Alexandroff compactification topology

Question

Let $(X, \mathcal{T})$ be a non compact topological space, $\infty \notin X$. We define

$$X^* := X \cup \{\infty\}$$

and $$\mathcal{T}^* := \{U \subseteq X^*\mid U \cap X \in \mathcal{T} \land (\infty \in U \implies X \setminus U \mathrm{\ compact)}\}$$

I want to prove that $(X^*, \mathcal{T}^*)$ is a topological space, and I am stuck at showing that the set is closed under arbitrary unions (I checked the other axioms and they worked out).

Here's my attempt:

Let $\mathcal{U} \subseteq \mathcal{T}^*$. Then $(\bigcup \mathcal{U}) \cap X = \bigcup_{U \in \mathcal{U}}(\underbrace{U \cap X}_{\in \mathcal{T}}) \in \mathcal{T}$

and if $\infty \notin \bigcup \mathcal{U}$ we are done. If $\infty \in \bigcup \mathcal{U}$, there exists $U \in \mathcal{U}$ such that $\infty \in U$, and hence $X \setminus U$ is compact, and $X\setminus(\bigcup \mathcal{U}) = \bigcap_{U \in \mathcal{U}}(X \setminus U) \subseteq X\setminus U$. So, if I can show that $X \setminus\bigcup \mathcal{U}$ is closed in $(X,\mathcal{T})$, compactness follows, but I can't show this.

I also tried $X \setminus \bigcup\mathcal{U} = \bigcap_{U \in \mathcal{U}}(X \setminus U)$ but I can't conclude anything from this since the intersection of compact sets is not necessarily compact.

Henno Brandsma · Accepted Answer · 2018-05-19T10:45:17.157

0

$X\setminus U$ is compact and closed (as $X \cap U$ is open in your definition) and so the intersection $\bigcap_{U \in \mathcal{U}}(X \setminus U) \subseteq X\setminus U$ is a closed subset of $X$ that is a subset of a compact set, hence compact. And so $\bigcup \mathcal{U}$ is open.

edited May 19 '18 at 10:45

answered May 19 '18 at 09:51

Henno Brandsma

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I don't understand. Do I need to add closedness to prove the set is a topology or not? – May 19 '18 at 09:53
@Math_QED yes, see also the alexandroff extension on Wikipedia. – Henno Brandsma May 19 '18 at 09:54
@Math_QED and then we get a closed subset of a compact space etc. and the proofs go through. – Henno Brandsma May 19 '18 at 09:55
Yes, I see. Thanks. Looks like my book is wrong here. – May 19 '18 at 09:56
@Math_QED Indeed. This is mostly used when $X$ is Hausdorff anyway and closedness is automatic then. Hence lots of people misremember. – Henno Brandsma May 19 '18 at 09:57
@Math_QED rereading your definition: it's actually correct as $X \cap U$ always has to be open, and so $X\setminus U$ is closed. – Henno Brandsma May 19 '18 at 10:36
So the book was correct? Thanks! – May 19 '18 at 11:39
@Math_QED It was. But I read it badly (and so did you) :) – Henno Brandsma May 19 '18 at 11:41
Yeah, I missed that X\U was closed, which immediately solves the question. – May 19 '18 at 11:42
@HennoBradsma Hi professor, here I posed a question that is similar to this which you answered, but in your answer there isn't some points that I didn't understeand: so could I ask your assistance, please? Excuse me for the trouble, thanks – Antonio Maria Di Mauro Mar 17 '20 at 22:09
@AntonioMariaDiMauro What don't you understand? I only address in this answer the question the OP asked, not a full proof of it being a topology. – Henno Brandsma Mar 17 '20 at 22:11

Alexandroff compactification topology

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