# One point (non-Hausdorff) compactification of compact space

A compactification of a space $$X$$ is an embedding $$f:X \to Y$$ so that (1) $$Y$$ is compact, (2) $$f(X)$$ is dense in $$Y$$. If furthermore, $$Y\setminus f(X)$$ is a single point, we say it is a one point compactification.

If we assume $$X$$ is a compact and Hausdorff space (thus LCH), does there exist a (non-Hausdorff) $$Y$$, such that there exists $$f:X \to Y$$ that is a one-point compactification of $$X$$?

If we require $$Y$$ to be Hausdorff, this is impossible see here. However, since compact does not imply closed in general, I'm not sure the case when we relax this condition. In particular, even if compactification exists, is it unique up to homeomorphism?

$$S^{1}$$ is an example, but I cannot think of a compactification. (Edit: I think taking $$S_1 \sqcup S_1$$ and quotient it out by a space so that every two points on the circle are combined besides 1 point might be an example, but I need to think more carefully, it might not work)

• Your example works, but it's not unique. Add a point and give it only one neighborhood: the whole space. Commented Dec 31, 2023 at 19:01

Let $$Y$$ be the set given by $$X$$ together with some other point $$\infty$$. Declare a set to be open if it's either open in $$X$$ or the whole space. Then the inclusion of $$X$$ into $$Y$$ is continuous and the closure of $$X$$ is all of $$Y$$ since $$\{ \infty \}$$ is not open. Moreover, $$Y$$ is trivially compact, as any open cover includes an open set containing $$\infty$$ but the only such set is $$Y$$ itself.

For what it's worth, in my opinion, the moral is that non-$$T_1$$ spaces are terrible, not anything interesting about compactifications.

• Your example could be made T1: let $\infty$ have cofinite neighborhoods. This again compactifies any T1 space as a dense subset of a T1 compact space. Commented Dec 31, 2023 at 19:03
• @stevenclontz good example, and my editorialization is not great. Commented Dec 31, 2023 at 19:08
• @StevenClontz Thank you for the example. Also, I guess even if we require T1, the one-point compactification is not unique to homeomorphism? Commented Dec 31, 2023 at 19:23
• Correct: this works for any T1 space, including a locally compact T2 space, and usually the result isn't T2, making it different from the standard compactification (unless the base space is discrete). Commented Dec 31, 2023 at 19:27
• @StevenClontz Sorry I don't quite understand. In this case, I require the original space to be compact and T2, so I don't think there is a standard one point compactification? Commented Dec 31, 2023 at 19:31

Assume the topological space $$(X,\tau)$$ is not compact. All its non-Hausdorff one-point compactifications $$Y=X\cup\{\infty\}$$ can be described, as show in Non-Hausdorff one-point compactifications.

There are basically three types:

1. 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. This is the largest topology on $$Y$$ that satisfies the requirements.

2. The open extension topology given by $$\tau_2=\tau\cup\{Y\}\;.$$ Starting with the topology on $$X$$, we have added a single nbhd of $$\infty$$, namely the whole space. This is the smallest topology on $$Y$$ that satisfies the requirements.

3. Any topology intermediate between $$\tau_1$$ and $$\tau_2$$. $$X$$ will also be open in $$Y$$ 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, and many others.

If $$X$$ is compact, the Alexandroff extension is not a compactification of $$X$$ because $$X$$ is not dense in $$Y$$. But, as mentioned in the other answers, the open extension topology is always a compactification (with $$X$$ open in $$Y$$).

Additional example: suppose $$X$$ is a singleton. Then there are exactly two one-point compactifications of $$X$$. The open extension topology is the Sierpinski space. The other one is the two point space with the indiscrete topology (the embedding is not open in this case).

1. No finite space $$X$$ has a $$T_1$$ one-point compactification $$Y=X\cup\{\infty\}$$: since $$X$$ is finite, $$X$$ must be closed in $$Y$$ and therefore cannot be dense.

2. Every infinite $$T_1$$ space $$X$$ has a $$T_1$$ one-point compactification: let $$Y=X\cup\{\infty\}$$, where neighborhoods of $$\infty$$ are co-finite.

3. $$T_1$$ one-point compactifications are not unique. Consider $$[0,1]$$, and let $$\infty\in[0,1]\cup\{\infty\}$$ have basic open neighborhoods of the form $$(a,1)\cup\{\infty\}$$. This is a strictly finer topology than the one described in (2).

4. Similar to (1), no compact space $$X$$ has a KC (Kompacts are Closed) compactification $$Y=X\cup\{\infty\}$$: since $$X$$ is compact, $$X$$ must be closed in $$Y$$ and therefore cannot be dense.

5. Every non-compact KC space $$X$$ has a $$k_2$$-Hausdorff (P171) one-point compactification: let $$Y=X\cup\{\infty\}$$, where neighborhoods of $$\infty$$ are co-compact (the Alexandrov extension). To see this, note $$\{\infty\}$$ is not open as $$X$$ is not compact, so $$X$$ is dense. $$k_2$$-Hausdorff was shown in this answer.

6. $$k_2$$-Hausdorff one-point compactifications are not unique. This answer shows two distinct $$k_2$$-Hausdorff compactifications for $$\mathbb R$$.