
MHB Master
#1
October 4th, 2018,
19:53
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
Last edited by Peter; October 5th, 2018 at 21:09.

October 4th, 2018 19:53
# ADS
Circuit advertisement

MHB Master
#2
October 4th, 2018,
22:01
Thread Author
Originally Posted by
Peter
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 by Peter; October 4th, 2018 at 22:13.