Zorn's lemma states that
If $X$ is a partially ordered set such that every chain in $X$ has an upper bound, then $X$ contains a maximal element.
In his proof Halmos firstly replaces the abstract partial ordering by the inclusion order. Then he considers a set of all chains in $X$ (let's call it ${X_C}$) and shows that it has these properties:
- Every subset of each set in ${X_C}$ is in ${X_C}$.
- The union of each chain of sets in ${X_C}$ is in ${X_C}$.
These properties are derived partially from the fact that we are looking at inclusion order.
Later on he proves that every non-empty collection of a non-empty set, satisfying these two properties, has a maximal element, thus proving that the collection ${X_C}$ also has a maximal element. Therefore $X$ has a maximal element.
The problem I see here is that he proved Zorn's lemma only for sets that satisfy the two properties (e.g. sets with inclusion order). If we imagine a partially ordered set whose every chain has an upper bound, which has an ordering that does not lead to such properties, then the proof doesn't include it.
I would like to know what is wrong with my reasoning.
Best Answer
Every partial order $(X,\leq)$ is isomorphic to a partial order $(Y,\subseteq)$. Simply define $Y_x=\{y\in X\mid y\leq x\}$, then $Y_x\subseteq Y_{x'}$ if and only if $x\leq x'$.
Now. Add to $Y$ all sets of the form $Y_C=\bigcup_{x\in C}Y_x$ where $C$ is a chain. Since $C$ has an upper bound, any $Y_C$ is a subset of some $Y_x$ (and possibly equal to one).
If $Y$ (with the added $Y_C$'s) has a maximal element, it has to be of the form $Y_x$ for some $x$, since any $Y_C$ must be below (or equal to) some $Y_x$, and we are looking at a maximal element. Therefore $x$ itself was maximal in $X$.