How Do We Prove a Subset of an Ordinal is Well-Ordered?

[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