The title is my question.


Alexandrov space here means finite dimensional Alexandrov space with curvature bounded below ,denoted by CBB.


Let $\gamma$ be a simple curve in a $n$ dimensional CBB $M$ with two end points in the manifold part of $M$. 


**Analysis**.
$\gamma$ can be covered by a finite collection $\mathcal U$ of cone neighborhoods, where the cone neighborhood here means a subspace of $M$ homeomorphic to $\mathbb{R}^m\times cone$. Since every cone nbhd in $\mathcal{U}$ must contain two manifold points. If we are able to find a new path contained in the manifold part of every cone nbhd in $\mathcal{U}$, then we done. So my question is reduced to the the restriction of the cone nbhd. However, since I don't know the the structure of the $cone$ part in the cone nbhd very well. I don't know how to get such a new path. Maybe the density of the manifold points in $M$ would help, but I don't know how to use this condition. 



Whether the manifold part of an Alexandrov space is connected?
Yes.


Assume $A$ is an $m$-dimensional Alexandrov space and $\Omega\subset A$ be the maximal open subset which is a topological $m$-manifold and $A'\subset A$ be the subset of all points with tangent space isometric to Euclidean space.


From Perelmans paper "Beginning of Morse...", we get that $A'\subset \Omega$.


The set $A'$ is a dense convex subset in $A$, this follows from Burago--Gromov--Perelman paper and from my paper on parallel translation.
(Convexity means that any geodesic with the ends in $A'$ lies completely in $A'$.) 
In particular $A'$ is connected; hence the result follows.


