Theorem (The Archimedean Property). \(\mathbb{N}\) is not bounded from above in \(\mathbb{R}\).
Proof. Let us assume by negation that \(\mathbb{N}\) is bounded from above in \(\mathbb{R}\).
Let us notice \(\mathbb{N}\ne\emptyset\) since \(1\in\mathbb{N}\).
Then by the upper bound theorem, \(\mathbb{N}\) has a supremum.
In particular, according to the characterization of the supremum,
\[\forall\varepsilon>0,\ \exists n\in\mathbb{N},\ s.t.\ \sup\left(\mathbb{N}\right)-\varepsilon<n\le\sup\left(\mathbb{N}\right)\]
Choosing \(\varepsilon=1\) provides such \(n_{0}\in\mathbb{N}\): \[\sup\left(\mathbb{N}\right)-1<n_{0}\le\sup\left(\mathbb{N}\right)\] \[\therefore\sup\left(\mathbb{N}\right)<n_{0}+1\underset{*}{\in}\mathbb{N}\] \((*)\) \(\mathbb{N}\) is closed under addition.
By the supremum definition we get the contradiction \[n_{0}+1\le\sup\left(\mathbb{N}\right)<n_{0}+1\]
Hence \(\mathbb{N}\) is not bounded from above in \(\mathbb{R}\). ∎