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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:
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. ... ... ?
*** EDIT ***
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
*** FINISH EDIT ***
Help will be appreciated ...
Peter
==========================================================================
It may help Physics Forums 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 Physics Forums 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 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. ... ... ?
*** EDIT ***
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
*** FINISH EDIT ***
Help will be appreciated ...
Peter
==========================================================================
It may help Physics Forums 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 Physics Forums 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
Attachments

Searcoid  1  Theorem 1.4.3 ... ... PART 1 ... .....png8.7 KB · Views: 282

Searcoid  2  Theorem 1.4.3 ... ... PART 2 ... ......png64.4 KB · Views: 283

Searcoid  1  Start of section on Ordinals ... ... PART 1 ... .....png55.5 KB · Views: 276

Searcoid  Definition 1.3.10 ... .....png39.8 KB · Views: 277

Searcoid  2  Definition 1.3.10 ... .....PART 2 ... ....png44.2 KB · Views: 265

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?temp_hash=bccb60496862d56962ecb212d49f4113.png64.4 KB · Views: 207

?temp_hash=bccb60496862d56962ecb212d49f4113.png55.5 KB · Views: 187

?temp_hash=bccb60496862d56962ecb212d49f4113.png39.8 KB · Views: 198

?temp_hash=bccb60496862d56962ecb212d49f4113.png44.2 KB · Views: 189