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?
 A: 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)
A: 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 $k\geq 2$.
