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It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this examplethis 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 $d(x,y)=0$.

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 $d(x,y)=0$.

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 $d(x,y)=0$.

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Anton Petrunin
Anton Petrunin
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It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov. One can make it bit simplerBelow 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 $\varepsilon_n>0$, but I am lazy to writeso 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 $\varepsilon_n$. A similar exampleConstruct 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 constructed inthe Euclidean metric. Then the $\mathbb R^2$$d$-length of $\gamma_n$ has order of $\varepsilon_n$. In particular $d(x,y)=0$.

It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov. One can make it bit simpler in your case, but I am lazy to write it. A similar example can be constructed in $\mathbb R^2$.

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 $d(x,y)=0$.

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Anton Petrunin
Anton Petrunin
  • 45k
  • 14
  • 135
  • 299

It can be zero.

Take the standard metric on $\mathbb R^3$ and the one given in this example by S. Ivanov. One can make it bit simpler in your case, but I am lazy to write it. A similar example can be constructed in $\mathbb R^2$.