 Solution

In this context a nowhere locally homogeneous metric, also called a bumpy metric, is a smooth Riemannian metric for which no two distinct nonempty open subsets are isometric with their induced metrics. Equivalently every local isometry between open subsets is the identity wherever defined: a nonidentity local isometry sends some point $x$ to a distinct point, and restriction to sufficiently small disjoint neighborhoods would violate the first formulation. This is the local-isometry meaning of the terminology here; degeneracy of periodic geodesics is a different use of the word bumpy.

Sunada's local isometry lemma states that on a compact smooth manifold without boundary of dimension $m\ge 2$ the nowhere locally homogeneous metrics contain a residual set in the space of smooth Riemannian metrics with its $C^{\infty }$ topology. In particular they are dense, by the Baire category theorem. A residual set is a countable intersection of open dense sets. The dimension assumption matters: every one-dimensional Riemannian metric is locally $d{s}^2$ in arclength coordinates and has local translations.

The jet bundle of maps $J^k(M,N)$ consists of equivalence classes $j_x^{k}f$ of smooth maps near $x$, where two maps agree to order $k$ at $x$ in coordinate charts. Its projection to $M\times N$ sends $j_x^{k}f$ to $(x,f(x))$. For a multi-index $\alpha$ in $m$ variables there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivatives of order $r$. Thus a coordinate chart consists of the source point, the target point, and $n$ coefficients for every derivative order $1$ through $k$. Summing the counts givesThis includes $J^0(M,N)=M\times N$. The fibre over $(x,y)$ is the space of truncated Taylor maps with fixed constant term $y$, locally modeled onFor $k=1$ its identification with $\mathrm{Hom}(T_{x}M,T_{y}N)$ is canonical. For higher $k$, changes of target coordinates mix derivatives of different orders, so this is a coordinate or connection-dependent description, not a canonical vector-bundle identification. The truncation $J^k\to J^{k-1}$ is an affine bundle modeled on the pullback of ${\mathrm{Sym}}^k(T^{\ast }M)\otimes TN$ over $M\times N$.

For the density assertion put $A={\overline{U}}_i$ and $B={\overline{U}}_j$. Their closures must be distinct. If $A=B$, the identity is an isometry for every Riemannian metric, and the requested complement is empty. Under the intended distinct-domain assumption, the smooth closed-ball hypothesis makes $A$ and $B$ regular closed domains, equal to the closures of their interiors. After interchanging them if necessary, there is a nonempty open set $W$ with compact closure in $\mathrm{int}(A)\setminus B$. Otherwise each interior would be contained in the other closure, forcing $A=B$.

Take any smooth Riemannian metric $g$. If ${\mathrm{vol}}_g(A)\ne {\mathrm{vol}}_g(B)$, it already lies outside ${\mathcal{S}}_{ij}$, because an isometry preserves the Riemannian volume form. If the volumes agree, choose a nonzero nonnegative smooth bump function $\eta$ supported in $W$ and setThese are positive definite Riemannian metrics, they agree with $g$ on $B$, and $g_t\to g$ in $C^{\infty }$ as $t\downarrow 0$, since every derivative of $g_t-g$ is $t$ times a fixed compactly supported smooth tensor. Their Riemannian volume forms satisfyHence ${\mathrm{vol}}_{g_t}(B)={\mathrm{vol}}_g(B)$ whereas ${\mathrm{vol}}_{g_t}(A)>{\mathrm{vol}}_g(A)$ for every $t>0$. In particularThe two domains cannot be isometric for $g_t$. Every $C^{\infty }$ neighborhood of $g$ therefore meets the complement of ${\mathcal{S}}_{ij}$, provingThis localized volume perturbation proves the requested density even when the two domains overlap, and works in every positive dimension. It does not by itself prove the stronger residual local-isometry statement: fixed isometric closures and arbitrary isometric open subsets are different conditions.