Three-point compactifications of $\mathbb{R}$? A few aeons ago I read a proof that the real line has no $n$-point compactifications for $n\ge 3$.  I don't remember it.  Could someone remind us all?
This says in effect that a function $f:\mathbb R\to\mathbb R$ cannot have more than two horizontal asymptotes.  We can put just one $\infty$ at both ends of the line, or $\pm\infty$ at opposite ends, but that's as far as this goes while $n$ remains finite.
 A: Corrected version (barring further mental hiccups):
Suppose that $X$ is an $n$-point compactification of $\Bbb R$, and let $p_1,\dots,p_n$ be the points at infinity. Let $U_1,\dots,U_n$ be open nbhds of $p_1,\dots,p_n$, respectively, having pairwise disjoint closures, and let $K=X\setminus\bigcup_{k=1}^nU_k$, a compact subset of $\Bbb R$. Let $V=\Bbb R\setminus K=\Bbb R\cap\bigcup_{k=1}^nU_k$, and let $\mathscr{V}$ be the family of order-components of $V$; since $K$ is compact, precisely two members of $\mathscr{V}$ are unbounded. The partition $\mathscr{V}$ of $V$ refines the partition $\{\Bbb R\cap U_k:k=1,\dots,n\}$, so at most two of the sets $\Bbb R\cap U_k$ are unbounded, and the result follows from this

Fact: If $p\in X\setminus\Bbb R$, and $U$ is a nbhd of $p$ in $X$, then $U\cap\Bbb R$ is unbounded.
Proof. If not, let $N\subseteq U$ be a compact nbhd of $p$. Then $N\cap\Bbb R$ is a compact subset of $\Bbb R$ and hence closed in $X$. But then $\{p\}=N\setminus(N\cap\Bbb R)$ is a nbhd of $p$, contrary to the hypothesis that $\Bbb R$ is dense in $X$. $\dashv$

This modern survey of Freudenthal’s theory of ends and the associated compactifications may be of interest.
