The set that we add to compactify is itself compact.

Question

Let $X$ be a topological space. Let $X_c$ a compactification of $X$. I want to know if $X_c\setminus X$ is always compact, or maybe more simply whether it is always closed in $X_c$. I think it is true and i have no counterexample. Indeed, if $X$ is itself compact then $X_c=X$ and $X_c\setminus X=\emptyset$ which is compact. Also if $X_c=X\cup \{\infty\}$ is the point compactificationn then $X_c\setminus X=\{\infty\}$ is closed in $X_c$. Also take any compact manifold $X_c$ with boundary $\partial X_c$ considered as the compactification of the manifold $X_c\setminus \partial X_c$ then the boundary $\partial X_c$ is closed in $X_c$. Do we have a general result about this or not? thank you for your help!

Answer 1

Do we have a general result about this or not?

General results require general definitions. We could look at the one given by the wikipedia:

A compactification of $X$ is a compact space $C$ together with a continuous mapping $f:X\to C$ such that $f$ is a homeomorphism onto the image and the image $f(X)$ is a dense subset of $C$.

With that definition we can easily find a counterexample: put $C=[0,1]$ and $X=[0,1]\cap\mathbb{Q}$ together with the inclusion.

Furthermore if we look at the Cech-Stone compactification then we have the following result:

Let $X$ be a Tychonoff space and $C$ its Cech-Stone compactification. Then $X$ is open in $C$ if and only if $X$ is locally compact.

For other types of compactifications you have to consider case-by-case.

1. It is true for the Alexandroff extension (a.k.a. the one-point compactification). Note that this requires $X$ to be Hausdorff, noncompact and locally compact. Otherwise the Alexandroff extension is not a compactification at all.

2. The boundary of a manifold is always a closed subset. It doesn't matter whether the manifold is compact or not.