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Zorn’s Lemma Part 1

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

The previous post about set theory is on this link. If you visit this website for the first time, you should read the previous posts.

(x) Theorem :

Any poset has the largest well ordered subset.

Proof :

Given any poset \left(P,\leq\right). Consider poset \left(Q,\ll\right) with Q is the set of all well ordered subset of poset \left(P,\leq\right) and \ll is a relation that defined as before. By Theorem in this post then all well ordered subset of poset \left(Q,\ll\right) has the smallest upper bound. By Theorem in this post Because of \left(Q,\ll\right) is a poset that every the well ordered subset has the smallest upper bound then Q has at least one maximum element. Suppose that the maximum element of Q is m, then m is also largest well ordered of poset P.

Next will be discussed about the theorem that became the basic of Zorn’s Lemma.

(y) Theorem :

For non-empty poset P that every the well ordered subset has upper bound then P has at least one maximum element.

Proof :

By Theorem (x) above then P has largest well ordered subset w. Let m be the upper bound w. The existence of m is guaranteed by hypothesis. Therefore w is the largest well ordered subset of P then m\in w. It because if m\notin w then w\cup\left\{ m\right\} is well ordered subset of P with w\subseteq w\cup\left\{ m\right\}. Contradiction with w is largest well ordered subset of P.

Furthermore, it will be shown that m is the maximum element of P. Take any x\in P. If m<x then w\cup\left\{ x\right\} is well ordered subset of P with w\subseteq w\cup\left\{ x\right\}. Contradiction with w is largest well ordered subset of P. Therefore, always apply that if m\leq x then x=m. In other words, m is the maximum element of P.

The next post will discuss about Zorn’s Lemma. Make sure you follow this ‘story’. Thanks for reading. May be useful.

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