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We Know that the topological space $Y$ is a compactification of the topological space $X$, if the space $Y$ is compact and hausdorff and $X$ is dense in $Y$. If for a positive integer $n$ we have a compactification $Y$ that $Y-X$ has only $n$ elements, we say that $Y$ is the $n$ point compactification of $X$. I have some Questions about the existence of $n$-point compactification of a space $X$.

If the space $X$ has an $n$-point compactification for $n>1$ is it true that $X$ is locally compact?

We Know that $\mathbb{R}$ has a one point compactification, homeomorphic with the circle. and it has a two point compactification, homeomorphic with the closed interval $[0,1]$. is it true that $\mathbb{R}$ has a three point compactification? if it exists,it is homeomorphic with which well Known topological space? and if the answer is negative, why?

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There cannot be an n-point compactification unless the space has at least n ends. –  Gjergji Zaimi May 2 '12 at 11:21
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Hi dear Gjergji . is "end" a topological notion? how can you define the notion of "end" in topological spaces? and what about the plane. we know that the one point compactification of $\mathbb{R}^2$ is homeomorphic with $\mathbb{S}^2$. but can we have a two point compactificatin of $\mathbb{R}^2$? –  Ali Reza May 2 '12 at 11:35

2 Answers 2

up vote 3 down vote accepted

Suppose a space $X$ has $k$ ends and an $n$ point compactification. We can show that $k\geq n$. Indeed, there are disjoint neighborhoods $A_1,\dots,A_n$ of each of these points at infinity. Now let $Y$ be the complement of $\cup A_i$. Then $Y$ is compact so under mild assumptions ($X$ is hemicompact) it is contained in a bigger compact subset of $X$, which we call $\bar{Y}$, whose complement in $X$ has exactly $k$ connected components. Then we form $\bar{A_i}$ as the intersection of $A_i$ and the complement of $\bar{Y}$. Since the $A_i$'s are disjoint any component can contain elements from at most one $A_i$ so we get $k\geq n$.

As an application, $\mathbb R$ has $2$ ends so it cannot have a $k$ point compactification for $k\geq 3$. Similarly $\mathbb R^m$ has one end for $m\geq 2$ so it cannot have a $k$ point compactification for $k\geq 2$.

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thank you dear friend, but is your deduction true for $[0,\infty)$ how many ends does have this space? –  Ali Reza May 2 '12 at 12:34
    
That space has one end, so it does not have 2 or higher point compactifications. –  Gjergji Zaimi May 2 '12 at 13:00

First of all, if $\gamma X$ is a compactification of $X$ and $\gamma X\setminus X$ is finite, then $X$ is locally compact. Namely, let $x\in X$. In $\gamma X$, $x$ has a neighborhood $U$ that is disjoint from the finite set $\gamma X\setminus X$. Since $\gamma X$ is normal, there is a neighborhood $V$ of $x$ whose closure is still a subset of $U$. Since $\gamma X$ is compact, the closure of $V$ is a compact neighborhood of $x$. Since $U$ is disjoint from $\gamma X\setminus X$, the closure of $V$ is a compact neighborhood of $x$ in $X$.

The definition of an "end" of a topological space can be found here: http://en.wikipedia.org/wiki/End_(topology)

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Thank you dear professor. is it true that when we change The finite case by countable case? i.e.($|Y-\mathbb{R}|$=$\aleph_0$). –  Ali Reza May 2 '12 at 12:42
    
Is there any compactification of $\mathbb{R}$ that the reminder space is countable? –  Ali Reza May 2 '12 at 12:44
    
@AliReza Olfati: I believe Magill (reference at end) has proved that a locally compact Hasudorff topological space has a "countable Hausdorff compactification" if and only if it has an $n$-point Hausdorff compactification for each integer $n \geq 1,$ which implies that $\mathbb R$ does not have a Hausdorff compactification with a countably infinite remainder. Kenneth D. Magill, Countable compactifications, Canadian Journal of Mathematics 18 (1966), 616-620. –  Dave L Renfro May 2 '12 at 20:53

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