Definition of compactification

Question

$(X',\tau')$ is a compactification of $(X,\tau)$ if

1. $(X',\tau')$ is compact
2. $(X,\tau)$ is a topological subspace of $(X',\tau')$
3. $X$ is dense in $X'$.

I was wondering why it is necessary to have the last condition. In my notes, I have written down that the third point is there to prevent 'trivial' compactifications such as $\bar{X}^{(X')}$, i.e. the closure of $X$ in $X'$. However, I don't see how this last one is compact.

So, my main question would be: why are we requiring that $X$ is dense in $X'$?

Answer 1

The space in your example is compact since it is a closed subspace of a compact space and thus itself compact.

The compactification should be universal in some sense i.e. there should be some uniqueness. If you leave out the last condition, consider $(0,1) \subseteq \mathbb{R}$. Then there would be very many different compactifications i.e. any closed and bounded set containing $(0,1)$ However, one would really like it to be $[0,1]$.

Answer 2

$X$ is dense in $X'$ means exactly that $\overline{X}^{(X')} = X'$. So it doesn't "prevent it"; it enforces it.

The idea of a compactification is to have a compact space $X'$ that contains $X$ as a subspace (or more precisely, a homeomorphic copy of it) and to prevent all sort of "irrelevant" new points in $X'\setminus X$, we make sure that all points of $X'$ "can be reached from $X$", which is the denseness part. So if $X$ is already compact and $X'$ is Hausdorff as well, then only $X$ itself is a compactification. We don't need to add points to it to make it compact (which is what the compactification "does" and the denseness of $X$ makes sure we can add lots of points but they're all "close to" $X$). The denseness is also used in a lot of unicity arguments involving compactifications.