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.
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.
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.