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I've been struggling with a concept concerning the Archimedean property proof. That is showing my contradiction that For all $x$ in the reals, there exists $n$ in the naturals such that $n>x$.

Okay so we assume that the naturals is bounded above and show a contradiction.

If the naturals is bounded above, then it has a least upper bound (supremum) say $u$

Now consider $u-1$. Since $u=\sup(\mathbb N)$ , $u-1$ is an element of $\mathbb N$. (here is my first hiccup, not entirely sure why we can say $u-1$ is in $\mathbb N$)

This implies (again not confident with this implication) that there exists a $m$ in $\mathbb N$ such that $m>u-1$. A little bit of algebra leads to $m+1>u$.

$m+1$ is in $\mathbb N$ and $m+1>u=\sup(\mathbb N)$ thus we have a contradiction.

Can anyone help clear up these implications that I'm not really comfortable with? Thanks!

Ben Grossmann
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Ben Anderson
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3 Answers3

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$u-1$ may not be an element of $\mathbb{N}$, but we can be certain that $u-1 < u$. Since $u$ is the supremum of $\mathbb{N}$ then $u-1$ cannot be an upper bound for $\mathbb{N}$. This means that there is an $n \in \mathbb{N}$ such that $u-1 < n$. But then $n+1 \in \mathbb{N}$ and $u = (u-1)+1 < n+1$, which is our contradiction! ($u$ is no longer an upper bound of $\mathbb{N}$.)

user642796
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I dont think you need the fact that $u\in N$(the fact is even not true) .And for the second difficulty the fact follows from the supremum property.As $u-1$ is not an upper bound so there exists a natural number greater than it.

Abhra Abir Kundu
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Proof: Assume for a contradiction that $\mathbb{N}$ is bounded above. Then $\sup\mathbb{N}$ exists by the Completeness Axiom, By the Approximation Property with $ε = 1/2$, there exists $k ∈ \mathbb{N}$ with $$\sup \mathbb{N} − \frac{1}{2} < k < \sup \mathbb{N} +\frac{1}{2} $$ But then $k + 1 ∈ \mathbb{N}$ and $k + 1 > \sup \mathbb{N} + 1$, a contradiction.

Michael Hardy
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user77904
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