When a space is NOT locally compact, does it have dense remainder in its compactification?

Question

Any space has to be dense in any of its compactifications (that is part of the definition).

Question 1) When the space is NOT locally compact, does that mean that also its remainder in any of its compactifications is always dense?

As an example, $l^2$ is dense in its Stone-Čech compactification and its remainder is also dense there. Maybe there are other examples, but this was first that came to my mind.

Question 2) Can this happen also for other than Stone-Čech compactifications, or not?

I am assuming Hausdorff compactifications. Thank you for providing any insights.

Answer 1

Question 1: No.

Let $X = [0,1] \setminus N$ with $N = \{1/n \mid n \in \mathbb N\}$. This space is not locally compact because $0$ does not have any compact neigborhood. It has $[0,1]$ as a compactification, but here it has $N$ as a remainder which is not dense.

Question 2: Yes.

Daniel Wainfleet has given an example: Take $X = [0,1] \cap \mathbb Q$. It has $[0,1]$ as a compactification different from Stone-Čech compactification, and the remainder is dense.

Answer 2

Let $Y$ be a compact $T_2$ space. Suppose $$D\subsetneqq C=\overline C=\overline D\subsetneqq Y.$$ Let $X$ be the subspace $D\cup (Y\setminus C).$

Then id$_X:X\to Y$ is a compactification of the non-compact space $X.$ Its remainder is $C\setminus D,$ which is disjoint from the non-empty set $Y\setminus C.$

So $Y\setminus C$ is a non-empty open subset of $Y$ which is disjoint from the remainder.

Example: With the usual topology, let $Y=[0,2]$ and $C=[0,1]$ and $D=C\cap \Bbb Q.$