I'm reading this paper. And I have some questions about Alexandrov upper angle.
$M$ is a complete geodesic metric space and $\mathbb M^2_\kappa$ is a complete simply connected Riemannian 2-manifold with constant sectional curvation $\kappa>0$. For all $x,y \in M$, let $xy$ denote a segment connecting $x$ and $y$. Given $p,x,y ∈ M$, denote $(\overline{p},\overline{x},\overline{y})$ a comparison triple of $(p,x,y)$, i.e.$\overline{p},\overline{x},\overline{y}\in \mathbb M^2_\kappa$, $|\overline{py}| = |py|$, $|\overline{px}| = |px|$ and $|\overline{xy}| = |xy|$.
For a hinge $H_p(x,y)$ (consists of the vertex $p$ and two sides $px$ and $py$) in $M$, the Alexandrov upper angle of $H_p(x, y)$ is then defined by $$\angle_p(x,y) := \limsup_{u \in{px},v\in{py}, u,v\to p} \angle_\overline{p}(\overline{u},\overline{v}) $$
I have two questions,
1.Why is $\angle_p(x,y)$ independent of curvature $\kappa$?
2.Why the upper angle always exists as a limit if we assume $M$ is a space of curvature $\geq \kappa$, i.e. for every point $p\in M$, there is a neighborhood of $p$ within which every hinge $H_p(x,y)$ satisfies$|uv|\geq |\overline{u} \overline{v}|$, whenever $ u\in px, v\in py $,$ \overline{u}\in \overline{px}, \overline{v}\in \overline{py} $ and $|pu|=|\overline{pu}, |pv|=|\overline{pv}|$?
If $p,\ x,\ y$ are points in $M$ assume that $\widetilde{\angle}^k [p\ xy]$ is an angle in a comparison triangle $\widetilde{\triangle}^k pxy$ in $k$-plane $M^2_k$. That is $\angle [p\ xy]$ is limit of $\widetilde{\angle} [p\ xy]$ First note that $k\mapsto\widetilde{\angle}^k [p\ xy]$ is nondecreasing (I can not prove this)
So we will use this fact : If $k<K$,
\begin{align} 0 &\leq \widetilde{\angle}^K [p\ xy] -\widetilde{\angle}^k [p\ xy] \\&\leq \widetilde{\angle}^K [p\ xy] +\widetilde{\angle}^K [x\ yp] +\widetilde{\angle}^K [y\ px] -\widetilde{\angle}^k [p\ xy]-\widetilde{\angle}^k[ x\ yp] -\widetilde{\angle}^k [y\ px] \\&= K \cdot {\rm area} \widetilde{\triangle }^K pxy -k\cdot {\rm area}\widetilde{\triangle}^k pxy \end{align}
If lengths $px,\ py$ go to $0$, then areas go to $0$ So we complete the proof