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

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 $k\geq 2$.

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)