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Ordinal by definition

  1. Aug 11, 2013 #1
    A set [itex]x[/itex] is well-ordered by [itex]<[/itex] if every subset of [itex]x[/itex] has a least element. Here [itex]<[/itex] is assumed a linear ordering, meaning that all members of a set can be compared, unlike with partial ordering.

    A set [itex]x[/itex] is transitive if it has property [itex]\forall y\;(y\in x\to y\subset x)[/itex].

    A set [itex]\alpha[/itex] is ordinal, if it is transitive and well-ordered by [itex]\in[/itex].

    The claim: If [itex]\alpha[/itex] is an ordinal, and [itex]\beta\in\alpha[/itex], then [itex]\beta[/itex] is ordinal too.

    A book says that this claim is clear "by definition", however I see only half of the proof by definition.

    We have [itex]\beta\in\alpha\to\beta\subset\alpha[/itex], and a subset of a well-ordered set is also well-ordered, so that part is clear by definition.

    We should also prove a claim [itex]\forall\gamma\;(\gamma\in\beta\to\gamma\subset\beta)[/itex]. How is this supposed to come from the definition? I only see [itex]\gamma\in\beta\to\gamma\in\alpha\to\gamma\subset\alpha[/itex].

    ---

    update: Oh I understood this now! No need for help. :cool: But I would like to complain that the book is playing fool on the reader. I wouldn't call that "by definition".

    ---

    second update: We assume [itex]\gamma\in\beta[/itex] and then

    [tex]
    \neg(\gamma\subset\beta)\to \exists\delta\;(\delta\in\gamma\land\delta\notin\beta)
    [/tex]
    [tex]
    \to\exists\delta\;\big(\delta\in\gamma\land(\beta\in\delta\lor \beta=\delta)\big)
    [/tex]
    [tex]
    \to\exists\delta\big(\underbrace{(\delta\in\gamma\land\beta\in\delta)}_{\to 0=1}\lor\underbrace{(\delta\in\gamma\land \beta=\delta)}_{\to 0=1}\big)\to 0=1
    [/tex]

    Does that look like "by definition"? :devil:
     
    Last edited: Aug 11, 2013
  2. jcsd
  3. Aug 11, 2013 #2

    verty

    User Avatar
    Homework Helper

    I think they call it by definition for this reason. Since ##\beta\subset\alpha##, ##\beta## is well ordered by ##\in##, as you pointed out. So for ##\beta' < \beta## in ##\alpha##, ##\beta'\in\beta##, and for ##\beta' > \beta##, ##\beta\in\beta'## which precludes ##\beta## containing any of these larger elements. But ##\beta\subset\alpha##, therefore ##\beta## is exactly the union of elements of ##\alpha## less than ##\beta##. But then ##\forall\gamma\in\beta \; (\gamma\subset\beta)## and ##\beta## is transitive.

    So in a sense, ##\beta## is defined in this way by those definitions.
     
    Last edited: Aug 11, 2013
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