The Well-Ordering Principle for the Natural Numbers

[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 1.2.10 ...

Theorem 1.2.10 reads as follows:

View attachment 6980
View attachment 6981Towards the end (second last line) of the above proof by Bloch, we read the following:

" ... ... We now have a contradiction to the fact that no element such as \(\displaystyle a + 1\) exists in \(\displaystyle G\). ... ... "
I do not understand this remark ... as above \(\displaystyle a + 1\) has earlier been proved to belong to \(\displaystyle G\) ..

Can someone explain the remark "We now have a contradiction to the fact that no element such as \(\displaystyle a + 1\) exists in \(\displaystyle G\)" in the context of the proof and explain just what is going on ... ...

Help will be much appreciated ...

Peter

 

[Evgeny.Makarov]

The element of $G$ that is smaller than any other element of $G$ is called the least element. (Formally is should be "a least element", but it's easy to show that if a least element exists, then it is unique.) The theorem claims that every nonempty set of natural numbers has the least element. The proof is by contradiction, and the very first assumption made is that the least element does not exist. It is with this assumption that the last two lines of the proof derive a contradiction. Namely, the end of the proof shows that $a+1\in G$ and $a+1\le x$ for all $x\in G$, i.e., that $a+1$ is the least element, which does not exist by assumption.

 

[Math Amateur]

The element of $G$ that is smaller than any other element of $G$ is called the least element. (Formally is should be "a least element", but it's easy to show that if a least element exists, then it is unique.) The theorem claims that every nonempty set of natural numbers has the least element. The proof is by contradiction, and the very first assumption made is that the least element does not exist. It is with this assumption that the last two lines of the proof derive a contradiction. Namely, the end of the proof shows that $a+1\in G$ and $a+1\le x$ for all $x\in G$, i.e., that $a+1$ is the least element, which does not exist by assumption.

Thanks for a very clear explanation Evgeny ...

i appreciate your help ...

Peter