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Road to Zorn’s Lemma Part 3

Hello, infinite-set reader. Set theory is back.

As I said in the previous post, today we will discuss about the theorem of any poset that the every non-empty well ordered subset has the smallest bound on the poset. Happy reading.

Theorem :

Given any poset \left(P,\leq\right) with the every non-empty well ordered subset has the smallest bound in P. Given fuction f:P\rightarrow P with x\leq f\left(x\right) for all x\in P. For any p\in P then W\left(p\right) is well ordered set such that :

    \[ sup\, W\left(p\right)\in W\left(p\right)\]

and if m=sup\, W\left(p\right) then apply :

    \[ f\left(m\right)=m\]

Proof :

To prove the above theorem is sufficiently proved that W\left(p\right)=C\left(p,m\right) because if applicable W\left(p\right)=C\left(p,m\right) then m\leq f\left(m\right) by fuction definiton of f. On the other hand , since m\in W\left(p\right) then by Theorem (ii) in this post applies f\left(m\right)\in W\left(p\right). Since m is sup\, W\left(p\right) then f\left(m\right)\leq m. Thus, we get f\left(m\right)=m.

It will be proven that W\left(p\right)=C\left(p,m\right) with m=sup\, W\left(p\right). Given any H\subseteq W\left(p\right) with H\neq\emptyset. Further taken any r\in H then by Theorem (y) in this post can be consider C\left(p,r\right). Furthermore, r\in H\cap C\left(p,r\right). Because of C\left(p,r\right) has the smallest element then H\cap C\left(p,r\right) also has the smallest element. Furthermore, inf\, H\cap C\left(p,m\right)=inf\, H. In other words, for any H\subseteq W\left(p\right) with H\neq\emptyset then H has the smallest element which resulted W\left(p\right) is well ordered set. Next, by Theorem (i) in this post then C\left(p,r\right)\subseteq W\left(p\right). Thus, we get p\in W\left(p\right) and if r\in W\left(p\right) then p\leq r. In other words, p is the smallest element of W\left(p\right).

It will then be proved that W\left(p\right) satisfy the p-chain C\left(p,m\right) axioms. It has been proved that W\left(p\right) is well ordered set. According to the hypothesis on the proof of this theorem, then W\left(p\right) has the largest element. Let’s say m=sup\, W\left(p\right). Defined V=W\left(p\right)\cup\left\{ m\right\}. Will be shown V is p-chain. Since W\left(p\right) is well ordered set then V also well ordered set with p is the smallest element and m is the largest element. Condition (i) of definition p-chain is fulfilled.

Next, take any z\in V with z\neq m then z\in W\left(p\right). Since z\neq m then z^{+}\in V. Note that z and z^{+} are the last two elements on p-chain C\left(p,z^{+}\right). The result is valid z^{+}=f\left(z\right) for all z^{+}\in V. Condition (ii) of definition p-chain is fulfilled.

Next, take any U\subseteq V with U\neq\emptyset then m is the upper bound U. There are two possibilities. The first possibility, if there is no r\in W\left(p\right) such that x\leq r for all x\in U then sup\, U=m\in V. Second possibility, if there is r\in W\left(p\right) such that x\leq r for all x\in U then sup\, U=r\in W\left(p\right)\subseteq V. Condition (iii) of definition p-chain is fulfilled. In other words V is a p-chain with m being its largest element.

Since W\left(p\right) is the set of all elements p-chain then V=W\left(p\right). Then we get W\left(p\right) is p-chain C\left(p,m\right).

You should reopen my previous post to get a good understanding. This is because my posts are related to each other. If you forget a little definition or theorem, you will be confused. I try to make this discussion easier by providing links. I hope this makes it easier for you to understand this discussion.

Thanks for reading. May be useful.

Road To Zorn’s Lemma Part 2

Hello, infinite-set reader. Set theory is back.

Before reading this post, you should read the previous post first. It deals with the definition of C\left(p,r\right) and W\left(p\right). As a reminder only if you forget. The post is still about the character of C\left(p,r\right) and W\left(p\right). Happy reading.

(x) Theorem :

If r,s\in W\left(p\right) then either of the following two applies, s\in C\left(p,r\right) or r<s.

Proof :

Defined S=C\left(p,r\right)\cap C\left(p,s\right) then S is well ordered set because S is subset of C\left(p,r\right) and C\left(p,s\right) both of which are well ordered set. Further, by definition p-chain condition (i) then S\neq\emptyset and by definition p-chain condition (iii) then x=sup\, S are in C\left(p,r\right) and also in C\left(p,s\right) which resulted x\in S. Next, assume s\notin C\left(p,r\right) then x\neq s. if x\neq r then by definition p-chain condition (ii) x^{+}=f\left(x\right)\in S. Contradictions with x=sup\, S. In other words, if s\notin C\left(p,r\right) then r=x\in S which resulted r\in C\left(p,s\right). Because of r\neq s then by definition p-chain condition (i) we get r<s. Even though s\in C\left(p,r\right) and r<s are two things that are contradictory then the assumption is rejected. Theorem is proven.

(y) Theorem :

If r\in W\left(p\right) then there is a unique C\left(p,r\right) such that C\left(p,r\right)=\left\{ s\in W\left(p\right)|s\leq r\right\}.

Proof : (more…)