Given a complete Riemannian manifold $M$ and point $p\in M$, denote $\mathrm{Cut}_p$ the cut locus of $p$ and $\mathrm{Cut}_p^1\subset \mathrm{Cut}_p$ the set of points $q$ which are connected to $p$ by more than one length minimising geodesic. According to a remark in Sakai's Riemannian geometry book (Rmk. 4.9), the latter forms a dense subset - but I don't understand why.
Question: Why is $\mathrm{Cut}_p^1\subset \mathrm{Cut}_p$ dense?
(I use density in this answer on MO. Comments on how to avoid this property to prove regularity of Riemannian distance function are also very welcome.)
As user Chee kindly pointed out to me in a comment on MO one can use the following result from Klingenberg's Riemannian geometry book (Theorem 2.1.12):
Theorem. On a complete Riemannian manifold $(M,g)$, if $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ for some $(p,v)\in TM$, then $\exp_p$ fails to be injective in every neighbourhood of $v$.
(Thus if injectivity fails infinitesimally, it also fails in every neighbourhood. This is of course false for a general smooth map.)
Now to conclude, take $x \in \mathrm{Cut}_p\backslash \mathrm{Cut}_p^1$, and note that $x=\exp_p(v)$ for some $v\in T_pM$ with $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ (cf. Petersen Lemma 5.78). Take a sequence of opens $U_1\supset U_2\supset\dots\subset T_pM$ with $\bigcap U_n=\{v\}$, then according to the theorem above $\exp_p$ fails to be injective on each $U_n$ and thus $\exists x_n\in \mathrm{Cut}_p^1\cap\exp_p{U_n}$. Clearly $x_n\rightarrow x$ and thus we have proved the density $\mathrm{Cut}^1_p\subset \mathrm{Cut}_p$.