Well Ordering Theorem

Hello, infinite-set reader. Set theory is back. Today we will learn about Well Ordering Theorem.

This post is the last post about Zorn’s Lemma. The next post will be different from the previous post. But in the end, all the discussion of set theory will be interrelated. My imagination, our discussion will lead to the discussion of cardinality. Pray I write diligently. Happy reading.

Well Ordering Theorem

(x) Theorem :

Every poset has largest simply ordered subset.

Proof :

GivenĀ  poset \left(P,\leq\right). Consider poset \left(Q,\subseteq\right) with Q is the set of all simply ordered subset of poset \left(P,\leq\right). By Theorem in the previous post then every simply ordered subset of poset \left(Q,\subseteq\right) have smallest upper bound. Since \left(Q,\subseteq\right) is poset that the every simply ordered subset has smallest upper bound then by Zorn’s Lemma, Q has at least one maximum element. Suppose that the maximum element of Q is r then r is also largest simply ordered of poset P.

As already mentioned, the following theorem explains that for any set, the set can be formed as a well ordered set. This theorem is often called the Well Ordering Theorem.

Theorem : (Well Ordering Theorem)

Every set can be well ordered.

Proof :

Take any set of E. Consider set W is the set of all subsets of E that can be well ordered. Then W is partially ordered by \ll. By Theorem (x) above then W has largest simply ordered subset S. Claim that \cup S is well ordered and \cup S=E.

For proof \cup S is well ordered analogue on proof of theorem in this link.

Next will be proved \cup S=E. It is clear that \cup S\subseteq E because for all x\in\cup S it is true that x\in E. Next, for proof E\subseteq\cup S, suppose not to apply so then it exists y\in E such that y\notin\cup S. Consider \left(\cup S\right)\cup\left\{ y\right\} then \left(\cup S\right)\cup\left\{ y\right\} can be well ordered with y is the largest/last element of \cup S. As a result S\cup\left(\cup S\right)\cup\left\{ y\right\} is simply ordered subset of W which contains proper subset S. Contradictions with S is largest simply ordered subset of W. Proven E can be well ordered.

Thanks for reading. See you again in the next post. If you have any criticism or suggestion, please leave a comment below or email me.

Add a Comment

Your email address will not be published. Required fields are marked *