Properties of the Ordinals ....

[Math Amateur]

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

 

[andrewkirk]

Hello Peter!

##\alpha## is an ordinal, hence is well-ordered, hence any subset of ##\alpha## has a minimum element.
We have also deduced that ##\beta\subseteq \alpha##on line 1, from the fact that ##\alpha## is an ordinal and that##\beta\in\alpha##.

Now consider an arbitrary subset ##C## of ##\beta##. Since ##\beta\subseteq\alpha## it follows that ##C\subseteq\alpha## and hence it must have a minimum element.

If you wanted to be especially rigorous, you could prove the transitivity of the subset property, which we used here, ie that

$$C\subseteq\beta\subseteq \alpha\Rightarrow C\subseteq\alpha$$

It's pretty easy, just using the definition of a subset.

 

[Math Amateur]

Thanks Andrew ...

I appreciate your help ...

Peter