We will prove that all norms are equivalent to $\norm {\, \cdot \,}_2$. By definition, two norms are equivalent on $\R^d$ {{iff}}: :$\forall \mathbf x \in \R^d : \exists m, M \in \R_{> 0} : m \norm {\mathbf x}_a \le \norm {\mathbf x}_b \le M \norm {\mathbf x}_a$ === "Less or equal" condition === Let $\set {\mathbf e_1 \dots \mathbf e_n}$ be a standard basis in $\R^d$. We have that each $\mathbf x \in \R^d$ is uniquely expressible as: :$\displaystyle \mathbf x = \sum_{i \mathop = 1}^d x_i \mathbf e_i$ where $x_i$ is a scalar. Then: {{begin-eqn}} {{eqn | l = \norm {\sum_{i \mathop = 1}^d x_i \mathbf e_i}       | o = \le       | r = \sum_{i \mathop = 1}^d \norm {x_i \mathbf e_i}       | c = Norm axiom: triangle inequality }} {{eqn | r = \sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i}       | c = Norm axiom: positive homogeneity }} {{eqn | r = \sqrt {\paren {\sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i} }^2 } }} {{eqn | o = \le       | r = \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \sqrt {\sum_{j \mathop = 1}^d \size {x_j}^2 }       | c = Cauchy-Schwarz Inequality }} {{eqn | o = \le       | r = \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \norm {\mathbf x}_2       | c = {{Defof|P-Norm|$p$-norm}} }} {{eqn | r = M \norm {\mathbf x}_2       | c = Define $\displaystyle M := \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2}$}} {{end-eqn}} By definition of norm, its image is the set of non-negative real numbers: $M \in \R_{\ge 0}$. Hence: :$\forall \mathbf x \in \R^d : \exists M \in \R_{\ge 0} : \norm {\mathbf x} \le M \norm {\mathbf x}_2$ {{qed|lemma}} === Existence of $m$ === Let $K := \set {\mathbf y \in \R^d : \norm {\mathbf y}_2 = 1}$ be a unit sphere in $\struct {\R^d, \norm {\, \cdot \,}_2}$. By Unit Sphere is Closed in Normed Vector Space, $K$ is closed in $\struct {\R^d, \norm{\, \cdot \,}_2}$. By definition, $K$ is bounded in $\struct {\R^d, \norm{\, \cdot \,}_2}$. Hence, by the Heine-Borel theorem, $K$ is a compact set. By Norm on Vector Space is Continuous Function, the map $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ is continuous from $\struct {K, \norm {\, \cdot \,}_2}$ to $\struct {\R_{\ge 0}, \size {\, \cdot \,}}$: :$\forall \mathbf y_1, \mathbf y_2 \in K : \size {\norm {\mathbf y_1} - \norm {\mathbf y_2}} \le \norm {\mathbf y_1 - \mathbf y_2} \le M \norm {\mathbf y_1 - \mathbf y_2}_2$ By Weierstrass theorem, $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ attains a minimum value $m$ for some $\mathbf y \in K$. Suppose $m = 0$. Then $\norm {\mathbf y} = 0$. By the norm axiom of positive definiteness, $\mathbf y = 0$. Then $\mathbf y \notin K$. Hence, $m \ne 0$. So $m > 0$. Furthermore: :$\forall \mathbf y \in \R^d : \norm {\mathbf y}_2 = 1 : \norm {\mathbf y} \ge m $ {{qed|lemma}} === "Greater or equal" condition === Suppose $\mathbf x = 0$. Then we have equality. Suppose $\mathbf x \ne 0$. Let $\displaystyle \mathbf y = \frac {\mathbf x} {\norm {\mathbf x}_2}$. We have that $\norm {\mathbf y}_2 = 1$ and $\mathbf y \in K$. Then: {{begin-eqn}} {{eqn | l = m        | o = \le        | r = \norm {\mathbf y}  }} {{eqn | r = \norm {\frac {\mathbf x} {\norm {\mathbf x}_2} }  }} {{eqn | r = \frac {\norm {\mathbf x} } {\norm {\mathbf x}_2} }} {{end-eqn}}  This implies that: :$m \norm {\mathbf x}_2 \le \norm {\mathbf x}$ Therefore: :$\forall \mathbf x \in \R^d : \exists m, M : m \norm {\mathbf x}_2 \le \norm {\mathbf x} \le M \norm {\mathbf x}_2$ By definition, all norms are equivalent to $\norm{\, \cdot \,}_2$. By Norm Equivalence is Equivalence, the equivalence relation $\norm {\, \cdot \,} \sim \norm {\, \cdot \,}_2$ is transitive. Thus, all norms are equivalent. {{qed}}