Ordinals .... Searcoid, Theorem 1.4.6 ....

[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 1.4.6 ...

Theorem 1.4.6 reads as follows:

View attachment 8467
View attachment 8468
My question regarding the above proof by Micheal Searcoid is as follows:

How do we know that \(\displaystyle \alpha\) and \(\displaystyle \beta\) are not disjoint? ... indeed ... can they be disjoint?

What happens to the proof if \(\displaystyle \alpha \cap \beta = \emptyset\)?
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:
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View attachment 8470Hope that helps ...

Peter

 

[mmwave]

Dear Peter,

Thank you for reaching out for help with understanding Theorem 1.4.6 in Michael Searcoid's book. Theorem 1.4.6 states: "If $\alpha$ and $\beta$ are ordinals, then $\alpha \cup \beta$ is an ordinal." This theorem is important because it shows that the union of two ordinals is itself an ordinal, which is a fundamental concept in abstract analysis.

To answer your question, we do not know for certain that $\alpha$ and $\beta$ are not disjoint. In fact, they could potentially be disjoint. However, the proof of Theorem 1.4.6 does not rely on the assumption that $\alpha$ and $\beta$ are not disjoint. The proof simply shows that if $\alpha$ and $\beta$ are ordinals, then their union $\alpha \cup \beta$ is also an ordinal.

If $\alpha \cap \beta = \emptyset$, the proof still holds. This is because the definition of an ordinal does not require that all of its elements be distinct. In other words, the elements of $\alpha$ and $\beta$ can overlap, but their union will still be an ordinal.

I hope this clarifies your question and helps you better understand Theorem 1.4.6. If you have any further questions, please don't hesitate to ask.