text: # Norm Equivalence is Equivalence
  
  ## Theorem
  
  Let $X$ be a vector space.
  
  Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be equivalent norms on $X$.
  
  Denote this relation by $\sim$:
  
  $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
  
  Then $\sim$ is an equivalence relation.
  
  ## Proof
  
  ### Reflexivity
  
  Let $\norm {\, \cdot \,}$ be a norm on $X$.
  
  Then for all $x \in X$ we have that:
  
  $\norm x = 1 \cdot \norm {x}$
  
  Therefore:
  
  $1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$
  
  Hence:
  
  $\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.
  
  $\Box$
  
  ### Symmetry
  
  Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
  
  Then:
  
  $\exists m, M \in \R_{> 0} : m \le M : \forall x \in X : m \norm x_b \le \norm x_a \le M \norm x_b$
  
  Consider two inequalities, obtained by division by $M$ and $m$:
  
  $\dfrac m M \norm x_b \le \dfrac 1 M \norm x_a \le \norm x_b$
  $\norm x_b \le \dfrac 1 m \norm x_a \le \dfrac M m \norm x_b$
  
  Notice, that:
  
  $\dfrac 1 M \norm x_a \le \norm x_b \le \dfrac 1 m \norm x_a$
  
  We have that $m \le M$ implies $\dfrac 1 m \ge \dfrac 1 M$.
  
  Define $C := \dfrac 1 m$ and $c := \dfrac 1 M$.
  
  Hence, $c \le C$ and:
  
  $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_a$
  
  $\Box$
  
  ### Transitivity
  
  Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$ and $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_c$.
  
  Then:
  
  $\exists m_{a b}, M_{a b} \in \R_{> 0} : m_{a b} \le M_{a b} : \forall x \in X : m_{a b} \norm x_b \le \norm x_a \le M_{a b} \norm x_b$
  $\exists m_{b c}, M_{b c} \in \R_{> 0} : m_{b c} \le M_{b c} : \forall x \in X : m_{b c} \norm x_c \le \norm x_b \le M_{b c} \norm x_c$
  
  Generate $2$ more inequalities by multiplying the second inequality by $m_{a b}$ and $M_{a b}$:
  
  $m_{a b} m_{b c} \norm x_c \le m_{a b} \norm x_b \le m_{a b} M_{b c} \norm x_c$
  $M_{a b} m_{b c} \norm x_c \le M_{a b} \norm x_b \le M_{a b} M_{b c} \norm x_c$
  
  From above it follows that:
  
   $\ds m_{a b} m_{b c} \norm x_c$ $\le$ $\ds m_{a b} \norm x_b$ as shown above $\ds$ $\le$ $\ds \norm x_a$ by hypothesis $\ds$ $\le$ $\ds M_{a b} \norm x_b$ by hypothesis $\ds$ $\le$ $\ds M_{a b} M_{b c} \norm x_c$ as shown above
  
  Define $k := m_{a b} m_{b c}$ and $K := M_{a b} M_{b c}$.
  
  Then $k \le K$ and:
  
  $k \norm x_c \le \norm x_a \le K \norm x_c$
  
  Therefore:
  
  $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_c$
  
  $\blacksquare$
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