Let $M \in \R^{m \times n} : m, n \in \N_{\mathop > 0}$ be a real matrix. Denote the $\paren {i, j}$-th entry of $M$ by $a_{ij}$. Note that the set of matrix elements of $M$ is a finite set of real numbers. We have that: :Real Numbers form Ordered Field :Non-Empty Finite Set has Greatest Element {{Help|Something of this sort}} Therefore, $M$ has the greatest element. === Norm Axiom $(\text N 1)$ === {{begin-eqn}} {{eqn | l = \norm M_\infty       | r = \max_{\begin {split}                    1 \mathop \le i \mathop \le m \\                   1 \mathop \le j \mathop \le n                    \end {split} }              \size {a_{ij} }       | c = Greatest Element is Supremum, {{defof|Max Operation}} }} {{eqn | o = \ge       | r = 0 }} {{end-eqn}} Equality is obtained for $M$ being a zero matrix. Suppose $\norm M_\infty = 0$. Then:  :$\displaystyle \forall i, j : 1 \le i \le m, 1 \le j \le n : \size {a_{ij}} = 0$ In other words, $M$ is a zero matrix. {{qed|lemma}} === Norm Axiom $(\text N 2)$ === {{begin-eqn}} {{eqn | l = \norm {\alpha \cdot M}_\infty       | r = \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} }              \size {\alpha m_{ij} }       | c = Greatest Element is Supremum }} {{eqn | r = \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} } \size \alpha \size {a_{ij} }       | c = Absolute Value of Product }} {{eqn | r = \size \alpha \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} } \size {a_{ij} } }} {{eqn | r = \size \alpha \norm M_\infty       | c = Greatest Element is Supremum }} {{end-eqn}} {{qed|lemma}} === Norm Axiom $(\text N 3)$ === Let $P, Q \in \R^{m \times n}$. Denote their $\paren {i, j}$-th matrix elements as $p_{ij}$ and $q_{ij}$ respectively. Fix $i,j \in \N : 1 \le i \le m, 1 \le j \le n$. We have that: {{begin-eqn}} {{eqn | l = \size {p_{ij} + q_{ij} }       | o = \le        | r = \size {p_{ij} } + \size {q_{ij} }       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} } \size {p_{ij} } +              \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} } \size {q_{ij} }       | c = {{defof|Max Operation}} }} {{eqn | r = \norm P_\infty + \norm Q_\infty       | c = Greatest Element is Supremum, {{defof|Supremum Norm}} }} {{end-eqn}} This holds for any $i,j$. Hence: {{begin-eqn}} {{eqn | l = \norm {P + Q}_\infty       | r = \max_{\begin {split}                   & 1 \mathop \le i \mathop \le m\\                   & 1 \mathop \le j \mathop \le n                   \end {split} } \size {p_{ij} + q_{ij} }       | c = Greatest Element is Supremum }} {{eqn | o = \le       | r = \norm P_\infty + \norm Q_\infty }} {{end-eqn}} {{qed|lemma}} All norm axioms are seen to be satisfied. Hence the result. {{qed}}