A description/proof of that Hausdorff maximal principle: chain in partially-ordered set is contained in maximal chain
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of partially-ordered set.
- The reader knows a definition of chain in set.
- The reader knows a definition of maximal element of set.
- The reader admits the Zorn's lemma: for any set such that for every chain, the union of the chain is a member of the set, the set has a maximal element.
Target Context
- The reader will have a description and a proof of the Hausdorff maximal principle that any chain in any partially-ordered set is contained in a maximal chain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any partially-ordered set, \(\langle S, R \rangle\), any chain, \(S_1 \subseteq S\), is contained in a maximal chain, \(S_2 \subseteq S\), where 'maximal chain' means a maximal element of the set of all the chains by the inclusion ordering, where there may be multiple maximal chains.
2: Proof
Let us take the set of all the chains in \(S\) that contain \(S_1\), with the inclusion ordering, denoted as \(S'\). \(S'\) is really a set as a subset of the power set of \(S\) (a formula for the subset axiom is required, but omitted here). \(S'\) is partially-ordered, because 1) for any \(p \in S'\), \(\lnot p \subset p\); 2) for any \(p_1, p_2, p_3 \in S'\) such that \(p_1 \subset p_2\) and \(p_2 \subset p_3\), \(p_1 \subset p_3\). For any chain in \(S'\), \(\{S_\alpha \vert \alpha \in A\} \subseteq S'\) where \(A\) is a possibly uncountable indices set, \(S'' := \cup_{\alpha \in A} S_\alpha\) is a member of \(S'\), because \(S''\) is a subset of \(S\); \(S_1 \subseteq S''\); \(S''\) is a chain in \(S\), because for any \(p_1, p_2 \in S''\), if \(p_1, p_2 \in S_\alpha\), \(p_1 \subseteq p_2\) or \(p_2 \subseteq p_1\), because \(S_\alpha\) is a chain, and if \(p_1 \in S_\alpha\) and \(p_2 \in S_\beta\) for \(S_\alpha \neq S_\beta\), \(S_\alpha \subseteq S_\beta\) or \(S_\beta \subseteq S_\alpha\), because \(\{S_\alpha\}\) is a chain, so, \(p_1, p_2 \in S_\alpha\) or \(p_1, p_2 \in S_\beta\) after all. So, by the Zorn's lemma: for any set such that for every chain, the union of the chain is a member of the set, the set has a maximal element, there is a maximal element, \(S_2 \in S'\). \(S_1 \subseteq S_2\). \(S_2\) is maximal also in the set of all the chains in \(S\), because if there was a chain, \(S_3 \subseteq S\), such that \(S_2 \subset S_3\), \(S_3\) would contain \(S_1\), so, \(S_3\) would be in \(S'\), and \(S_2\) would not be maximal in \(S'\), a contradiction.
