# Existence of a compactification of $\mathbb{R}$ with $\aleph_0$ remainder

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 $2^{\aleph_0}$?

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Typo: remainder, not reminder. – user5117 May 2 '12 at 17:39
Would it be possible for you to explain briefly why there is no compactification with $n$ points? – Joel David Hamkins May 3 '12 at 0:27
Hello Dear Joel David. you can search in the box of my questions. before this question, I posed a question titled by "A question about some special compactifications of $\mathbb{R}$". if it doesn't satisfy you, I can explain it for you. I think I have another interesting proof for your question. (best wishes) – Ali Reza May 3 '12 at 7:21

## 2 Answers

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 $|W|=2^{\mathfrak{c}}$.

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Thank you very much dear Anton Petrunin. your proof is prfect and resonable. But is there any compactification of $\mathbb{R}$ with cardinality $\mathfrak{c}$? – Ali Reza May 3 '12 at 7:33
Yes, there are compactifications of $\mathbb R$ whose remainder has cardinality $\mathfrak c$. Consider, for example, the topologist's sine curve. More precisely, consider the graph of $y=\sin(1/x)$ for $0<x\leq 1$ together with the vertical segment from $-1$ to 1 on the $y$-axis. The graph of $y=\sin(1/x)$ for $0<x<1$ is a copy of $\mathbb R$, and I've added one point $(1,\sin 1)$ to compactify it at one end and the vertical segment, of cardinality $\mathfrak c$, to compactify it at the other end. – Andreas Blass May 3 '12 at 13:34

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.

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Hi dear ramiro. thank you very much.But I don't have any thing about this article. please give more information about this article. I have to see how it proved by Charalambous. – Ali Reza May 2 '12 at 17:33
Are you sure Charalambous proved that $X$ has a compactification with a countable infinite remainder? The paper does not say so (but I haven’t studied the proof to see whether it gives that). This interpretation is suspect, because if $X$ is already compact, it satisfies both conditions, but obviously has no compactification with nonempty remainder. – Emil Jeřábek May 2 '12 at 18:11
Dear ramiro. your answer doesn't satisfy me. I think the answer of my question must be negative! please defend your Answer in a reasonable way, without refering to this mentioned theorem. – Ali Reza May 2 '12 at 18:15