Existence of a compactification of $\mathbb{R}$ with $\aleph_0$ remainder We know that the space $\mathbb{R}$ has compactifications with one point remainder, and two point remainder. but there is no compactification of $\mathbb{R}$ with three point remainder and the same holds for every finite natural number $n$ greater than 3.
We know that the Stone-Čech compactification of $\mathbb{R}$ has infinite remainder.
(i.e.$|\beta\mathbb{R}-\mathbb{R}|=2^\mathfrak{c}$
I have the same question for  other infinite cardinals less than $2^\mathfrak{c}$ as follows:
A. Is there any compactification $X$ of $\mathbb{R}$ with the property that $|X-\mathbb{R}|=\aleph_0$?
B. Can we improve our question to cardinals less than $2^{\aleph_0}$?
 A: EDIT: As Emil pointed out, I misinterpreted a statement in Charalambous article. Corrections are in boldface:
I don´t know if The answer to A is yes. For a separable metric space X (e.g. $\mathbb{R}$) the following are equivalent (see Compactifications with countable remainder by Charalambous in PAMS 1980):

*

*X is Cech-complete and rim-compact.


*X has a compactification with at most countable remainder.
$\mathbb{R}$ is Cech-complete because it is complete, and it is rim-compact because it is locally compact.
So $\mathbb{R}$ has a compactification with at most countable remainder, but you already knew that.
A: I guess you are interested in Hausdorff compactifications. (It is easy to construct a non-Hausdorff compactification with 3-point remainder.)
Set $W=\beta\mathbb R\backslash \mathbb R$;
note that $W$ has two connected components.
If $Z$ is an other compactification of $\mathbb R$ then $V=Z\backslash \mathbb R$
is an image of $W$.
Therefore $V$ has at most two connected components.
It follows that cardinality of $V$ may be 1, 2, or something between $\mathfrak{c}$ and $|W|=2^{\mathfrak{c}}$.
