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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.

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    Does your definition of "compactification" imply Hausdorff? Otherwise you can take the one point compactification and simply copy the extra point any finite number of times. (Open sets around two such points then intersect in at least a non-empty open set in $\mathbb{R}$ so the space is not Hausdorff.)2012-11-10

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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.

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    So in a sense the central idea of this argument is the fact that if you partition a linearly ordered set into intervals, then at most two of the intervals can be unbounded. $\qquad$2017-09-13