# Proving that well ordering principle implies Zorn's Lemma.

I am trying to prove that well ordering principle implies Zorn's Lemma. I think that I'm close but don't quite know to make the last step of my proof. Here is what I wrote so far:

Given that on every set, a well ordering can be defined, we should prove that Given a partially ordered set $A$, if every increasing chain in $A$ has a maximal element, Then $A$ has a maximal element.

Proof: Take $A$ partially ordered by $R$. We know that there exists a well ordering $S$ on $A$. Let $k$ be the smallest ordinal s.t. $k=|A|$ and let, $k^+=k+1$. Define by transfinite induction, a function, $g:k^+ \rightarrow A$ as follows:

1. $g(0)$ is the first element in $A$ by $S$.

For any , $\alpha < k^+$:

1. If $\alpha$ is a successor ordinal, s.t. $\alpha = \beta + 1$, then, define $g(\alpha)$, the first element (by $S$), $a \in A$ such that $g(\beta) <_{R} a$

2. if $\alpha$ is a limit ordinal, then, The set $\{g(\beta);\beta<\alpha\}$, is linearly ordered by $R$. Therefor it has an upper bound. From all the uppers bounds, we will take the first (By $S$) to be $g(a)$.

3. $g(k^{+})$ is linearly ordered. So, by the lemma assumption, it has an upper bound in $M \in A$.

4. We claim that $M$ is a maximal element of $A$. Because, if there would be $x >_{R} M$ in $A$, by the construction of $g$, $g(k^{+})$ would contain an element which ia greater or equall (by $R$) to $x$, contradicting the fact that $M$ is an upper bound of $g(k^{+})$.

The step which I'm not sure of is step 5. I am not sure weather the fact that $|k|=|A|$ and that $g(k^{+})$ is isomorphic to $k$ are enough. What do you think?

Thank you! Shir

Some issues:

1. Zorn's lemma is about partial ordered sets where every chain has an upper bound, not a maximal element. It can be very easy to arrange a chain without a maximal element in a partial ordered set which has an infinite chain to begin with.

2. Using $\kappa^+$ to denote $\kappa\setminus\{\varnothing\}$ is a horrible choice of notation, since $\kappa^+$ denotes the successor cardinal of $\kappa$.

3. You need to slightly modify the construction of the chain. Suppose that $\{g(\beta)\mid\beta<\alpha\}$ were defined, then $g(\alpha)$ is the least (in the well-order) such that $\{g(\beta)\mid\beta\leq\alpha\}$ is a chain. If no such choice is possible, then $\{g(\beta)\mid\beta<\alpha\}$ is a maximal chain already.

4. You didn't have to use contradiction here.

• I have thought about it and still have a problem. I don't think that the chain is neccesarly maximal. For instance, if $\alpha$ is a successor ordinal and $\alpha = \beta + 1$, I took, $g(\alpha)$ to be the first element $a$ by $S$ s.t. $g(\beta) < a$. but, as I understand the construction, there could be an element $g(\beta)<_{R}x<_{R}a$ in $A$. – topsi Nov 18 '13 at 10:55
• @Shir: Yes, you are right, you need to slightly modify your approach. I'll slightly modify the answer. – Asaf Karagila Nov 18 '13 at 11:57
• I have tried to complete the proof, and eddited what I wrote above. what do you think?? Thanks! – topsi Nov 18 '13 at 14:50