Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Hausdorff maximal principle
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Proof 2== The [[Bourbaki–Witt theorem]], together with the [[Axiom of choice]], can be used to prove the Hausdorff maximal principle. Indeed, let <math>P</math> be a nonempty poset and <math>X\mathrel{\mathop:}=\{C\subseteq P\,:\, C\ \text{is a chain}\}</math> be the set of all totally ordered subsets of <math>P</math>. Notice that <math>X\neq \emptyset</math>, since <math>P\neq \emptyset</math> and <math>\{x\}\in X</math>, for any <math>x\in P</math>. Also, equipped with the inclusion <math>\subseteq</math>, <math>X</math> is a poset. We claim that every chain <math>\mathcal{C}\subseteq X</math> has a [[supremum]]. In order to check this out, let <math>S</math> be the union <math>\bigcup_{C\in \mathcal{C}}C</math>. Clearly, <math>C\subseteq S</math>, for all <math>C\in \mathcal{C}</math>. Also, if <math>U</math> is any upper bound of <math>\mathcal{C}</math>, then <math>S\subseteq U</math>, since by definition <math>C\subseteq U</math> for all <math>C\in \mathcal{C}</math>. Now, consider the map <math>f\colon X\to X</math> given by <math>f(C)\mathrel{\mathop:}=\begin{cases}C, &\text{if}\ C\ \text{is maximal}\\ C\cup \{g(P\setminus C)\}, &\text{if}\ C\ \text{is not maximal}\end{cases}</math> where <math>g</math> is a [[choice function]] on <math>\{P\setminus C\}</math> whose existence is ensured by the Axiom of choice, and the fact that <math>P\setminus C\neq \emptyset</math> is an immediate consequence of the non-maximality of <math>C</math>. Thus, <math>C\subseteq f(C)</math>, for each <math>C\in X</math>. In view of the Bourbaki-Witt theorem, there exists an element <math>C_0\in \mathcal{C}</math> such that <math>f(C_0)=C_0</math>, and therefore <math>C_0</math> is a maximal chain of <math>P</math>. In the case <math>P=\emptyset</math>, the empty set is trivially a maximal chain of <math>P</math>, as already mentioned above. <math>\square</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)