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As the title, why do people define the segment of a well-ordering set? What's the main usage of the segment of a well-ordering set?

Well, I'm looking the book of Kunen, set theory. In chapter 1.6, he introduced the definition of pred, or we call the segment. I want to know what's his purpose of bringing this definition?

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Can you narrow this down a bit? Depending on the context, there are many different reasons to look at segments of well-ordered sets, just as there are many different reasons to look at intervals in any linearly ordered set. –  Brian M. Scott Jan 12 '12 at 3:37
    
Well, I'm looking the book of Kunen, set theory. In chapter 1.6, he introduced the definition of pred, or we call the segment. I want to know what's his purpose of bringing this definition? –  Paul Jan 12 '12 at 4:10
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I honestly don't recall the $\mathrm{pred}$ notation being used in Kunen's text after the proof of Theorem 1.7.6 (which asserts that every well-ordered set is isomorphic to a unique ordinal). I think that the proofs of this theorem as well as Theorem 1.6.3 (which provides a natural "linear ordering" of the (isomorphism types of) well-ordered sets) are the main uses of the notation. But these theorems are clearly more important that the specific notation. –  Arthur Fischer Jan 12 '12 at 5:51
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If you're a beginner set theorist, I think that there might be better starting points than Kunen. –  Asaf Karagila Jan 12 '12 at 7:41
    
@Arthur: Quite by accident I just stumbled across it in Section III.5 and in Theorem IV.5.6. –  Brian M. Scott Jan 12 '12 at 9:44

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Ken’s immediate reason for introducing the pred operator at that point is to be able to state Theorem I.6.3, which is fundamental to any understanding of well-orders: if $\langle X,\le\rangle$ and $\langle Y,\preceq\rangle$ are well-orders, then either they are isomorphic, or one is isomorphic to a proper initial segment of the other. In the very next section he goes on to construct the ordinals $-$ transitive sets $\alpha$ such that $\langle \alpha,\in\rangle$ is a well-order $-$ and shows that every well-order is isomorphic to one of them. This improves on Theorem I.6.3, because if $\alpha$ and $\beta$ are ordinals, then either they are equal (not just isomorphic), or one of them is a proper initial segment of the other (not just isomorphic to one). Moreover, each ordinal is a (necessarily proper!) initial segment of the proper class ON of ordinals. Thus, the notion of initial segment is intimately bound up with the concept of a well-order, which in turn is fundamental to set theory.

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It's very enlightening for a beginner of learning set theory! –  Paul Jan 12 '12 at 6:50

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