# Well ordering theorem and Zorn's lemma implies the axiom of choice.

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.

• Yes, there are many questions which contain these exact proofs. It's an important skill for a mathematician to be able to read their own proofs and be critical. Is there a point where you think that you're missing something, compared to the proofs that already exist on the site? Mar 30 at 16:51
• Union of functions isn't necessarily a function. It is under certain conditions...
– user239203
Mar 30 at 16:57
• You also need to show $F$ is nonempty in order to apply Zorn... this is trivial in this case, and maybe even follows from the chain-union closure argument if you're careful about it, but it should be addressed. Mar 30 at 18:29

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.
• Isnt this true because $T^\ast$ is the union of the choice functions in T? Mar 30 at 18:30
• @spaceisdarkgreen Is it because the functions in T are in a chain. So for every $f_i \in T f_i \subseteq T^\ast$? Mar 30 at 18:41
• @Elise Yeah, it's because they're in a chain, but $f_i\in T\to f_i\subseteq T^*$ would hold regardless since $T^*:=\cup T,$ so that doesn't have anything to do with it. The reason a union of functions could fail to be a function is cause some of the functions in the union could take incompatible values, e.g. $(1,2)\in f_1$ and $(1,3)\in f_2.$ But this can't happen in a chain since we'd have $f_1\subseteq f_2$ or $f_2\subseteq f_1$ in which case one of the two would have to contain both incompatible points, which it can't since it's a function. Mar 30 at 18:55