Ordinals ... Searcoid, Theorem 1.4.6 .

  • #1
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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 1.4.6 ...

Theorem 1.4.6 reads as follows:


?temp_hash=8f9cff7393f9c0d811f87ad59134570a.png

?temp_hash=8f9cff7393f9c0d811f87ad59134570a.png





My question regarding the above proof by Micheal Searcoid is as follows:

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

What happens to the proof if ##\alpha \cap \beta = \emptyset##?



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:



?temp_hash=8f9cff7393f9c0d811f87ad59134570a.png

?temp_hash=c13d2e58ed130749d22a8e1847b2f2a4.png





Hope that helps ...

Peter
 

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  • Searcoid - 1 -  Start of section on Ordinals  ... ... PART 1 ... .....png
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  • Searcoid - 2 -  Start of section on Ordinals  ... ... PART 2 ... ......png
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Answers and Replies

  • #2
andrewkirk
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They can be disjoint. If they are disjoint then ##\alpha\cap\beta=\emptyset## which is the ordinal 0. The proof still works.

To get a feel for this, consider the case where say ##\alpha=0=\emptyset##. If you follow the proof through with this, you'll see that it still works and that either ##\beta=\alpha=0## or ##0=\alpha\subset\beta##. The empty set is a subset of every later ordinal, because it is a subset of every set. More generally, every ordinal is a subset of every greater ordinal.
 
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  • #3
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Thanks for the help Andrew ...

Peter
 

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