Ordinals ... Searcoid, Corollary 1.4.5 ... ...

  • #1
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Main Question or Discussion Point

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 the Corollary to Theorem 1.4.4 ...

Theorem 1.4.4 and its corollary read as follows:


?temp_hash=9906c1123e18fde7daf9d16d07355c4b.png




Searcoid gives no proof of Corollary 1.4.5 ...

To prove Corollary 1.4.5 we need to show ##\beta \in \alpha \Longleftrightarrow \beta \subset \alpha## ... ...



Assume that ##\beta \in \alpha## ...

Then by Searcoid's definition of an ordinal (Definition 1.4.1 ... see scanned text below) we have ##\beta \subseteq \alpha## ...

But it is supposed to follow that ##\beta## is a proper subset of ##\alpha## ... !

Is Searcoid assuming that ##\beta \neq \alpha##? ... ... if not how would it follow that \beta \subset \alpha ... ?



Hope someone can help ...

Peter




============================================================================


It may help readers of the above post if the start of the section on ordinals was accessible ... so I am providing that text as follows:



?temp_hash=9906c1123e18fde7daf9d16d07355c4b.png




Hope that helps ...

Peter
 

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  • #2
andrewkirk
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For the forward direction ##\Rightarrow## of the equivalence ##\Leftrightarrow##, having ##\beta=\alpha## together with ##\beta\in\alpha## would give us ##\alpha\in\alpha##, which would contradict Theorem 1.4.3.

For the reverse direction we just use ##\beta\subset\alpha\to \beta\subseteq\alpha## and proceed from there.
 
  • #3
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Thanks Andrew ...

Just another question ... the Corollary goes on to state that, in particular, if ##\alpha \neq 0## then ##0 \in \alpha## ... can you help with the proof of this ... ?


Thanks again,

Peter
 
  • #4
andrewkirk
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0 is the empty set, which is a subset of every set, so in particular ##0\subseteq\alpha##. Putting this together with ##\alpha\neq 0## gives ##0\subset\alpha##. Applying the corollary to that with the arrow pointing left gives us ##0\in\alpha##.
 
  • #5
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Thanks Andrew,

Peter
 

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