Diameter of $\left\{(x,y) : x^2+y^2 = 1\right\}$

Question

Maybe it's a extremely trivial question, but working in the metric space $\mathbb{R}^2$, I've tried to find the diameter $\operatorname{diam}(T) = \sup\{d(x,y): x,y \in T\}$ of the set $T = \left\{(x,y) : x^2+y^2 = 1\right\}$.

The first attempt I made was trying to prove that $\operatorname{diam}(T) = 1$. So, we know that, given two points $x = (x_1, x_2)$ and $y = (y_1, y_2)$ in $T$, we have $d(x,y) = \sqrt{2-2 x_1 y_1 - 2 x_2 y_2}$; considering $x$ in the set of all distances, we need to show that $\sqrt{2-2 x_1 y_1 - 2 x_2 y_2} \leqslant 1$, for all such $x$. Unfortunately, from here I don't know how to proceed.

There is a better way to do it?

Answer 1

The diameter is $2$.

As $d((0, 1), (0, -1)) = 2$, the diameter is at least $2$. (To answer your question: I also do not know how to proceed there – the statement you tried to prove is wrong …)

For $x, y \in T$ we have by the triangle inequality $$d(x, y) \le d(x, 0) + d(0, y) = \sqrt{x_1^2 + x_2^2} + \sqrt{y_1^2 + y_2^2} = 1 + 1 = 2,$$ hence the diameter is at most $2$.

Answer 2

Hint:

First, $d(x,y)=\sqrt{2-2x_1 x_2-2y_1y_2}=\sqrt{2-2(\color{red}{x_1 y_1+x_2y_2})}$.

Second, by Cauchy-Schwarz inequality, $$|x_1 y_1 + x_2 y_2| \le \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}=1 \implies-1\le \color{red}{x_1 y_1 + x_2 y_2} \le 1.$$

Answer 3

$(\cos (a),\sin (a))$ and $(\cos (b),\sin (b)) $ are two points of T.

the square of the distance between them is $$D^2=(\cos (a)-\cos (b))^2+(\sin (a)-\sin (b))^2=$$ $$2-2\cos (a-b)=4\sin^2 (\frac {a-b}{2}) $$ $$\le 4$$

thus $$D\le 2$$

the equality holds for $b=0$ and $a=\pi$.