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I Addition/Multiplication of Natural Numbers - Bloch Th. 1.2.7

  1. Jul 15, 2017 #1
    I am reading Ethan D. Bloch's book: The Real Numbers and Real Analysis ...

    I am currently focused on Chapter 1: Construction of the Real Numbers ...

    I need help/clarification with an aspect of Theorem 1.2.7 (1) ...

    Theorem 1.2.7 reads as follows:



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    ?temp_hash=10af9cc36ca677ab540d906429ca0a63.png



    In the above proof of (1) we read the following:


    " We will show that ##G = \mathbb{N}##, which will imply the desired result. Clearly ##G \subseteq \mathbb{N}##. ... ... ... "


    Before he proves that ##1 \in G##, Bloch asserts that ##G \subseteq \mathbb{N}## ... what is his reasoning ...?

    It does not appear to me ... from the order in which he says things that he is saying

    ##1 \in G## ... therefore ##G \subseteq \mathbb{N}## ...

    Can we immediately conclude that ##G \subseteq \mathbb{N}## without relying on ##1 \in G## ... ... ?

    (Bloch does the same in a number of places in this chapter ... )


    Hope someone can help ... ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Jul 16, 2017 #2

    andrewkirk

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    This follows directly from the definition of ##G##, which is defined as the set of elements ##z## of ##\mathbb N## that satisfy the required cancellation property:
    $$G=\{z\in\mathbb N\ |\ z\textrm{ satisfies the cancellation property}\}$$
    In other words, ##G## is defined as the subset of ##\mathbb N## obtained by applying the criteria that elements must have the cancellation property. IIRC, in Zermelo-Frankel set theory axioms, this is using the Axiom Schema of Specification, which says we can make a subset of an existing set by applying a criterion to filter the elements.
     
  4. Jul 16, 2017 #3


    Thanks Andrew ... appreciate the help ...

    But ... what if no ##z## satisfy the criteria for membership of ##G## ... and ##G = \emptyset## ... ?

    Peter
     
  5. Jul 16, 2017 #4

    andrewkirk

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    Then the proof won't be able to be completed. But the statement that ##G\subseteq \mathbb N## will still be valid, because the empty set is a subset of every set.

    Perhaps your concern is not about the claim that ##G## is a subset of ##\mathbb N## but rather about whether it is non-empty. He proves that further down, first by showing that ##1\in G##.

    As a bit of trivia, a handy way to define the empty set, given a set ##A##, is to write it as
    $$\{x\in A\ |\ x\neq x\}$$
     
  6. Jul 16, 2017 #5

    Yes, it was about Bloch claiming that "clearly" ##G \subseteq \mathbb{N}## before he had proved that ##G## contained some elements ... I would have been happy if he had proved ##1 \in G## ... and then claimed ##G \subseteq \mathbb{N}## ... ...

    ... ... indeed for me ... we should be aware that all elements of ##G## are also elements of ##\mathbb{N}## - which we have .... AND ... we need to know that G is not empty ...


    Mind you ... you now have me thinking that ##G \subseteq \mathbb{N}## anyway since ##\emptyset \subseteq \mathbb{N}## ... ...

    Peter
     
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