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I Axiom of Infinity & Garling, Th. 1.7.4 & the successor set

  1. Nov 4, 2016 #1
    I am reading D. J. H. Garling: "A Course in Mathematical Analysis: Volume I Foundations and Elementary Real Analysis ... ...

    I am currently focused on Garling's Section 1.7 The Foundation Axiom and the Axiom of Infinity ... ...

    I need some help with Theorem 1.7.4 ... and in particular with the notion of a successor set ##Z^+## ... ...


    ... ... the relevant text from Garling is as follows:



    ?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png



    In the above text we read the following:

    " ... ... Suppose that ##S## is a successor set. Let

    ##Z^+ = \cap \{ B \in P(S) : B \text{ is a successor set } \} ## ... "



    Note also that Garling defines a successor set as follows:

    " ... ... A set ##A## is called a successor set if ##\emptyset \in A## and if ##a^+ \in A## whenever ##a \in A## ... ... "


    and

    Garling defines ##a^+## as follows:

    " ... ... If ##a## is a set, we define ##a^+## to be the set ##a \cup \{ a \}## ... ... "



    Now my problem is that I do not understand the definition of ##Z^+## ... ... in each example I construct I seem to get ##Z^+ = \emptyset## ... ... and this cannot be right ...


    For example ...

    Suppose that a successor set S is such that:

    ##S = \{ \emptyset , a , a \cup \{ a \} , \{ a , b \}, \{ a , b \} \cup \{ \{ a , b \} \} \}##

    ... then we have ...

    ##Z^+ = \cap B_i ## where ##B_i \in P(S)## and each ##B_i## is a successor set ...

    then we have ..

    ##B_1 = \{ \emptyset, a , a \cup \{ a \} \}##

    ##B_2 = \{ \emptyset, \{ a , b \} , \{ a , b \} \cup \{ \{ a , b \} \} \}##

    and

    ##B_3 = S##


    Indeed,

    ##B_1, B_2, B_3## seem to me to be the only subsets of ##P(S)## that are successor sets


    and


    ##\cup B_i = \emptyset##


    BUT ... surely this cannot be right ...


    Can someone clarify this issue and show me how ##Z^+## is meant to be constructed ...

    Hope someone can help ...

    Peter


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

    In order to enable readers to get a better understanding of Garling's notation and approach I am providing the first two pages of Section 1.7 ... as follows:



    ?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png
    ?temp_hash=ac8c3c6753bcd64546b6d02e773f06ff.png
     
  2. jcsd
  3. Nov 4, 2016 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    [itex]B_1[/itex] is not a successor set. Maybe you are confused by the use of the variable [itex]a[/itex] in the definition of successor set. The definition means "for any set [itex]a[/itex], if [itex]a[/itex] is in [itex]A[/itex], then [itex]a^+[/itex] is in [itex]A[/itex]".

    So if [itex]A[/itex] is a successor set, then [itex]0[/itex] is in [itex]A[/itex], and so is [itex]0^+[/itex], and so is [itex]0^{++}[/itex], etc. So every successor set has an infinite number of elements.
     
  4. Nov 5, 2016 #3
    Thanks Steven ... Based on what you have said, I'll now try to reformulate my example ...

    I am assuming ##S## should be as follows ... ...

    ##S = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ , a, a^+, a^{++}, a^{+++}, \ ... \ ... \ ... \ , \{ a , b \}, \{ a , b \}^+, \{ a , b \}^{++}, \{ a , b \}^{+++}, \ ... \ ... \ ... \ \} ##


    Then Garling defines ##Z^+## as follows:


    ##Z^+ = \cap B_i## where ##B_i \in P(S)## and each ##B_i## is a successor set ...


    ... ... then we find ... ...


    ##B_1 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ \}##


    ##B_2 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ a, a^+, a^{++}, a^{+++}, \ ... \ ... \ ... \ \}##


    ##B_3 = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ , \{ a , b \}, \{ a , b \}^+, \{ a , b \}^{++}, \{ a , b \}^{+++}, \ ... \ ... \ ... \ \}##

    and

    ##B_4 = S##


    ... so ... ...


    ##B_1, B_2, B_3, B_4 ## now seem to me to be the only subsets of ##P(S)## that are successor sets


    and we find that ...


    ##\cap B_i = \{ \emptyset , \emptyset^+, \emptyset^{++}, \emptyset^{+++}, \ ... \ ... \ ... \ \}##


    Is that correct?

    Peter


    NOTE - just by the way, I think that what Garling is calling a "successor set" is often called an "inductive set" ... can you confirm that this is the case?
     
    Last edited: Nov 5, 2016
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