313: Hausdorff Maximal Principle: Chain in Partially-Ordered Set Is Contained in Maximal Chain

<The previous article in this series | The table of contents of this series | The next article in this series>

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
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

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, $⟨S,R⟩$, 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 $S^{\prime }$. $S^{\prime }$ is really a set as a subset of the power set of $S$ (a formula for the subset axiom is required, but omitted here). $S^{\prime }$ is partially-ordered, because 1) for any $p\in S^{\prime }$, $\mathrm{\neg }p\subset p$; 2) for any $p_1,p_2,p_3\in S^{\prime }$ such that $p_1\subset p_2$ and $p_2\subset p_3$, $p_1\subset p_3$. For any chain in $S^{\prime }$, $\{S_{\alpha }|\alpha \in A\}\subseteq S^{\prime }$ where $A$ is a possibly uncountable indices set, $S^{″}:={\cup }_{\alpha \in A}{S}_{\alpha }$ is a member of $S^{\prime }$, 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 $S_{\alpha }\ne S_{\beta }$, $S_{\alpha }\subseteq S_{\beta }$ 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, $S_2\in S^{\prime }$. $S_1\subseteq S_2$. $S_2$ 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 $S^{\prime }$, and $S_2$ would not be maximal in $S^{\prime }$, a contradiction.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>