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Properties of the Ordinals ...

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 need some help in fully understanding Theorem 1.4.3 ...

Theorem 1.4.3 reads as follows:



Searcoid - 1 -  Theorem 1.4.3 ... ... PART 1 ... .....png
Searcoid - 2 -  Theorem 1.4.3 ... ... PART 2 ... ......png


In the above proof by Searcoid we read the following:

"... ... Then \(\displaystyle \beta \subseteq \alpha\) so that \(\displaystyle \beta\) is also well ordered by membership. ... ...


To conclude that \(\displaystyle \beta\) is also well ordered by membership, don't we have to show that a subset of an ordinal is well ordered?

Indeed, how would we demonstrate formally and rigorously that \(\displaystyle \beta\) is also well ordered by membership. ... ... ?



Help will be appreciated ...

Peter



==========================================================================


It may help MHB readers of the above post to have access to the start of Searcoid's section on the ordinals ... so I am providing the same ... as follows:



Searcoid - 1 -  Start of section on Ordinals  ... ... PART 1 ... .....png




It may also help MHB readers to have access to Searcoid's definition of a well order ... so I am providing the text of Searcoid's Definition 1.3.10 ... as follows:




Searcoid - Definition 1.3.10 ... .....png
Searcoid - 2 - Definition 1.3.10 ... .....PART 2 ... ....png





Hope that helps,

Peter
 
Last edited:

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 need some help in fully understanding Theorem 1.4.3 ...

Theorem 1.4.3 reads as follows:







In the above proof by Searcoid we read the following:

"... ... Then \(\displaystyle \beta \subseteq \alpha\) so that \(\displaystyle \beta\) is also well ordered by membership. ... ...


To conclude that \(\displaystyle \beta\) is also well ordered by membership, don't we have to show that a subset of an ordinal is well ordered?

Indeed, how would we demonstrate formally and rigorously that \(\displaystyle \beta\) is also well ordered by membership. ... ... ?



Help will be appreciated ...

Peter



==========================================================================


It may help MHB readers of the above post to have access to the start of Searcoid's section on the ordinals ... so I am providing the same ... as follows:








It may also help MHB readers to have access to Searcoid's definition of a well order ... so I am providing the text of Searcoid's Definition 1.3.10 ... as follows:











Hope that helps,

Peter

I have been reflecting on the above post on the ordinals ...


Maybe to show that that \(\displaystyle \beta\) is also well ordered by membership, we have to demonstrate that since every subset of \(\displaystyle \alpha\) has a minimum element then every subset of \(\displaystyle \beta\) has a minimum element ... but then that would only be true if every subset of \(\displaystyle \beta\) was also a subset of \(\displaystyle \alpha\) ...

Is the above chain of thinking going in the right direction ...?

Still not sure regarding the original question ...

Peter
 
Last edited: