Homeomorphism between compact subsets and their diameters

Question

Let $V_1$ be a Banach space and let $A_1 \subset V_1$ be a compact subset of $V_1$

Let $V_2$ be a Banach space and let $A_2 \subset V_2$ be a compact subset of $V_2$

Let $f: A_1 \to A_2 $ be a homeomorphism

Let $x_1,y_1 \in A_1$ such that $$\left\| x_1-y_1 \right\|=\sup_{v,u \in A_1} \left\| v-u \right\|$$

Let $x_2,y_2 \in A_2$ such that $$\left\| x_2-y_2 \right\|=\sup_{v,u \in A_2} \left\| v-u \right\|$$

I would like to know if is true that $$ f(x_1)=x_2 \text{ and } f(y_1)=y_2$$

thanks

Answer 1

No, take $V_1=V_2=\mathbb{R}^2$ and $A_1=A_2=D$ a unit closed disk. Define $f:D\to D$ to be a rotation (any that is not the identity).

Take $x_1=x_2,y_1=y_2$ to be any antipodal points (they maximize the distance on the disk). By the choice of $f$ obviously $f(x_1)\neq x_1$. Additionally $f(x_1)\neq y_1$ if $f$ is not a rotation by $\pi$.

This is not true even if you weaken the condition, i.e. for $x_1, y_1$ maximizing the distance $f(x_1), f(y_1)$ do not have to maximize the distance. The example can be given by taking $x_1, y_1$ to be antipodal on the disk in $\mathbb{R}^2$ and then contracting (a little bit) the disk along the line connecting $x_1, y_1$ to produce a homeomorphic space in which $f(x_1), f(x_2)$ do not maximize the distance.

Answer 2

If this were true, any self-homeomorphism of $S^1$ in $\Bbb R ^2$ would have to send antipodal points to antipodal points. Even worse: By picking the points conveniently you can use it to "prove" that every such self-homeomorphism is the identity.