A property of complete metric spaces makes them length (path or inner) metric spaces, Clarification of a proof

Question

In the book "Metric Structures for Riemannian and Non-Riemannian Spaces", by Misha Gromov, I found a proof of the following statement (of Theorem 1.8. restated here more concentrated)

Let $(X,d)$ be a complete metric space with the property that for all $x,y \in X$ and all $\varepsilon > 0$ there is a point $z \in X$ with $$ \sup( d(x,z), d(z,y) ) \le \frac{1}{2} d(x,y) + \varepsilon $$ then $(X,d)$ is a path (length or inner) metric space.

The proof is to show that for any pair of points in X there exists a continuous curve joining them and then the length of this curve can be arbitrarily minimized to the distance between the two points. In the book we have

• What I don't understand is how it comes to that product at all. I could show that LHS is smaller that a sum instead of a product, and by a good choice of $\varepsilon_k = \varepsilon_0/2^{2k}$, for some $\varepsilon_0 > 0$, the RHS becomes simply $\frac{1}{2^n}(d(x,y) + \varepsilon_0)$, see below.

• Also the extension of $f$ to the other real numbers on $[0,1]$ is not clear to me. Is it like this ? For each real number $x$ (not dyadic rational) take a Cauchy sequence $(x_j)$ of the dyadic rationals convergent to $x$ and show that $(f(x_j))$ is a Cauchy sequence in $X$, and by completeness of $(X,d)$ it has a limit $y$ which one defines as $f(x)$. Then to show this doesn't depend on the Cauchy sequence chosen one takes another Cauchy sequence $(x_j')$ and combines it with first sequence in $(z_j)$ and shows that $(f(z_j))$ is also Cauchy and since its subsequence $(f(x_j)$ converges to $y$ so does all other Cauchy sequences do.

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My approach:

So to find an upper bound on the sum of distances between points constructed by the near midpoint property I proceed as follows ($k$ below is $n$ above and $j$ below is $k$ above, sorry for the confusion)

Answer 1

For $(1)$: Let's first follow the author's pattern. Begin with $\delta=d(x,y)$. We know for any chosen $\kappa>0$ there is $z_{1/2}$, $d(x,z_{1/2})\le\frac{1}{2}\delta+\kappa$. The author sees fit to pick $\kappa=\frac{1}{2}\delta\epsilon_1$ here; then we have $d(x,z_{1/2})\le\frac{1}{2}\delta+\frac{1}{2}\delta\epsilon_1$ (and the same for $d(y,z_{1/2})$). Ok, with the new "$d(x,y)$" term being $d(x,z)$ we can find $z_{1/4}$ with $d(x,z_{1/4})\le\frac{1}{2}d(x,z)+\kappa\le\frac{1}{2}(\delta/2+\epsilon_1\delta/2)+\kappa$ where we are again free to choose $\kappa>0$; the author wants to choose $\kappa=\frac{1}{2}\epsilon_2\cdot(\delta/2+\epsilon_1\delta/2)$. Note: that means we can now write $d(x,z_{1/4})\le\left(\frac{1}{2}+\epsilon_2\right)\left(\frac{1}{2}\delta+\epsilon_1\cdot\frac{1}{2}\delta\right)$ (and the same bounds for e.g. $d(z_{1/4},z_{1/2})$). To continue the induction, I know I can find $z_{1/8}$ with $d(x,z_{1/8})\le\frac{1}{2}d(x,z_{1/4})+\kappa$. How do I choose $\kappa$? I choose it to be $\frac{1}{2}\epsilon_3\cdot\text{ the previous bound }$. Then: $$d(x,z_{1/8})\le\frac{1}{2}\left(\left(\frac{1}{2}+\epsilon_2\right)\left(\frac{1}{2}\delta+\epsilon_1\cdot\frac{1}{2}\delta\right)\right)+\frac{1}{2}\epsilon_3\cdot\underset{\text{the previous bound}}{\underbrace{\left(\left(\frac{1}{2}+\epsilon_2\right)\left(\frac{1}{2}\delta+\epsilon_1\cdot\frac{1}{2}\delta\right)\right)}}\\=\left(\frac{1}{2}+\epsilon_3\right)\left(\frac{1}{2}+\epsilon_2\right)\left(\frac{1}{2}\delta+\epsilon_1\cdot\frac{1}{2}\delta\right)$$Ok. I can then find $z_{1/16}$ and pick $\kappa=\frac{1}{2}\epsilon_4\cdot\text{ the previous bound on $d(x,z_{1/8})$}$. Then $d(x,z_{1/16})\le(\frac{1}{2}+\frac{1}{2}\epsilon_4)\cdot\text{ the previous bound }=(\frac{1}{2}+\frac{1}{2}\epsilon_4)(\frac{1}{2}+\frac{1}{2}\epsilon_3)(\frac{1}{2}+\frac{1}{2}\epsilon_2)(\frac{1}{2}\delta+\epsilon_1\frac{\delta}{2})$. Et cetera.

It's now hopefully clear we can choose the $z_{m/2^n}$ such that $d(z_{m/2^n},z_{(m+1)/2^n})\le\left(\frac{\delta}{2}+\frac{\epsilon_1\delta}{2}\right)\left(\frac{1}{2}+\epsilon_2\right)\cdots\left(\frac{1}{2}+\epsilon_n\right)$ for all $n$, and then it's easy to bound this by $\frac{\delta}{2^n}\prod_{j=1}^\infty(1+\epsilon_j)$.

Your pattern of choosing each $k$th "$\kappa$" to be $\epsilon\cdot 2^{-2k}$ where $\epsilon>0$ is some fixed initial constant works just fine. Under these choices, it's easy to prove by induction that, if we denote the $k$th bound as $\delta_k$ - starting from $\delta_0=\delta=d(x,y)$ - we have: $$\delta_k=2^{-k}\cdot d(x,y)+\epsilon\cdot\sum_{j=k+1}^{2k}2^{-j}$$Then we can always have that: $d(x,z_{1/2^n}),d(z_{1/2^n},z_{2/2^n}),\cdots,d(z_{(2^n-1)/2^n},y)$ are all bounded by $2^{-n}\cdot(d(x,y)+\epsilon)$ using a geometric series.

For $(2)$: it's easy to deduce from the bound on $f$ that $f:\Bbb Q'\to X$ is uniformly continuous, where $\Bbb Q'$ denotes the metric space of dyadic rationals in $[0,1]$; since $[0,1]$ is the metric completion of $\Bbb Q'$, uniform continuity is well known to be a sufficient condition to lift $f$ to a unique continuous extension $f:[0,1]\to X$ (if $X$ is complete). This is to do with the fact that uniformly continuous maps preserve Cauchy sequences; your proposed argument can be made to work, but crucially rests on the uniform continuity.

What you also need to do, which you have have not mentioned in your post, is determine the continuity of this extended map. It’s not also not immediately immediate that the length of the resultant lift $f$ is bounded by $\delta\prod_{j=1}^\infty(1+\epsilon_j)$, you have to do some fiddling and use the fact $f$ is uniformly continuous.