From Wikipedia: In metric spaces, the completeness is a property of the metric. It is not a property of the topological space itself. If you move on to an equivalent metric (that is a metric which induces the same topology), the completeness can get lost. Regarding two equivalent norms on a normed vector space, however, one of them is complete if and only if the other one is complete. Therefore, in the case of normed vector spaces, the completeness is a property of the norm topology, which does not depend on the specific norm.

Is this a mistake? Because if $d$ and $D$ are equivalent metrics then $x_n$ is Cauchy in $D$ if and only if $x_n$ is Cauchy in $d$. Therefore $X$ is complete in $d$ if and only if complete in $D$?

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    $\begingroup$ There is no mistake. Equiuvalent metrics gives the same topology. But there are non equivalent metrics that can give the same topology. $\endgroup$ – Norbert Mar 15 '13 at 9:56

The metric induced by a norm has a very strong dependence on the underlying linear space, whereas your everyday metrics may have complete disregard for the underlying linear space.

The simplest example would be $\mathbb{R}$. We know that the standard metric is induced by a norm (the absolute-value function is the prototypical norm), but we can construct metrics on $\mathbb{R}$ that are not complete (but then which are not induced by any norm). An example of this is: $$\rho ( x , y ) = | \arctan (x) - \arctan (y) |.$$ Under this metric the sequence $\langle n \rangle_{n =1}^\infty$ is a Cauchy sequence, but it clearly does not converge. But note, also, that $$\rho ( ax , ay ) \neq |a| \rho ( x ,y )$$ so that this metric cannot be induced by a norm on the linear space $\mathbb{R}$.

There is a stronger notion than just generating the same topology. Two metrics $d , \rho$ on a set $X$ are called Lipschitz equivalent if there are positive constants $\alpha , \beta$ such that $$\begin{gather}\rho ( x , y ) \leq \alpha\, d ( x , y ) \\ d ( x , y ) \leq \beta\, \rho ( x , y )\end{gather}$$ for all $x,y \in X$. It is easy to see that any two Lipschitz equivalent metrics induce the same topology, and also that they have the same convergent sequences. Beyond this it is not too difficult to show that two Lipschitz equivalent metrics have the same Cauchy sequences. This then says that given two Lipschitz equivalent metrics on a set $X$, they are either both complete, or both incomplete. We can show that any two norm metrics on a linear space which generate the same topology are Lipschitz equivalent, which gives the result mentioned in the Wikipedia article.

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    $\begingroup$ Does $\rho(x,y) = |arctan(x) - arctan(y)|$ induce the same topology than $|x-y|$? $\endgroup$ – blue Mar 15 '13 at 10:35
  • $\begingroup$ @blue: Yes, it does. This basically follows from the fact that $\arctan$ is a homeomorphism from $\mathbb{R}$ onto $( \frac{-\pi}{2} , \frac{\pi}{2} )$. $\endgroup$ – user642796 Mar 15 '13 at 10:48
  • $\begingroup$ Excuse me, and what about $d$ and $D = d / (d + 1)$? They induce same topology. Do they also have same Cauchy sequences? We only have Lipschitz constant on one side. $\endgroup$ – blue Mar 15 '13 at 14:26
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    $\begingroup$ @blue: They would have the same Cauchy sequences. If $d^\prime = d / ( d + 1 )$, I am sure you noticed that $d^\prime \leq d$, but note also that given $0 < \epsilon < 1$ we have $d^\prime ( x , y ) < \epsilon$ iff $d ( x , y ) < \epsilon / ( 1 - \epsilon )$. We can then show that the identity function is uniformly continuous between $( \mathbb R , d )$ and $( \mathbb R , d^\prime )$, and vice versa. From here we can show that they have the same Cauchy sequences. $\endgroup$ – user642796 Mar 15 '13 at 15:10

Wikipedia provides an example: "the real numbers [...] are complete but homeomorphic to the open interval $(0,1)$, which is not complete".


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