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Proper Subsets of Ordinals ... ... Another Question ... ... Searcoid, Theorem 1.4.4 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Micheal Searcoid's book: "Elements of Abstract Analysis" ... ...

I am currently focused on understanding Chapter 1: Sets ... and in particular Section 1.4 Ordinals ...

I have another question regarding the proof of Theorem 1.4.4 ...

Theorem 1.4.4 reads as follows:



Searcoid - Theorem 1.4.4 ... ....png






In the above proof by Searcoid we read the following:

"... ... Moreover, since \(\displaystyle x \subset \alpha\), we have \(\displaystyle \delta \in \alpha\). But \(\displaystyle \beta \in \alpha\) and \(\displaystyle \alpha\) is totally ordered, so we must have \(\displaystyle \delta \in \beta\) or \(\displaystyle \delta = \beta\) or \(\displaystyle \beta \in \delta\) ... ... "


My question is regarding the three alternatives \(\displaystyle \delta \in \beta\) or \(\displaystyle \delta = \beta\) or \(\displaystyle \beta \in \delta\) ... ...


Now ... where \(\displaystyle (S, <)\) is a partially ordered set ... \(\displaystyle S\) is said to be totally ordered by \(\displaystyle <\) if and only if for every pair of distinct members \(\displaystyle x, y \in S\), either \(\displaystyle x < y\) or \(\displaystyle y < x\) ... ..


So if we follow the definition exactly in the quote above there are only two alternatives .... \(\displaystyle \delta \in \beta\) or \(\displaystyle \beta \in \delta\) ... ...


My question is ... where does the \(\displaystyle =\) alternative come from ... ?

How does the \(\displaystyle =\) alternative follow from the definition of totally ordered ... ?


Help will be appreciated ...

Peter
 

steenis

Well-known member
MHB Math Helper
Jul 30, 2016
250
Now ... where \(\displaystyle (S, <)\) is a partially ordered set ... \(\displaystyle S\) is said to be totally ordered by \(\displaystyle <\) if and only if for every pair of distinct members \(\displaystyle x, y \in S\), either \(\displaystyle x < y\) or \(\displaystyle y < x\) ... ..
Peter
Please, read this definition very very carefully, and ask yourself: what if the pair of members is/are not distinct ?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Please, read this definition very very carefully, and ask yourself: what if the pair of members is/are not distinct ?



Thanks Steenis ...

See that key term is "distinct"... if not distinct then members are equal ... enough said ...

Thanks for your help ...

Peter