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Every nonempty closed boundary subset invariant under a non-elementary Kleinian group contains the Kleinian limit set. Such a subset cannot be a singleton. Its hyperbolic convex hull contains an interior point, whose orbit lies in the hull; in the Klein ball model the boundary of this hull is contained in the original closed boundary subset.

Let $F\subseteq S_{\infty }^2$ be nonempty, closed and $\Gamma$-invariant. We first show it cannot be a singleton when $\Gamma$ is a non-elementary Kleinian group.

Suppose every element fixes one boundary point, conjugated to $\infty$. Each boundary map then has the affine form $z\mapsto az+b$. If all elements have $|a|=1$, their action in upper half-space keeps the height of $o$ fixed. An orbit sequence escaping compact interior sets must have its horizontal coordinate tending to infinity along any convergent boundary subsequence, so its only possible boundary limit is $\infty$. If some element has $|a|\ne 1$, move its second fixed point to $0$, making it $f(z)=az$. For any $g(z)=cz+d$, its conjugates are $f^{n}g{f}^{-n}(z)=cz+a^{n}d$. Choose $n\to +\infty$ or $-\infty$ so $a^n\to 0$. If $d\ne 0$, these are distinct group elements converging to $z\mapsto cz$, contradicting discreteness. Thus $d=0$ for every group element: all fix both $0$ and $\infty$, and their only possible interior-orbit boundary limits are these two points. In both cases the Kleinian limit set has at most two points. This contradicts non-elementarity. Therefore $F$ has at least two distinct points.

Use the Klein ball model, in which hyperbolic geodesics are straight chords, and take the closed Euclidean convex hull $C$ of $F$ in the closed ball. It contains an interior point $o$, since the open chord between two distinct points of $F$ lies in the open ball. The interior part of this hull is hyperbolically convex and invariant under $\Gamma$: isometries preserve geodesic segments, so they preserve the hyperbolic convex hull of the invariant set. Thus $\Gamma o\subseteq C$.

The only points of $C$ on the boundary sphere are the points of $F$. Indeed, if $\xi \in S_{\infty }^2\setminus F$, compactness of $F$ gives $\xi \cdot z\le c<1$ for every $z\in F$, because $\xi \cdot z=1$ on the unit sphere only when $z=\xi$. The same linear inequality holds throughout $C$, excluding $\xi$ from it. Conversely $F\subset C$, so $C\cap S_{\infty }^2=F$.

Every boundary accumulation point of $\Gamma o$ consequently lies in $F$. Base-point independence from part (b) identifies that accumulation set with $\Lambda (\Gamma )$, proving minimality of the Kleinian limit set:This proves the stated containment for every nonempty closed invariant set, not merely for sets containing an arbitrarily chosen orbit on the boundary.