Definition: $|A|\le |B| \iff$ there exists an injection from $A$ to $B$.
Theorem: For any two sets $A$ and $B$, $|A|\le |B|$ or $|B|\le |A|$.
My attempt:
Assume $|B| \not\le |A|$, I will prove $|A|\le |B|$.
Let $X=\{f:A'\to B \mid A'\subseteq A \text{ and } f\text{ is injective}\}$. We define a partial order $<$ on $X$ by $$f_1<f_2 \iff f_1\subseteq f_2$$
Since $f:\emptyset\to B$ is injective, $f\in X$. Thus $X\neq\emptyset$. For any chain $Y$ from $X$, let $f^\ast=\bigcup Y$.
- $f^\ast$ is a mapping
For $(a,b_1),(a,b_2)\in f^\ast$, there are $f_1,f_2 \in Y$ such that $(a,b_1)\in f_1$ and $(a,b_2)\in f_2$. Since $Y$ is a chain, we can safely assume $f_1 < f_2$. It follows that $f_1 \subseteq f_2$. Thus $(a,b_1),(a,b_2)\in f_2$ and consequently $b_1=b_2$ by the fact that $f_2$ is a mapping.
- $f^\ast$ is injective
Assume $(a_1,b),(a_2,b)\in f^\ast$. Then there exists $f_1,f_2 \in Y$ such that $(a_1,b) \in f_1$ and $(a_2,b) \in f_2$. Since $Y$ is a chain, we can safely assume $f_1 < f_2$. It follows that $f_1 \subseteq f_2$. Thus $f_1(a_1)=f_2(a_1)$ and consequently $f_2(a_2)=b=f_1(a_1)=f_2(a_1)$. Hence $f_2(a_2)=f_2(a_1)$ and consequently $a_2=a_1$ by the fact that $f_2$ is injective. It follows that $f^\ast$ is injective.
To sum up, $f^\ast$ is injective and thus $f^\ast \in X$. Furthermore, $f^\ast$ is an upper bound of chain $Y$. Thus $X$ satisfies the requirement of Zorn's Lemma and hence has a maximal element $\bar{f}:\bar{A} \to B$. Let $\bar{B}=\operatorname{ran}\bar{f}$. Then $\bar{f}:\bar{A} \to \bar{B}$ is bijective.
I claim that $\bar{B}\subsetneq B$. If not, $\bar{B} = B$. Then $\bar{f}:\bar{A} \to B$ is bijective and thus $\bar{f}^{-1}:B \to \bar{A} \subseteq A$ is bijective. This contradicts our very first assumption that $|B| \not\le |A|$. Hence $\bar{B}\subsetneq B$ and thus $\exists b'\in B\setminus \bar{B}$. It follows that $b'\notin \bar{B}=\operatorname{ran}\bar{f}$.
I claim that $\bar{A}=A$. If not, $A\setminus \bar{A} \neq \emptyset$. Thus there exits $a'\in A\setminus \bar{A}$. It follows that $a'\notin \bar{A} = \operatorname{dom}\bar{f}$. We define $F:\bar{A}\cup\{a'\} \to \bar{B}\cup\{b'\}$ by $F(a)=\bar{f}(a)$ for all $a\in\bar{A}$ and $F(a')=b'$. It's clear that $F$ is a bijection from $\bar{A}\cup\{a'\}$ to $\bar{B}\cup\{b'\}$ and that $\bar{f}\subsetneq F$. Then $F$ is an injection from $\bar{A}\cup\{a'\}$ to $B$ and $\bar{f}\subsetneq F$. This contradicts the maximality of $\bar{f}$. Hence $\bar{A}=A$.
To sum up, $\bar{f}:A \to B$ is injective and thus $|A|\le |B|$.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
