How to prove $(0,1) \times \mathbb{R} \, , \, (0,2) \times \mathbb{R}$ are not isometric?

Question

I am trying to prove in an elementary way that $X_1=(0,1) \times \mathbb{R} \, , \, X_2=(0,2) \times \mathbb{R}$ (with the standrad euclidean metric inherited from $\mathbb{R}^2$) are not isometric as metric spaces.

One way to do this is to prove they are not isometric as Riemannian manifolds, and then use Myers–Steenrod theorem. To see they are not isometric as Riemannian manifolds, note that in $X_2$ there are pairs of orthogonal geodesic segments, one with length arbitrarily close to $2$ (horizontal) and the other (vertical) arbitrarily long.

This is impossible in $X_1$.

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I would like to find a more elementary argument, which does not rely on Riemannian geometry.

I think one possible approach is to show $X_2$ contains squares* with sides arbitrarily close to $2$ and $X_1$ does not.

This seems obvious, however a rigorous argument may not be trivial.

*Define a square with side $L$ in a metric space as: $\{x_0,x_1,x_2,x_3 \}$ such that: $d(x_i,x_{i+1})=L \, \forall i \in \mathbb{Z}_3$ and $d(x_0,x_2)=d(x_1,x_3)=L\sqrt{2}$

Answer 1

Instead of squares I would have taken circles. But come to think of it, we might simply check for the existenc of an equilateral triangle of side length $2$.

• In $(0,2)\times\Bbb R$, the points $(a,-1)$, $(a,1)$, $(\sqrt 3+a,0)$ are such a triangle (provided $0<a<2-\sqrt 3$)
• In $(0,1)\times \Bbb R$, there is no such triangle: Suppose $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ are such a triangle and wlog. $y_1\le y_2\le y_3$. Then $y_3\le y_1+2$ and hence for one $i\in\{1,3\}$ we have $|y_2-y_i|\le 1$. Then $(x_i-x_2)^2+(y_i-y_2)^2=4$ implies $|x_i-x_2|\ge\sqrt 3>1$, contradiction.