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?



A question about some  special compactifications of $\mathbb{R}$
Suppose a space $X$ has $k$ [ends](https://en.wikipedia.org/wiki/End_%28topology%29) 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](https://en.wikipedia.org/wiki/Hemicompact_space)) 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 $k\geq 2$.


