This is a follow-up to this question regarding one-space compactifications.
First recall a few definitions. An embedding is a continuous injective map $c:X\to Y$ that gives a homeomorphism from $X$ to its image. A compactification of $X$ is an embedding of $X$ as a dense subset of a compact space $Y$. By identifying $X$ with its image in $Y$ one just needs to extend $X$ to a bigger set $Y$ and give $Y$ a compact topology such that the subspace topology on $X$ coincides with the original topology on $X$.
Let $(X,\tau)$ be a noncompact topological space. A compactification $X^*=X\cup\{\infty\}$
obtained by adding a single point will be called a one-point compactification. (Wikipedia seems to reserve that term for the case of $X$ locally compact Hausdorff. In that case there is a unique Hausdorff one-point compactification of $X$. But I'll use the term more generally for convenience.)
I am interested in finding the range of all possible one-point compactifications of $X$. So find all the ways to give a compact topology to $X^*$ that induces the original topology on $X$. The fact that $X$ is dense in $X^*$ follows automatically because $X$ is not compact, so it cannot be closed in a compact space.
General fact: It is shown in the linked question at the top that every open nbhd of $\infty$ must be the complement in $X^*$ of a closed compact subset of $X$.
I can think of the following different cases of one-point compactifications:
-
The Alexandroff extension of $X$ with topology
$$\tau_1=\tau\cup\{(X\setminus C)\cup\{\infty\}:C\text{ is compact and closed in }X\}\;.$$
Starting with the topology on $X$, we have added as many open nbhds of $\infty$ as possible.
The inclusion map of $X$ into $X^*$ is an open embedding. This is the largest topology on $X^*$ with an open embedding. -
The open extension topology given by
$$\tau_2=\tau\cup\{X^*\}\;.$$
Starting with the topology on $X$, we have added a single nbhd of $\infty$, namely the whole space.
The inclusion map is an open embedding. This is the smallest topology on $X^*$ with an open embedding. -
Any topology intermediate between $\tau_1$ and $\tau_2$. $X$ will also be open in $X^*$ and the topology will contain all of $\tau$. For nhbds of $\infty$ one has to pick of suitable subfamily of the complements of all closed compact subsets of $X$. For example, the complements of all finite closed compact subsets, or the complements of all countable closed compact subsets.
These should cover all cases of one-point compactification with an open embedding.
($X$ is open in $X^*$ if and only if the topology on $X^*$ contains all of $\tau$.)
Now is every one-point compactfication of $X$ always an open embedding? Are there examples where $X$ is not open in $X^*$? What are conditions on $X$ that ensure the only one-point compactifications are the ones above?
Side note: One is usually not interested in taking a one-point compactification of a space $X$ that is already compact. But I still find the following observation worth noting. If $X$ is compact, the Alexandroff extension of $X$ will not be a compactification, since $\infty$ is then an isolated point and hence $X$ is not dense in $X^*$. But the open extension topology on $X$ plus one point is always a one-point compactification (with $X$ open in $X^*$).
If $X$ is a one-point space, it has exactly two one-point compactifications. The open extension topology is the Sierpinski space. The other is the two point space with the indiscrete topology, in which case the embedding is not open.
Best Answer
If $X$ is not compact, then every one-point compactification of $X$ is an open embedding. Indeed, suppose $X$ is not open in a one-point compactification $X^*=X\cup\{\infty\}$. This means there is some $x\in X$ such that every open neighborhood of $x$ in $X^*$ contains $\infty$. But then every open neighborhood of $x$ contains a neighborhood of $\infty$, and thus its complement is a compact subset of $X$. Since $X$ is embedded in $X^*$, this means that every open neighborhood of $x$ in $X$ has compact complement. But this implies $X$ is compact, since every open cover of $X$ includes a set which contains $x$, and then the complement of that set is covered by finitely many other sets in the cover.