Hallo, I have the following question:
Let $(X,d)$ be a complete metric space. Is then $(X,\operatorname{dist})$ also complete? Here by $\operatorname{dist}$ I mean the metric induced by $d$ by: $\operatorname{dist}(x,y)=\inf(L(\gamma))$, where the infimum is taken over all paths joining $x,y$ and $L(\gamma)=\sup(\sum\_{i=1}^{n-1}d(\gamma(t\_{i}),\gamma(t\_{i+1})))$, where the supremum is taken over all partitions $P=(0=t\_{1}< ... < t\_{n}=1)$ of $[0,1]$, where $\gamma$ is defined. Intuitively it should be but I dont know how to prove it. If its true, how can one prove it?


cheers
denis



complete metric space
The intrinsic metric induced by a complete metric space is always a complete extended metric. To prove this, we shall write $d^{\sharp}$ for the extended metric $dist(x,y)=\inf(L(\gamma))$. 


Also, I want to remark that $d^{\sharp}$ is in general not a metric since we may have $d^{\sharp}(x,y)=\infty$ even when $X$ is path-connected. For instance, if we take some fractal $X$ such the boundary of the Koch snowflake, then $dist(x,y)=\infty$ for every pair of points $x,y\in X$. 


However, $d^{\sharp}$ is always an extended metric (By an extended metric I mean that $d^{\sharp}$ is a metric in every way except for the fact that we may have $d(x,y)=\infty$). Notions such as Cauchy sequences and completeness still hold if we use extended metrics instead of metrics. Furthermore, if we let $d'(x,y)=Min(d^{\sharp}(x,y),1)$, then $d'$ is a metric with the same uniform structure as $d^{\sharp}$(and hence the same Cauchy sequences, topology,...).


We shall now prove that $d^{\sharp}$ is complete using standard arguments.
To prove that $d^{\sharp}$ is complete assume that $(x\_{n})\_n$ is a Cauchy sequence with respect to the extended metric $d^{\sharp}$. Then since $d(x,y)\leq d^{\sharp}(x,y)$ for each $x,y\in X$, the sequence $(x\_{n})\_{n}$ is Cauchy with respect to the metric $d$. Since $(X,d)$ is a complete metric space the sequence $(x\_{n})\_{n}$ converges to some point $x\in X$ with respect the metric $d$. Since $(x\_{n})\_{n}$ is Cauchy with respect to $d^{\sharp}$, we may take a subsequence $(y\_{n})\_{n}$ of $(x\_{n})\_{n}$ such that $d^{\sharp}(y\_{n},y\_{n+1})<\frac{1}{2^{n}}$ for all $n$. In this case, there is some path $\gamma\_{n}:[0,1]\rightarrow X$ such that $\gamma\_{n}(0)=y\_{n},\gamma\_{n+1}(1)=y\_{n+1}$ and
$L(\gamma\_{n})<\frac{1}{2^{n}}$. If we let $\gamma:[0,\infty]\rightarrow X$ be the function where $\gamma(r+n)=\gamma\_{n}(r)$ for $n\geq 0$ and $r\in[0,1]$ and $\gamma(\infty)=x$, then $\gamma$ is a continuous function. However, the restricted function $\gamma|\_{[k,\infty]}$ is a path from the point $y\_{k}$ to $x$ with $L(\gamma|\_{[k,\infty]})<\sum\_{n=k}^{\infty}\frac{1}{2^{k}}=\frac{1}{2^{k-1}}$. Therefore $d^{\sharp}(y\_{k},x)<\frac{1}{2^{k-1}}$, so the sequence $(y\_{n})\_{n}$ also converges to $x$ with respect to the generalized metric $d^{\sharp}$.


