I am curently studying the well ordering theorem's(WOL) and zorn's lemmas's(ZL) equivalence with the axiom of choice(AOC). I have constucted the proofs of WOL and ZL implying AOC as below:
Well ordering theorem implies the axiom of choice.
Proof: Let $S$ be a collection of non empty sets and by the well ordering theorem there exists a linear relation $\leq$ such that ($\cup S$,$\leq$) is well ordered. Consequently, there is a least element $m$ for every $ s \in S$. Then, the function $F:S \rightarrow \cup S$ where $F(s)=m \in s $ for every $s \in S$ is a choice function which chooses the least element each time.
Zorn's lemma implies the axiom of choice
Let $A$ be the collection of non empty sets and $F$ be the collection of choice functions $f$ such that the domain of $f$, dom($f) \subseteq A$ and $f(a) \in a,$ for all $a \in A$. Define the following partial order: $f_1 \leq f_2$ if and only if $f_1 \subseteq f_2$.
Thus, $(F,\subseteq)$ is a poset and $T$ be a linearly ordered subset (a chain) of the poset. Then $T^\ast$ be the union of functions $f$ in $T$, i.e., $T^\ast = \cup T$. Here $T^\ast$ is a function as the union of functions is a function and also the upper bound of $T$ such that $T^\ast \in F$. Then by the Zorn's lemma there exists a maximal function $f_{max}$ in $(F,\subseteq)$.Suppose that the domain of $f_{max} \neq X$. Then there is some element $x \in X$ where $x \notin dom(f_{max}$). Define a choice function $g$ on $x$. Then let $f^\ast = f_{max} \cup g$ which is a contradiction since $f_{max}$ is the maximal element.
I realise there are many questions requesting help on this particular subject. However, i would greatly appreciate any comments on any missing details or mistakes in my proofs above. Thank you very much.
The proofs are fine, but if the proofs is supposed to show that the student understood Zorn's lemma, then I'd expect to see the proof that $T^*$ is in $F$, rather than just stating that it is.