TITLE: Infimum of a finite number of distances in the plane
QUESTION [9 upvotes]: Suppose one has a finite number of distances $d_1,\ldots,d_k$ on the Euclidean plane all of which metricize the usual Euclidean topology.
Define for each pair of points $x$ and $y$ in the plane
$$d(x,y) = \inf\left\lbrace d_{i_1}(x_0,x_1) + \cdots d_{i_l}(x_{l-1},x_l) \right\rbrace$$
where the infimum is taken over all finite sequences with $x_0 = x,x_1,\ldots,x_l = y$ and all possible choices of $i_1,\ldots, i_l$.
Question: Is it possible for $d(x,y)$ to be $0$ for a pair of distinct points $x \neq y$?
Also, I'm interested in the same question replacing $\mathbb{R}^2$ by an arbitrary Polish space.

REPLY [11 votes]: This is possible even on the real line.
There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor functions. See, for example, Gelbaum and Omsted, "Counterexamples in analysis" (1964), Chapter 8, Example 30.
This function has the following property: for every $\varepsilon>0$ there is a collection of disjoint intervals $[a_i,b_i]\subset[0,1]$ with total length greater than $1-\varepsilon$ and total variation of $f$ less than $\varepsilon$:
$$
\sum (b_i-a_i)>1-\varepsilon, \qquad \sum (f(b_i)-f(a_i)) < \varepsilon .
$$
Now define metrics $d_1$ and $d_2$ on $[0,1]$ as follows: $d_1$ is the standard metric and $d_2$ is the pull-back of the standard metric by $f$, i.e. $d_2(x,y)=|f(x)-f(y)|$. You can go from 0 do 1 through points $a_1,b_2,a_2,b_2,\dots$ using the distance $d_2$ between $a_i$ and $b_i$ and $d_1$ between $b_i$ and $a_{i+1}$. Thus $d(0,1)<\varepsilon$ for every $\varepsilon>0$.

REPLY [5 votes]: The infimum can be zero as pointed out by Anton Petrunin.  Here's a construction on the interval $[0,1]$.
Consider a sequence of piecewise linear functions $f_n:[0,1] \to [0,1]$ each of which is strictly increasing defined by (see figure):  

$f_0(x) = x$ for all $x$.
$f_{n+1}$ coincides with $f_n$ on diadic numbers of denominator $2^{-n}$.
$f_{n+1}\left(\frac{2k+1}{2^{n+1}}\right) = f_n\left(\frac{k}{2^n}\right) + \frac{1}{3^{n+1}}$ for integer $k$.

Let $f = \lim\limits_{n \to +\infty}f_n$ and $d_1$ be the pullback of the standard Euclidean distance on $[0,1]$ under $f$.   One has that
$$d_1(0,1/2) = \frac{1}{3}$$
$$d_1(0,1/4) = d_1(1/2,3/4) = 1/9$$
$$d_1(0,1/8) = d_1(1/4,3/8) = d_1(1/2,5/8) = d_1(3/4,7/8) = 1/27$$
etc...
One can construct in similar fasion a distance $d_2$ satisfying 
$$d_2(1/2,1) = \frac{1}{3}$$
$$d_2(1/4,1/2) = d_2(3/4,1) = 1/9$$
$$d_2(1/8,1/4) = d_2(3/8,1/2) = d_2(5/8,3/4) = d_2(7/8,1) = 1/27$$
etc...
By considering diadic partitions one sees that the infimum is $0$ when one can use the distances $d_1$ and $d_2$.