# Complete spaces and finite dimensional spaces

We know that if every Cauchy sequence is convergent we call the space is complete . We have a theorem states every finite dimensional space should be complete . Let consider $(\mathbb R,d)$ where $d(x,y)=|\arctan(x)-\arctan(y)|$ it is obvious that $\mathbb R$ is finite dimensional but this space is not complete. Where is the error?

• The statement is that every finite-dimensional normed vector space is complete. This is a metric but it's not a norm. – Qiaochu Yuan Dec 13 '17 at 2:41
• What is a Cauchy sequence with respect to this metric that does not converge? – Ovi Dec 13 '17 at 2:44
• @Ovi. The sequence $(n)_{n\in \Bbb N}.$ The function $f(x)=\arctan x$ is a homeomorphism from $(\Bbb R,d)$ to $((-\pi /2, \pi /2), d)$,where $d(u,v)=|u-v|.$ – DanielWainfleet Dec 13 '17 at 3:16
• @DanielWainfleet Thank you – Ovi Dec 13 '17 at 3:18

Remember that completeness is a property of metric spaces, but every normed space is a metric space, with metric $\rho(x,y)=\|x-y\|$. It is with regards to this metric that we talk about completeness of normed spaces.
In your particular example, you have given a metric on the set $\mathbb R$, but no norm, so you cannot apply the theorem you are thinking of (It would be a useful exercise for you to inspect the proof of the theorem to see where the normed structure is used.) To show that the metric space $(\mathbb R,d)$ is not complete we need to find a Cauchy sequence that does not converge in this metric space. There are many sequences that will do the job. The simplest is the sequence $(x_n)=(n)$. If you have not already done so, I leave it to you to show that $(x_n)$ is Cauchy, but does not converge in $(\mathbb R,d)$, but feel free to ask for hints if you get stuck.
In light of what I said earlier about the theorem concerning the completeness of finite dimensional vector spaces, the incompleteness of $(\mathbb R,d)$ thus means that the metric $d$ cannot come from a norm on $\mathbb R$. It might be interesting to see why not.