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I'm trying to prove that a given function is a metric on some set. I'm confused now because the maths is not not adding up.

Let $X=${$x \in \mathbb{R^2}:\lvert x \rvert =1$}. Given $x,y \in X$, define $\theta$ to be the counter-clockwise angle between $x$ and $y$. $0\leq\theta<2\pi$. Define $\sigma:X\times X \rightarrow \mathbb{R}$, $\sigma(x,y)=min${$\theta,2\pi-\theta$}

Claim: $\sigma$ is a metric on $X$

I started by proving that $\sigma(x,y)=0 \iff x=y$

Proof: "$\Rightarrow$" Let $\sigma(x,y)=0$ then $\min${$\theta,2\pi-\theta$}$=0$

Case 1. Suppose $\theta = \min${$\theta,2\pi-\theta$}. Then $\theta=0$ so obviously $x=y$.

Case 2. Suppose $2\pi-\theta=\min${$\theta,2\pi-\theta$}. Then $2\pi-\theta=0$. We then see that this means $\theta =2\pi$. This is a contradiction since $\theta<2\pi$ by definition of $\theta$. ($2\pi-\theta$ cannot be the minimum?)

I know that it's possible for $2\pi-\theta=\min${$\theta,2\pi-\theta$}. So what am I doing wrong?

Adeeb
  • 733

1 Answers1

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It is indeed possible that $2\pi - \theta$ is the minumum. But in the particular case where $\sigma(x,y)=0$ you just showed it is not possible, as this would imply $\theta=2\pi$. So when $\sigma(x,y)=0$ it is not possible that $2\pi-\theta$ is the minimum.

fidbc
  • 909