Is it true that a metric over $\mathbb{R}^n$ (or $\mathbb{C}^n$) always makes a complete metric space?

Question

I'm studying for my calculus II exam and I was thinking about the statement in the title.

My "sloppy" proof is something like: Over $\mathbb{R}^n$ (or $\mathbb{C}^n$) the Bolzano-Weierstrass theorem is always true so the completeness of the metric space should be independent of the specific metric that I define.

I couldn't come up with a better proof nor did I find a similar statement on my book.

Could you please help me clarify this problem?

Answer 1

No, it is not true. Consider, for instance, the following metric on $\Bbb R$:$$d(x,y)=\bigl|\arctan(x)-\arctan(y)\bigr|.$$With respect to this metric, the sequence $1,2,3,4,\ldots$ is a Cauchy sequence. But it doesn't converge.