Is there any metric on R with which it is incomplete.

Question

I read the definition of complete metric space. A metric space (X,d) is complete if every cauchy sequence in X converges in X. So, from this definition it looks like if a metric space is not complete and it has some cauchy sequence in it then only possibility is it does not contain limit of that sequence. That is that particular point is missing from that space. For example, consider (0,1] with usual metric and sequence 1/n is cauchy but do not have limit in this space. I also read proof that space of real numbers R is complete with usual metric. But, is there exist some other metric on R such that it will not be complete? I am confusing because I can't imagine a cauchy sequence in R which do not have its limit in R. please help

Answer 1

If $(X,d)$ is a metric space and $Y$ is any set such that there is a bijection $f: Y \to X$ the we can define a metric $D$ on $Y$ by $D(y_1,y_2)=d(f(y_1),f(y_2))$. It is easy to check from the definition that $(Y,D)$ is complete iff $(X,d)$ is complete. Thus, if you take any incomplete metric space $(X,d)$ with the same cardinality as $\mathbb R$ then you can use the existence of a bijection $f: Y \to X$ to get a metric on $Y=\mathbb R$ which is not complete. Example: Use $(0,1)$ with the usual metric and use the bijection $f:Y\to X$ defined by $f(y) =\frac y {1+|y|}$ or the bijection $g:Y\to X$ defined by $g(y) =\frac 2 {\pi} \arctan y$.