Is an ellipsoid surface, with shortest-path length used as metric, a metric space?
Is an ellipsoid surface, with shortest path length used as metric, a metric space?
1 Answers
Yes of course.
In any metric space in which any two points can be connected by a rectifiable curve (i.e curves $c:[0,1] \to X$ of finite length $\ell(c)$), define the shortest length metric by $p(x,y) = \inf_{c:x \to y} \ell(c(x,y)),$ where the infimum is taken over all rectifiable curves connecting $x$ to $y$. I claim that it is a metric, I'm not claiming that it induces the right topology, though.
Since we have $\ell(c) \geq d(x,y)$, we have $p(x,y) \gt 0$ whenever $x \neq y$. Symmetry is obvious since the length of a rectifiable curve is invariant under reparameterization $t \mapsto 1-t$.
The triangle inequality follows immediately from the the definition of the infimum: Choose paths connecting $x$ to $y$ and paths connecting $y$ to $z$ whose length realizes the infimum up to $\varepsilon/2$. Their concatenation is a path of length at most $p(x,y) + p(y,z) + \varepsilon$ and connects $x$ to $z$ and since $\varepsilon$ was arbitrary we have $p(x,z) \leq p(x,y) + p(y,z)$.
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0@Ronaldo: To be fair, I just didn't have the patience of going into this argument, so I *"cheated"* my way around it :) Thanks, it certainly is a good and helpful point you're making! – 2011-05-13