Explanation for why $\mathbb R^2 \setminus \{0\}$ with the Euclidean metric is not a geodesic metric space?

Question

A geodesic metric space is a metric space $X$ where for any two points $x, y \in X$ there exists a geodesic segment, i.e., an isometry $\gamma: [a,b] \to X$ where $\gamma(a)=x$ and $\gamma(b)=y$.

I read that $\mathbb R^2 \setminus \{0\}$ with the Euclidean metric is not a geodesic metric space as there is no geodesic segment between $(-1,0)$ and $(1,0)$.

I guess I don't understand the concept of geodesic. Why is the top of half of the unit circle not a geodesic segment between $(-1,0)$ and $(1,0)$?

It surely seems like we can find an isometry $\gamma: [a,b] \to \mathbb R^2 \setminus \{0\}$ where $\gamma(a)=(-1,0)$ and $\gamma(b)=(1,0)$. Why is this not possible?

Answer 1

Perhaps that you are missing the fact that, by definition, an isometry preserves distance. The distance between $(1,0)$ and $(-1,0)$ is $2$ and and length of the half-circle that you mentioned is $\pi$, which is greater than $2$.

Answer 2

Assume you have such a $\gamma$. The (continuous) curve $\gamma$ passes somewhere through the $y$-axis, say at $\gamma(\tau) = (0,\sigma)$, where $\tau\in (a,b)$ and $\sigma\neq 0$. Then $$ \sqrt{1+\sigma^2} = |(1,\sigma)| = |(0,\sigma)-(-1,0)| = |\gamma(\tau)-\gamma(a)| = \tau - a $$ and $$ \sqrt{1+\sigma^2} = |(1,-\sigma)| = |(1,0)-(0,\sigma)| = |\gamma(b)-\gamma(\tau)| = b - \tau. $$ Hence, $$ b-a = (b - \tau) + (\tau - a) = 2\sqrt{1+\sigma^2}. $$ But $b-a = |\gamma(b)-\gamma(a)| = |(1,0)-(-1,0)| = 2$, which leads to $\sigma=0$, a contradiction.