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 if and only if $\struct {R,\norm {\,\cdot\,}_2}$ is complete.

Proof

By Cauchy Sequence Equivalence, for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_1$ if and only if $\sequence {x_n}$ is a Cauchy sequence in $\norm{\,\cdot\,}_2$.

By Convergent Equivalence, for all sequences $\sequence {x_n}$ in $R$:

$\sequence {x_n}$ converges in $\norm{\,\cdot\,}_1$ if and only if $\sequence {x_n}$ converges in $\norm{\,\cdot\,}_2$.

Hence:

every Cauchy sequence in $\norm{\,\cdot\,}_1$ converges in $\norm{\,\cdot\,}_1$ if and only if every Cauchy sequence in $\norm{\,\cdot\,}_2$ converges in $\norm{\,\cdot\,}_2$.

The result follows.

$\blacksquare$