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

1. Jul 15, 2017

### Math Amateur

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) ...

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

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2. Jul 16, 2017

### andrewkirk

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.

3. Jul 16, 2017

### Math Amateur

Thanks Andrew ... appreciate the help ...

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

Peter

4. Jul 16, 2017

### andrewkirk

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
$$\{x\in A\ |\ x\neq x\}$$

5. Jul 16, 2017

### Math Amateur

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