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.


4 Answers 4


It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov. Below I give simplification of his example which works in your case.

Simplified example. Choose two points $x$ and $y$ in $\mathbb R^2$ and construct a sequence of arcs $\gamma_n$ between them. For each $n$ choose small $\varepsilon_n>0$, so that $\varepsilon_n\to 0$ very fast. Choose disjoint $\varepsilon_n$-intervals on $\gamma_n$, so that it cover all $\gamma_n$ except set of lenght $\varepsilon_n$. Construct metric $d_1$ so that it is very cheap to go along each such interval, but compensate it by making very expancive to get to such interval, so you can not use it as a shortcut.

Let $d_2$ be the Euclidean metric. Then the $d$-length of $\gamma_n$ has order of $\varepsilon_n$. In particular $d(x,y)=0$.

  • $\begingroup$ It would be very kind of you to explain the example a bit more fully. $\endgroup$ Apr 2, 2013 at 13:38
  • $\begingroup$ @Joel, Sergei Ivanov did it very well $\endgroup$ Apr 2, 2013 at 13:47

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 $\varepsilon>0$.

  • 2
    $\begingroup$ Neat. I think I'll put this on an exam, perhaps with a hint. $\endgroup$ Apr 2, 2013 at 15:52

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):

  1. $f_0(x) = x$ for all $x$.
  2. $f_{n+1}$ coincides with $f_n$ on diadic numbers of denominator $2^{-n}$.
  3. $f_{n+1}\left(\frac{2k+1}{2^{n+1}}\right) = f_n\left(\frac{k}{2^n}\right) + \frac{1}{3^{n+1}}$ 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 $d_2$.

alt text


Unless I'm missing something, the answer is almost trivially no:

For all i, $d_i$ takes values in $\mathbb{R}^+$, so the $d(x,y)=0$ if and only if there is a decomposition such that for all $j$, $d_{i_j}(x_{j-1},x_j)=0$, which happens if and only if $x_{j-1}=x_j$. Therefore, if $d(x,y)=0$, then $x=x_0=x_1=\dots=x_l=y$ (and vice-versa).

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    $\begingroup$ The infimum can be zero even if all sums in the set are positive. $\endgroup$ Apr 2, 2013 at 11:29
  • $\begingroup$ you're right, I did indeed miss something... $\endgroup$
    – Denis
    Apr 2, 2013 at 12:24

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