Is the tangent cone of a totally convex subset again totally convex? $X$ be an Alexandrov's space with lower curvature bound and $C$ be a totally convex subset, i.e. for any $x,y \in C$ and any geodesic $\gamma$ (that is a locally shortest path) connecting $x$ and $y$ we have $\gamma \subseteq C$. For $p \in C$ the tangent cone $K_pC \subset K_p X$ is thus well defined. My question is:
Is $K_pC$ totally convex as well?
It is not hard to see that $K_pC$ is convex in the sense that any unique shortest connection between points in $K_pC$ also lies within $K_pC$, solving this problem for example in the riemannian case. (In fact let $v,w \in K_pC$ together with a unique shortest geodesic $\gamma$ connecting the two points. Using the scaling invariance of the problem together with $(K_pC,0) = \lim_{\lambda \to \infty} (\lambda C,p)$ one may approximate $\gamma$ by geodesics contained in $C$.
But i think in general it might not be possible to approximate arbitrary geodesics like this).
 A: (Too long for a comment)
The question is interesting and it might be hard.
From the comments: The totally convex subset $C$ usually appears as a sublevel set of a locally Lipschitz convex function (I do not know other sources of totally convex subsets).
If $C$ is a sublevel set of a convex function for a not mimimal value $a$
then so is $K_pC$, in particular $K_pC$ is totally convex.
Related stuff. Instead of tangent cone you might consider the same question for a (noncollapsing) Gromov--Hausdorff convergence $A_n\to A_\infty$. (In particular you may think that $A=A_n=A_\infty$ for all $n$ and $C_n$ is a sequence of totally convex sets.) Here some relevant statements which might be useful. 


*

*Any minimizing geodesic in $A_\infty$ can be approximated by minimizing geodesics in $A_n$. (Any minimizing geodesic can be approximated by unique minimizing geodesic, which is approximated by minimizing geodesic in $A_n$.)

*If $A_n$ are Riemannian then any geodesic in $A_\infty$ can be approximated by geodesic in $A_n$. (You approximate a minimizing piece and then extend the approximation.) The general case would follow if the geodesic in Alexandrov space without boundary have infinite extension with probability 1 (this is not known now). 

*You might consider version of definition of totally convex set with quasigeodesics instead of geodesics. In this case the answer is NO; take $A_n=A_\infty$ to be a 2-dimensional cone and the sets $C_n$ which which lie on distance $\ge 1$ from the tip, but for its limit $C_\infty$ there is a quasigeodesic which pass through the tip.

