For a complete non-compact Riemannian manifold with sectional curvature positive, it is diffeomorphic to $\mathbb{R}^n$ by known result. Choose a point $p$ on the manifold, is it possible that the distance function $d(p, \cdot)$ has a sequence of critical points going to the infinity of the manifold?
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$\begingroup$ Sectional curvature is positive or nonpositive? $\endgroup$– Mahdi - Free PalestineCommented Aug 6, 2017 at 11:12
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$\begingroup$ strictly positive sectional curvature. $\endgroup$– mmaatthhCommented Aug 6, 2017 at 11:45
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Yes, it is a classical result.
Let $q_1,\dots,q_n$ be a sequence of critical points such that $$|q_{n+1}-p|\ge 2\cdot |q_n-p|.$$ By Toponogov's comparison, $$\measuredangle [p\,^{q_i}_{q_j}]\ge \tfrac\pi3.$$ Hence we get a bound on $n$.
answered Aug 6, 2017 at 18:36
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$\begingroup$ What about $3$-dim complete Riemannian manifold with $Rc> 0$? Schoen-Yau had proved it is diffeomorphic to $\mathbb{R}^3$, is it also true that all critical points with respect to a fixed point $p$ is in a compact set?@Anton Petrunin $\endgroup$– mmaatthhCommented Aug 7, 2017 at 0:00