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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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    Do you mean that any *continuous* function has at most two horizontal asymptotes?2012-11-09
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    I doubt that it matters whether it's continuous. Suppose $f(x) = 1/x$ if $x\in \mathbb Q$ and $-1/x$ otherwise. There's _one_ horizontal asymptote. Suppose $f(x) = 1/x$ if $x\in \mathbb Q$ and $10-1/x$ otherwise. There are _no_ horizontal asymptotes. How could discontinuities result in more than two horizontal asymptotes?2012-11-09
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    Yeah, I suppose you're right. I forgot that asymptotes are $\lim\limits_{x\to\pm\infty}$ rather than an accumulation point at $\pm\infty$.2012-11-09
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    could you tell me how you are defining horizontal asymptote? I don't know of a way to formulate them that doesn't seem to implicitly assume a 2-point compactification.2012-11-09
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    @Logan : Say you have an $n$-point compactification $\mathbb{R}\cup\{\infty_1,\ldots,\infty_n\}$. If $\lim\limits_{x\to\infty_k} f(x)=L_k$, then that's a horizontal asymptote.2012-11-10
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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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    What do you mean by regular? $T_3$ comes to mind, but there's no reason why a compact, $T_3$, closed subset of $\mathbb{R}$ should have to be an interval.2012-11-09
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    @Logan: $U\subseteq X$ is a regular open set iff $\operatorname{int}\operatorname{cl}U=U$; $F\subseteq X$ is regular closed iff $\operatorname{cl}\operatorname{int}F=F$. This is standard terminology.2012-11-09
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    Why must a compact, regular closed subset of $\mathbb R$ be an interval? What about, for example, the union of two disjoint, closed intervals?2012-11-10
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    @Andreas: Because I got ahead of myself and forgot to write down all of the details. It may be a little before I get back to it; I’m tied up at the moment.2012-11-10
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    What does "order-components of $V$" refer to?2012-11-14
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    @Seirios: If $\langle X,\le\rangle$ is a linear order, and $A\subseteq X$, the order-components of $A$ are the maximal order-convex subsets of $A$. A set $C\subseteq X$ is order-convex iff whenever $a\le b\le c$ and $a,c\in C$, then $b\in C$.2012-11-14
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    @BrianM.Scott: Thank you.2012-11-14
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    @Seirios: You’re welcome.2012-11-14
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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