Addition and Multiplication of Natural Numbers - Bloch Th. 1.2.7 .... ....

[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) ...

Theorem 1.2.7 reads as follows:
https://www.physicsforums.com/attachments/6976
https://www.physicsforums.com/attachments/6977
In the above proof of (1) we read the following:" We will show that \(\displaystyle G = \mathbb{N}\), which will imply the desired result. Clearly \(\displaystyle G \subseteq \mathbb{N}\). ... ... ... "Before he proves that \(\displaystyle 1 \in G\), Bloch asserts that \(\displaystyle G \subseteq \mathbb{N}\) ... what is his reasoning ...?

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

\(\displaystyle 1 \in G\) ... therefore \(\displaystyle G \subseteq \mathbb{N}\) ...

Can we immediately conclude that \(\displaystyle G \subseteq \mathbb{N}\) without relying on \(\displaystyle 1 \in G \)... ... ?Hope someone can help ... ...

Peter

 

[Opalg]

In the above proof of (1) we read the following:" We will show that \(\displaystyle G = \mathbb{N}\), which will imply the desired result. Clearly \(\displaystyle G \subseteq \mathbb{N}\). ... ... ... "Before he proves that \(\displaystyle 1 \in G\), Bloch asserts that \(\displaystyle G \subseteq \mathbb{N}\) ... what is his reasoning ...?

The definition of $G$ is of the form $G = \{z\in \Bbb{N} \mid \ldots \}$. This says that each element $z$ of $G$ is an element of $\Bbb{N}$, in other words $G\subseteq\Bbb{N}$.

 

[Math Amateur]

The definition of $G$ is of the form $G = \{z\in \Bbb{N} \mid \ldots \}$. This says that each element $z$ of $G$ is an element of $\Bbb{N}$, in other words $G\subseteq\Bbb{N}$.

Thanks Opalg ... appreciate the help ...

But ... what if no z satisfy the criteria for membership of G ... and G = \(\displaystyle \emptyset\) ... ?Peter

 

[Opalg]

Thanks Opalg ... appreciate the help ...

But ... what if no z satisfy the criteria for membership of G ... and G = \(\displaystyle \emptyset\) ... ?Peter

That's not a problem. The empty set is a subset of every set. So even if $G = \emptyset$, it's still true that $G\subseteq \Bbb{N}$.

 

[Math Amateur]

That's not a problem. The empty set is a subset of every set. So even if $G = \emptyset$, it's still true that $G\subseteq \Bbb{N}$.

Thanks again Opalg ...

That certainly resolves that issue ...

Peter