MHB Properties of the Ordinals ....

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The discussion centers on understanding Theorem 1.4.3 from Michael Searcoid's "Elements of Abstract Analysis," specifically regarding the well-ordering of ordinals. The theorem states that if β is a subset of α, then β is also well-ordered by membership, prompting questions about the formal demonstration of this property. Participants suggest that to prove β is well-ordered, one must show that every subset of β has a minimum element, which hinges on the relationship between subsets of α and β. The conversation reflects a deeper inquiry into the definitions and implications of well-ordering within the context of ordinals. Clarifying these concepts is essential for a rigorous understanding of the theorem.
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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:
View attachment 8451
View attachment 8452In the above proof by Searcoid we read the following:

"... ... Then $$\beta \subseteq \alpha$$ so that $$\beta$$ is also well ordered by membership. ... ... To conclude that $$\beta$$ 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 $$\beta$$ 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:
View attachment 8453

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:

View attachment 8454
View attachment 8455Hope that helps,

Peter
 

Attachments

  • Searcoid - 1 - Theorem 1.4.3 ... ... PART 1 ... .....png
    Searcoid - 1 - Theorem 1.4.3 ... ... PART 1 ... .....png
    1.8 KB · Views: 113
  • Searcoid - 2 - Theorem 1.4.3 ... ... PART 2 ... ......png
    Searcoid - 2 - Theorem 1.4.3 ... ... PART 2 ... ......png
    13.6 KB · Views: 105
  • Searcoid - 1 - Start of section on Ordinals ... ... PART 1 ... .....png
    Searcoid - 1 - Start of section on Ordinals ... ... PART 1 ... .....png
    32.5 KB · Views: 112
  • Searcoid - Definition 1.3.10 ... .....png
    Searcoid - Definition 1.3.10 ... .....png
    9 KB · Views: 100
  • Searcoid - 2 - Definition 1.3.10 ... .....PART 2 ... ....png
    Searcoid - 2 - Definition 1.3.10 ... .....PART 2 ... ....png
    9.1 KB · Views: 104
Last edited:
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Peter said:
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 $$\beta \subseteq \alpha$$ so that $$\beta$$ is also well ordered by membership. ... ... To conclude that $$\beta$$ 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 $$\beta$$ 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 $$\beta$$ is also well ordered by membership, we have to demonstrate that since every subset of $$\alpha$$ has a minimum element then every subset of $$\beta$$ has a minimum element ... but then that would only be true if every subset of $$\beta$$ was also a subset of $$\alpha$$ ...

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

Still not sure regarding the original question ...

Peter
 
Last edited:

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