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I am a first-semester BSc. Engineering student and I have been studying the Archimedean Principle as part of my coursework. Recently, I came across Jay Cummings' proof for this principle in his book 'Real Analysis - A Long Form Mathematics Textbook'. While studying the proof, I encountered a difficulty and would appreciate some clarification.

The Archimedean Principle states that,

'Iff a and b are real numbers with a > 0, then there exists a natural number n such that na > b.'

In 'Real Analysis'; the author proves the Archimedean Principle using a proof by contradiction.

A summarized version of his proof is as follows:

Since $a, b \in \mathbb R;$ then, $a/ b \in \mathbb R$,
Let $x = a/ b ,(x \in \mathbb R)$,
Assume
for contradiction, that $x$ is an upper bound of $\mathbb N$.

The author is attempting to prove that,
Statement: $(\forall x \in \mathbb R)$, $(\exists n \in \mathbb Z)$, such that $n> x$. (For any real number $x$, there exists an integer $n$ such that $n> x$.)

Since $\mathbb N \in \mathbb R$ and $\mathbb R$ is complete, $sup(\mathbb N)$ exists.
Let $sup(\mathbb N)= \alpha$,
Then, $\alpha - 1$ is
not an upper bound of $\mathbb N$.
That is, $\exists m \in (\mathbb Z,\mathbb N)$ such that, $m> \alpha -1$.
Consequently, $m+ 1> \alpha$.

#Contradiction $(sup(\mathbb N)= \alpha$. But also, $m+ 1> \alpha$ and $(m+ 1)\in \mathbb N)$

Therefore, we can conclude that $x$ is not an upper bound of $\mathbb N$.
By substituting $b/ a$ for $x$ in Statement we get,
$\forall b/ a \in \mathbb R$; $\exists n \in \mathbb Z$, such that $n> b/ a$.
Therefore it follows that, $\exists n \in \mathbb N$, such that na> b. (There exists a natural number n, such that na> b.
)

However, I have a question regarding the proof.

Since we are considering the set $\mathbb N$, where the last element is $sup(\mathbb N)$ and the element before that is $sup(\mathbb N) - 1$ all the numbers between these two are decimals and are not included in $\mathbb N$. On the other hand, $m$ belongs to $\mathbb N$, so $m \le sup(\mathbb N)$. Therefore, how can there exist an integer $m$ such that, $m> sup(\mathbb N)- 1$?

Doesn't this imply that there exists an integer $ m$; such that, $sup(\mathbb N)- 1< m \le sup(\mathbb N)$, while $sup(\mathbb N)- 1$ and $sup(\mathbb N)$ are the two closest integers?

$\mathbb N$ = {$1, 2, 3, \dots, sup(\mathbb N)- 2, sup(\mathbb N)- 1, sup(\mathbb N)$}

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    $\begingroup$ $\Bbb N$ is not an element of $\Bbb R$, $\Bbb N$ is a subset of $\Bbb R$. $\endgroup$
    – jjagmath
    Commented Jul 16, 2023 at 3:40

1 Answer 1

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"Doesn't this imply that there exists an integer $ m$; such that, $sup(\mathbb N)- 1< m \le sup(\mathbb N)$, while $sup(\mathbb N)- 1$ and $sup(\mathbb N)$ are the two closest integers?"

Yes, in fact, since $\alpha-1$ is not an upper bound of $\mathbb{N}$ it means that there exists at least one number from $\mathbb{N}$ that is larger than it, let it be $m$ like shown in the proof. As you said $m\le \sup{(\mathbb{N})},m\in \mathbb N$ and this means that $m=\sup{(\mathbb N)}$.

I hope you understand the rest of the proof from here.

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