# complete metric space

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

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sorry I do not see where this is answered. where exactly? – denis May 28 '13 at 6:51
can you point it out, please. – denis May 28 '13 at 6:55
is the statement in the question above true or not? If yes, how could one prove it ? – denis May 28 '13 at 7:10
What is left unsaid in the answer and the wikipedia link is that the original metric and the "intrinsic metric" can be pathologically different. For example, if $X$ is the Cantor set or any other totally disconnected metric space then dist completely degenerates to the unique $0,\infty$ extended metric. – Lee Mosher May 28 '13 at 13:57

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 $d^{\sharp}$.

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You say $\mathrm{dist}(x,y) = \infty$ for every pair of distinct points on the Koch snowflake. Surely this can't be true? Suppose $x$ was introduced by the $n^\mathrm{th}$ iteration of the construction, and $y$ by the $m^\mathrm{th}$, with $m < n$. Then both are in the $n^\mathrm{th}$ iteration, which is not fractal at all, so there should be a reasonable path between them. Adding more points can only provide more (shorter) paths (although I suspect the shortest path is actually already in stage $n$). – Ben Millwood Jun 25 '13 at 15:52
Ben. I don't think I understand your objection. I am talking about the limit of the Koch snowflake. – Joseph Van Name Jun 25 '13 at 18:25
Of course. But the Koch snowflake is only badly-behaved at its perimeter: the interior surely has perfectly ordinary paths with perfectly ordinary lengths in it. (Just to make sure we are talking about the same thing, here is a picture commons.wikimedia.org/wiki/… ) – Ben Millwood Jun 27 '13 at 13:39
I was talking about the boundary of the Koch snowflake. I edited the answer, so hopefully it is now more clear. – Joseph Van Name Jun 27 '13 at 22:14
Aha, okay. I find that much more believable. – Ben Millwood Jun 28 '13 at 16:08