Norm is Complete Iff Equivalent Norm is Complete

Theorem
Let $R$ be a division ring.

Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be equivalent norms on $R$.

Then:
 * $\struct {R,\norm {\,\cdot\,}_1}$ is complete $\struct {R,\norm {\,\cdot\,}_2}$ is complete.

Proof
By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.

By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:
 * $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$.

Hence:
 * every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$.

The result follows.