There is a (unique) compactification $X^*$ where $X^*-X=\{w\}$ of $X$ if $X$ is locally compact

Question

I've been trying to prove the following proposition:

Let $X$ be a locally compact Hausdorff space. Then, there is a (unique) compactification $X^*$ where $X^*-X=\{w\}$, with $w$ a point of $X^*$.

(For now, I'm not interested in proving uniqueness, only existence).
As far as I understand, we can define compactification as follows:

Definition: Let $X$ be a nonempty set. If $Y$ is a compact space,$X$ is a subset of $Y$ and $X$ is dense in $Y$, then we say that $Y$ is a compactification of $X$.

Question 1: Is this the most general definition?

So, I must find a set $X^*=X\cup \{w\}$ such that $X^*=\overline{X}$.

Question 2: How exactly can I go about it?

EDIT: I know that the chosen topology for $X^*$ is of the form $$ \tau^*=\tau \cup \{X^*-C : C\text{ is compact and } C\subseteq X \} $$ and I want to know how can I motivate this topology. (This can replace my second question).

What I thought: $\{w\}$ can't be open in $X^*$ since $X^*$ is the closure of $X$, and if $\{w\}$ is open there is one point of $X^*$ which has a neighborhood with no points of $X$ (contradiction).
$X$ must have the subspace topology. Hence, given an open set $U$ of $X$, it must be $U=X\cap U^*$ for some open set $U^*$ of $X^*$. One way this can happen is if $U$ is open in $X^*$. The other one is if $U\cup \{w\}$ is open in $X^*$. So, for each open set $U$ of $X$, either $U$ is open in $X^*$ or $U\cup \{w\}$ is (or both).
I'm not seeing how one excludes these second ones, and also how one uses the compactness of $X^*$ to motivate the $\{X^*-C : C\text{ is compact and } C\subseteq X \}$ part of the topology.

Answer 1

This is a well-known construction called the Alexandroff extension.

Answer 2

The most general definition says that a compactifation of a space $X$ is a pair $(Y, e)$ where $Y$ is compact and $e: X \rightarrow Y$ is an embedding, which means that $e$ defines a homeomorphism between $X$ and $e[X]$ as a subspace of $Y$, such that $e[X]$ is dense in $Y$. Often $Y$ is assumed to be Hausdorff as well; this will depend on your textbook.

In this case we do need to assume that $Y$ is Hausdorff to get unicity etc. Also $X$ is assumed to be locally compact and Hausdorff, so it's completely regular and it makes sense to also have the compactification be that as well.

In your case you only use the identity map, instead of general embedings $e$. This extra exactness is needed if we want to discuss equivalence of compactifications, because these have to respect the embeddings.

For the one-point compactification we define $Y = X \cup \{w\}$ where $w \notin X$. We also take $e(x) =x $ and define the topology on $Y$ as

$$\mathcal{T}_Y = \mathcal{T}_X \cup \{ \{w\} \cup O: O \subset X: X\setminus O \text{ is closed and compact in } X \}\text{.}$$

As $X$ is Hausdorff, compactness alone of $X \setminus O$ is enough (in Hausdorff spaces a compact set is closed, closedness is automatic); I wrote for the most general case where neither $X$ nor $Y$ are necessarily $T_2$.

To see this obeys the axioms for being a topology, see @egreg's "parallel" answer. It involves a bit of case checking,but it is not too hard.

Now check the requirements: $Y$ is compact and $e:X \rightarrow X$ is an embedding and $X$ is dense in $Y$, that $|Y\setminus X| = 1$ is true by construction.

$Y$ is compact is clear: if $\mathcal{U}$ is an open cover of $Y$, then $w$ is covered by some $U_w \in \mathcal{U}$. By the definition of the topology on $Y$, $U_w = \{w\} \cup O_w$, with $X \setminus O_w$ compact. The points of $X \setminus O_w$ are covered by finitely many $\mathcal{U}' \subset \mathcal{U}$, and then $Y$ is covered by those $\mathcal{U'}$ and $U_w$ together, as $ Y = \{w\} \cup O \cup (X\setminus O_w) = \{w\} \cup X$. So every open cover of $Y$ has a finite subcover.

$Y$ is Hausdorff: let $p,q \in Y$ with $p \neq q$. If $p , q$ are both in $X$, they have disjoint open neighbouurhoods $O_p$ and $O_q$ in $X$ by assumption, and these $O_p$ and $O_q$ are still open and disjoint in $Y$. If one of them equals $w$, say (it doesn't matter) $q =w$, then let $U_p$ be an open set of $X$ such that $\overline{U_p}$ is compact (or let $C_p$ be a compact neighbourhood of $p$, depending on how your text defines local compactness, and set $U_p = \operatorname{int}(C_p)$). Then $U_p$ is open in $Y$ and $U_q = \{w\} \cup (X \setminus \overline{U_p})$ is open in $Y$ (as $X \setminus (X \setminus \overline{U_p}) = \overline{U_p}$ is compact). And they are clearly disjoint.

If $O$ is open in the relative topology $\mathcal{T}_Y$ restricted to $X \subset Y$, then $O = O' \cap X$ for some $O \in \mathcal{T}_Y$. This is the definition of a subspace topology. If $O' \in \mathcal{T}_Y$, then $O = O'$ and $O \in \mathcal{T}_X$. Or if $O' = \{w\} \cup U$ with $U \subset X$ with $ \setminus U$ compact. Then as noted, by $T_2$-ness, $X \setminus U$ is closed and so $X \setminus (X \setminus U) = U$ is open in $X$ and $O =O' \cap X = U \in \mathcal{T}_X$.

So the topology $X$ inherits from $Y$ is exactly the original $\mathcal{T}_X$; we didn't add new open sets to $X$ when we constructed $Y$ (this is the reason that for general $X$ we need the complement of $U$ to be both closed and compact: compact for the above compactness proof, closed for showing $X$ has no new open sets.) So $e(x) = x$ from $X$ to $Y$ is indeed an embedding (openness of $e$ has been shown above). That $X$ is dense in $Y$ is clear when $X$ is not already compact as $w \in \overline{X}$, every neighbourhood of $w$ is of the form $U = \{w\} \cup O$ and intersects $X$, as otherwise $X = X\setminus O$, which cannot be as the latter is compact. Note that if $X$ were compact we could take $O = \emptyset$ and $w$ would be an isolated point of $Y$ and $X$ would not be dense. This is a degenerate case (we don't have a proper compactification in that case).

So $Y$ satisfies the requirements for the Aleksandrov-compactification.

Answer 3

You want that $X^*=X\cup\{e\}$ is endowed with a topology that induces the same topology you're given on $X$ and that $X^*$ is compact.

Let's look for some necessary conditions about this topology.

Suppose the open set $U$ contains $e$. Then $X^*\setminus U$ must be compact, because it is closed in $X^*$. Conversely, if $K$ is compact in $X$, then it will be so in $X^*$ and therefore $X^*\setminus K$ must be open.

Again, if $U$ is open in $X^*$ and $e\in U$, then $U\cap X$ must be open in $X$. Since we want that $X^*$ is Hausdorff, $\{e\}$ is closed; therefore $X$ is open in $X^*$ and consequently, every open set in $X$ is also open in $X^*$.

So far we have:

1. every open set in $X$ must be open in $X^*$
2. the complement of every compact set in $X$ must be open in $X^*$.

Let's see whether these sets form a topology.

Let $(K_i)$ be a family of compact sets in $X$ and consider $U_i=X^*\setminus K_i$. Then $$ \bigcup_i U_i = X^*\setminus\Bigl(\bigcap_i K_i\Bigr) $$ is the complement of a compact set.

It easily follows that the family of sets we are dealing with is closed under arbitrary unions.

Finite intersections are dealt with in a similar fashion: the union of a finite number of compact sets is compact.

It's clear that $X$ is dense in $X^*$, because each nonempty open set in $X^*$ intersects non trivially $X$.

We now only need to check the topology is Hausdorff. If $x,y\in X$, $x\ne y$, we are obviously able to find two open disjoint sets containing $x$ and $y$, respectively.

So we remain with $x\in X$ and $e$. If $U$ is an open set containing $x$ and $V$ is an open set containing $e$, it follows that $U\subseteq X^*\setminus V$, so $X^*\setminus V$ is a compact neighborhood of $x$ in $X$.

In particular, $X$ must be locally compact. Conversely, if $X$ is locally compact, take a compact neighborhood $K$ of $x$ in $X$ and set $V=X^*\setminus K$. Then $K$ and $V$ are disjoint neighborhoods of $x$ and $e$, respectively, so $X^*$ is Hausdorff.

Note that this also proves uniqueness.