$\mathbb{R}$ -trees are CAT(0) space

Question

An $\mathbb{R}$-trees is a metric space $(X,d)$ such that

1. there is a unique geodesic segment (denoted $[x,y]$ ) joining each pair of points $x,y\in X$ ;

2. if $[x,y]\cap[y,z]=\{y\}$ , then $[x,y]\cup[y,z]=[x,z]$.

$\mathbb{R}$-trees are CAT(0) space but I can't prove it!

How can I Prove $\mathbb{R}$-trees are CAT(0) space?

Thanks for helping

Answer 1

Take a triple $ x,y,z \in T $, where $ T $ is your tree. Suppose that in that tree they appear in the alphabetical order. Then the $ \mathbb{R}^2 $ comparison triangle has to have sides of length $ d(x,y), d(y,z), d(x,z) $, but from the definition of a tree it holds that $ d(x,z) = d(x,y) + d(y,z) $. Hence, the comparison triangle is degenerated

You can see from this point that the $ \text{CAT}(0) $ inequality is in fact satisfied as an equality