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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.

Michael Hardy
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  • Do you mean that any continuous function has at most two horizontal asymptotes? – Asaf Karagila Nov 09 '12 at 22:43
  • 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? – Michael Hardy Nov 09 '12 at 22:47
  • Yeah, I suppose you're right. I forgot that asymptotes are $\lim\limits_{x\to\pm\infty}$ rather than an accumulation point at $\pm\infty$. – Asaf Karagila Nov 09 '12 at 22:51
  • 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. – Logan Stokols Nov 09 '12 at 23:30
  • @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. – Michael Hardy Nov 10 '12 at 05:53
  • 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.) – WimC Nov 10 '12 at 06:57

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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.

Community
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Brian M. Scott
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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. – Logan Stokols Nov 09 '12 at 23:43
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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. – Brian M. Scott Nov 09 '12 at 23:51
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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? – Andreas Blass Nov 10 '12 at 00:30
  • @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. – Brian M. Scott Nov 10 '12 at 00:34
  • What does "order-components of $V$" refer to? – Seirios Nov 14 '12 at 18:35
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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 $b\in C$. – Brian M. Scott Nov 14 '12 at 21:04
  • @BrianM.Scott: Thank you. – Seirios Nov 14 '12 at 21:30
  • @Seirios: You’re welcome. – Brian M. Scott Nov 14 '12 at 21:34
  • 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. $\qquad$ – Michael Hardy Sep 13 '17 at 19:16