a partial function, $b=c$ . This shows that $f$ is a partial function. Next, suppose $(a,c),(b,c)\in f$ . By the same argument used to show that $f$ is a function, we see that $a=b$ , so that $f$ is injective. Therefore $f\in P$ . Thus, by Zorn's lemma, $P$ has a maximal element $g$ . We want to show that $g$ is defined on all of $B$ . Now, $g$ can not be surjective, or else $g$ is a bijection from $\operatorname{dom}(g)$ onto $A$ . Then $g^{-1}: A\to B$ is an injection, contrary to the assumption. Therefore, we may pick an element $a\in A-\operatorname{ran}(g)$ . Now, if $\operatorname{dom}(g)\ne B$ , we may pick an element $b\in B-\operatorname{dom}(g)$ . Then the partial function $g^*: \operatorname{dom}(g)\cup \lbrace b\rbrace \to A$ given by

Since $g^*$ is injective by construction, $g^*\in P$ . Since $g^*$ properly extends $g$ , we have reached a contradiction, as $g$ is maximal in $P$ . Therefore the domain of $g$ is all of $B$ , and is our desired injective function from $B$ to $A$ .

Statement 2 implies WOP:

Let $A$ be a set and let $h(A)$ be its Hartogs number. Since $h(A)$ is not embeddable in $A$ , by statement 2, $A$ is embeddable in $h(A)$ . Let $f$ be the injection from $A$ to $h(A)$ is injective. Since $h(A)$ is an ordinal, it is well-ordered. Therefore, as $f(A)$ is well-ordered, and because $A\sim f(A)$ , $A$ itself is well-orderable via the well-ordering on $f(A)$ .

Since Zorn's lemma and the well-ordering principles are both equivalent to AC in ZF, the theorem is proved.