[proofplan] We argue by contradiction from the [least upper bound property](/page/Supremum_and_Infimum) of $\mathbb{R}$. If $\mathbb{N}$ had an upper bound, then its supremum $c$ would exist. The number $c - 1$ cannot still be an upper bound, so some natural number lies above it; taking its successor gives a natural number larger than $c$, contradicting that $c$ is an upper bound. [/proofplan] [step:Assume that $\mathbb{N}$ has a real upper bound] Suppose, for contradiction, that $\mathbb{N}$ is bounded above in $\mathbb{R}$. Since $1 \in \mathbb{N}$, the set $\mathbb{N}$ is non-empty. Therefore the hypotheses of the Least Upper Bound Axiom are satisfied: $\mathbb{N} \subset \mathbb{R}$ is non-empty and bounded above. [/step] [step:Take the supremum of $\mathbb{N}$ in $\mathbb{R}$] By the Least Upper Bound Axiom, there exists a [real number](/page/Real_Numbers) $c \in \mathbb{R}$ such that \begin{align*} c = \sup \mathbb{N}. \end{align*} Thus $c$ is an upper bound for $\mathbb{N}$, and every upper bound $b \in \mathbb{R}$ for $\mathbb{N}$ satisfies $c \leq b$. [/step] [step:Use the successor of a natural number to contradict maximal boundedness] The number $c - 1$ is not an upper bound for $\mathbb{N}$. Indeed, if $c - 1$ were an upper bound, then the least-upper-bound property of $c$ would give $c \leq c - 1$, impossible in the ordered field $\mathbb{R}$. Hence there exists $n \in \mathbb{N}$ such that $n > c - 1$. Since $\mathbb{N}$ is closed under the successor operation, $n + 1 \in \mathbb{N}$. Adding $1$ to the inequality $n > c - 1$ gives \begin{align*} n + 1 > c. \end{align*} This contradicts the fact that $c$ is an upper bound for $\mathbb{N}$. [guided] We now use the defining minimality of the supremum. Since $c = \sup \mathbb{N}$, two facts hold: first, $c$ is an upper bound for $\mathbb{N}$; second, no smaller real number can be an upper bound for $\mathbb{N}$. Consider the real number $c - 1$. It is strictly smaller than $c$. If $c - 1$ were also an upper bound for $\mathbb{N}$, then $c$ would not be the least upper bound. More formally, because $c$ is the least upper bound, every upper bound $b \in \mathbb{R}$ of $\mathbb{N}$ must satisfy $c \leq b$. Applying this to the hypothetical upper bound $b = c - 1$ would give \begin{align*} c \leq c - 1, \end{align*} which is impossible in the ordered field $\mathbb{R}$. Therefore $c - 1$ is not an upper bound for $\mathbb{N}$. By the definition of “not an upper bound,” there exists $n \in \mathbb{N}$ such that \begin{align*} n > c - 1. \end{align*} The natural numbers are closed under successor, so $n + 1 \in \mathbb{N}$. Adding $1$ to both sides of the strict inequality preserves the inequality in $\mathbb{R}$ and gives \begin{align*} n + 1 > c. \end{align*} Thus we have found an element $n + 1 \in \mathbb{N}$ that is larger than $c$. This contradicts the fact that $c$ is an upper bound for $\mathbb{N}$. [/guided] [/step] [step:Conclude the unbounded and quantified formulations] The contradiction shows that $\mathbb{N}$ is not bounded above in $\mathbb{R}$. The quantified formulation follows directly. If $x \in \mathbb{R}$ and no $n \in \mathbb{N}$ satisfied $n > x$, then $x$ would be an upper bound for $\mathbb{N}$, contradicting the result just proved. Therefore, for every $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ with $n > x$. [/step]