A sequence in a non-compact, connected, complete Riemannian manifold

I have come across a question, which I am not convinced I have the right answer to.

Let (M,g) be a connected, non-compact, complete Riemannian manifold, $p\in M$.

a) Show that there exists a sequence $(x_i)_{i\in \mathbb{N}}$ with $d(p,x_i)\overset{i\to \infty}{\longrightarrow} \infty$

b) Show that there exists $X_i\in T_pM$ with $|| X_i ||=1$ so that $x_i = exp_p(d(p,x_i)X_i)$

where $d:M\times M \to \mathbb{R}$ is the distance function.

For a) I thought, that since $M$ is non-compact, there exists a diverging geodesic $\gamma$ with $\gamma(0) = p$. Because $M$ is complete its length is $\infty$. Is that correct reasoning? In class we have not talked about non-compact manifolds.

For $b)$ I thought about using the fact that $exp_p$ is a radial isometry and sends straight lines through $0\in T_pM$ to geodesics through $p\in M$.

• Ok. It makes the second part easier, and my statement is wrong anyway : $\mathbb R$ is a counterexample. The right way to state it is that like $\mathbb R^n$, every closed and bounded set is compact in a complete Riemannian manifold. – астон вілла олоф мэллбэрг Sep 19 '18 at 12:44
Unfortunately no, you can't use that for (a) because it is circular reasoning (how do you propose to prove the existence of the ray $\gamma$ otherwise?). But (b) follows from standard equivalent definitions of completeness (e.g. Hopf-Rinow).
• Argue by contradiction. If the distance from p is all at most some finite R then $M$ is the image of the closed ball $\overline{B(0;R)}\subset T_pM$ under $\exp_p$ so would be compact. – user10354138 Sep 19 '18 at 11:08
• Not quite. You want to use some version of Hopf-Rinow to say there is a minimizing geodesic connecting $p$ to $x_i$ then look at the derivative (i.e. direction) at $p$. – user10354138 Sep 19 '18 at 11:14