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

To read this post, you should read the previous post first. It deals with the definition of that we will use. Happy reading.

**Theorem :**

Given poset with is the set of all well ordered subsets of poset and is a relation that defined as before. For any simply ordered subset of poset then always have the smallest upper bound.

**Proof :**

Given simply ordered subset of . Suppose then . Next, let’s say is non-empty simply ordered subset of . Consider :

It will be proved that is a well ordered subset of . Take any with then between the set , there is at least one that the element is on . Suppose the set is . Since is well ordered set then has the smallest element. Suppose the smallest element is . It will be proved that is also the smallest element for .

Take any with . Will be proven . Because of then there is such that . On the other hand, since is simply ordered set then or . If then and if then it is not possible to apply . It because, if then which results in a contradiction with is the smallest element of . In other words, apply for all . So for any , has the smallest element. We get is well ordered set which resulted . Thus, we get cause for any apply .

Based on the description above, for the next well ordered subset of poset on above is called the greatest ordered subset of . So is called the greatest ordered subset of if is not the initial segment of a well ordered subset of .

Thanks for reading. May be useful.

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