Homeomorphism classes of upper-half spaces without 'boundary' points. Let $$H=\\{(x,y)\in\Bbb R^2: y\geq 0\\}$$ i.e. the upper-half space. Let $A$ and $B$ be two finite subsets of the $x$-axis. Then if $|A|=|B|$, clearly, $H-A$ is homeomorphic to $H-B$. However, if $A$ and $B$ are two finite subsets which are not in bijection, how would one show that $H-A$ and $H-B$ are not homeomorphic? I think this is fairly intuitive, for if we take the orbits of points in $H-A$ and $H-B$ under their self-homeomorphism groups and then separate them into components, we will get sets of different cardinalities. But formalizing this is messy. I was wondering if someone could come up with something sleeker.

This is true for all upper half-spaces $H = \Bbb H^k$. By invariance of domain, $\partial M$ is a homeomorphism invariant of $M$. So if $H-A$ and $H-B$ were homeomorphic, then $\partial H - A$ would be homeomorphic to $\partial H - B$. Suppose $|A| = n, |B| = m$. Then $\partial H - A$ is homotopy equivalent to a wedge of $n$ copies of the $(k-1)$-sphere, and $\partial H - B$ is homotopy equivalent to a wedge of $m$ copies of the $(k-1)$-sphere. So you can tell their homotopy types apart by looking at their homology, which is distinct.