# A question about some special compactifications of $\mathbb{R}$

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?

• There cannot be an n-point compactification unless the space has at least n ends. – Gjergji Zaimi May 2 '12 at 11:21
• 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

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 $$k\geq 2$$.

• 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: https://en.wikipedia.org/wiki/End_(topology)

• 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