Jtdylktuy

Well-Ordering Principle implies Zorn's Lemma

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Let $(A,preccurlyeq)$ be a partially ordered set in which every chain has an upper bound.

By Well-Ordering Principle, there is a well-ordering $preccurlyeq'$ on $A$. Let $V$ be the class of all sets and $rm Ord$ be the class of all ordinals. First, we define function $f:mathcal{P}(A)setminus{emptyset} to A$ by $f(X)=min X$ (with regard to $preccurlyeq'$).

Next, we define function $G:V to V$ by $G(x)=f({ain A mid forall tin {rm ran}(x):t prec a})$ if $x$ is a function and ${ain A mid forall tin {rm ran}(x):t prec a} neq emptyset$, and $G(x)=A$ otherwise. By Transfinite Recursion Theorem, there is a function $F: {rm Ord} to V$ such that $F(alpha)=G(F restriction alpha)$ for all $alpha in {rm Ord}$.

It is not hard to verify (by Hartogs number) that $F(alpha)= A$ for some ordinal $alpha$. Let $lambda=min{alpha in {rm Ord} mid F(alpha)= A}$. Then ${rm ran}(Frestriction lambda)$ is clearly a chain in $(A,preccurlyeq)$ and has an upper bound $u in A$. If $u prec bar a$ for some $bar ain A$, we have $bar ain{ain A mid forall tin {rm ran}(Frestriction lambda):t prec a} neq emptyset$ and thus $F(lambda) in A$. This contradicts the fact that $F(lambda)=A$. So $u$ is the maximal element of $A$.

The proof itself is correct (oops), but saying "clearly a chain" is not a good thing to do, at least state what you are using here like you did with Hartogs number.

Let me suggest a more intuitive proof:

First, the intuition: assuming that Zorn's lemma is false, every chain is bounded but there is no maximal element. Now we create a sequence $x_beta$ to be strictly increasing sequence. Now if we build the sequence $langle x_betamid beta<alpharangle$ the set ${x_betamid beta<alpha}$ is a chain, so it has an upper bound, $y$, and because $y$ is not maximal element there is $x_alpha$ such that $yprec x_alpha$, so add $xalpha$ to the sequence, you can keep going and exhaust the ordinal like that, which is impossible.

Now, formal proof for the above:

Let $V,(A,preccurlyeq),f$ be defined as yours, assume that $A$ doesn't have maximal element, let $x_0in A$, and $H:Vto V$ defined as followed:

If $g$ is not order preserving function from some ordinal to $A$ then $H(g)=x_0$(the Do not care case).

If $g$ is a function from $betain Ord$ to $A$ and order preserving, then $g[beta]$ is a chain, let $U$ be the set of upper bounds of $g[beta]$, then take $f(U)$, now let $U'$ be the set of elements in $A$ that are greater than $f(U)$(with respect to $prec$), by assumption $U'ne emptyset$, so $H(g)=f(U')$.

Now, by the theorem there exists unique $F:Ordto A$ such that $F(beta)=H(Frestrictionbeta)$.

Use induction to prove that $Frestriction gamma:gammato A$ is order preserving for all $gamma$, so $Frestriction gamma:gammato A$ is injective, set $gamma=mbox{Hartogs’ number}$ and you got a contradiction.