Does homeomorphism of finite dimension topological vector space implies completeness Let $X$ be finite dimension topological vector space,that linear homeomorphic to complete vector space $\mathbb{R}^n$ with Euclidean topology.  Prove $X$ is complete also.
We know $\mathbb{R}^n$ is complete, and $X$ is finite dimension.To show the completeness I do as follows:
First since $X$ is metrilizable(due to $\mathbb{R}^n$ is metrilizable), we just need to check Cauchy sequence converge. Just treat $(x_n)\in X$ likes $(\bar{x}_i)\in \mathbb{R}^n$ which complete the proof.
Is my proof correct?Linear is vital in the proof correct?
 A: Here is a counterexample, which highlights the fact that completeness is not independent of the metric.
Take $X = \mathbb R$, which is a 1-dimensional topological vector space. Use the homeomorphism $f : \mathbb R \to (-\pi/2,\pi/2) $ given by $f(x) = \tan^{-1}(x)$ to transport the metric from $(-\pi/2,\pi/2)$ back to $\mathbb R$, obtaining $d(x,y) = \left|f(x) - f(y)\right|$. Any sequence $(x_n)$ that approaches $+\infty$ is nonconvergent in $\mathbb R$, but it is a Cauchy sequence because $(f(x_n))$ approaches $\pi/2$ and therefore $(f(x_n))$ is a Cauchy sequence.
What this counterexample highlights is that the metric space notion of completeness is not quite what one wants for purposes of studying topological vector spaces; this may be where your confusion lies.
Take a look at the wikipedia page on topological vector spaces, particularly the section on metrizability and the section on completeness, and you will see that the topological vector space notion of completeness is a somewhat different and a somewhat subtler than the metric space notion.
