Question

We now that the space $\mathbb{R}$ has compactifications with one point reminder, and two point reminder. but there is no compactification of $\mathbb{R}$ with three point reminder and the same holds for every finite natural number $n$ greater than3. We Know that the stone-cech compactification of $\mathbb{R}$ has infinite reminder. (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 or question to cardinals less than $2^{\aleph_0}$?

Answer 1

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 cardintality of $V$ may be 1, 2, or something between $\mathfrak{c}$ and $|W|=2^{\mathfrak{c}}$.

Answer 2

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):

1) X is Cech-complete and rim-compact.

2) 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.