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I'm practicing for my topology exam, for this I want to show that the one point compactification is in fact a topology. The definition I am working with:

Let $X$ be a topological space. The $\textit{one point compactification}$ of $X$ is the topological space $\hat{X} = X \sqcup \{*\}$, where a subset $U$ is open if either $U$ is an open subset of $X$ or $* \in U$ and $X\setminus U$ is compact and closed.

I have a solution, which seems long and untidy, so I'm looking for advice how to streamline the argument. Also, if you find any mistakes, let me know.

$\textit{Proof}$

Denote the topology of $\hat{X}$ by $\mathcal{T}$. We check the axioms:

  1. $\emptyset \subseteq X$, $\emptyset$ open in $X$ $\Rightarrow \emptyset \in \mathcal{T}$. Further, $X \setminus \hat{X} = \emptyset$, which is closed and compact, therefore $\hat{X} \in \mathcal{T}$.

  2. Let $A_i \in \mathcal{T}$ for $i \in I$. If $A_i \subseteq X$ for all $i$, then $\bigcup_{i}A_i \subseteq X$ is open as a union of open sets, and so $\bigcup_{i}A_i \in \mathcal{T}$. If $* \in \bigcup_{i}A_i$, we need to show that $\left(\bigcup_{i}A_i\right)^c = \bigcap_{i}A_i^c$ is closed and compact in $X$. It is closed in $X$ as an (arbitrary) intersection of sets that are closed in $X$. If $* \notin A_j$, then $A_j$ is open in X, so $A_j^c$ is closed. If $* \in A_j$, then $A_j^c$ is closed in $X$ by the definition of $\mathcal{T}$. To see that $\bigcap_{i}A_i^c$ is compact, let $k$ be such that $* \in A_k$, so $A_k^c$ is closed and compact. Now $\bigcap_{i}A_i^c \subseteq A_k$, i.e. a closed subset of a compact set, therefore also compact.

  3. Let $A_1, A_2 \in \mathcal{T}$. If $* \notin A_1$ and $* \notin A_2$, then $* \notin A_1 \cap A_2$ and $A_1 \cap A_2$ is open in $X$ as a finite intersection of open sets. If $* \in A_1$ and $* \in A_2$, then $* \in A_1 \cap A_2$ so we need to check that $\left(A_1 \cap A_2\right)^c = A_1^c \cup A_2^c$ is closed and compact in $X$. It is closed as the finite union of closed sets. For compactness, let $U_i, i \in I$ be an open cover of $A_1^c \cup A_2^c$. It's also an open cover of $A_1^c$ and $A_2^c$. Since both sets are compact, they admit finite subcovers, say $A_1^c \subseteq \bigcup_{j = 1}^{n}U_j =: O_1$ and $A_2^c \subseteq \bigcup_{k = 1}^{m}U_k =: O_2$. Then $O_1 \cup O_2$ is a finite subcover of $A_1^c \cup A_2^c$ which is therefore compact. For the case where $*$ is in only one of the sets, w.l.o.g. $* \in A_1$. Then, $* \notin A_1\cap A_2$, so we need to show that $A_1 \cap A_2$ is open in X. Let $\tilde{A_1}:= A_1 \setminus \{*\} = A_1 \cap X$, then $A_1 \cap A_2 = \tilde{A_1} \cap A_2$. Further, $A_2$ is open by definition and $\tilde{A_1}$ is open in X since $\tilde{A_1}^c = \left(A_1 \cap X \right)^c = A_1^c \cup \emptyset = A_1^c$, which is closed. So $A_1 \cap A_2$ is open as a finite intersection of open sets.

Thanks a lot for checking!

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  • $\begingroup$ This looks good to me! $\endgroup$ Commented Jan 18, 2021 at 13:10

1 Answer 1

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You're on the right track, you just need to clean up a bit. You were right thinking it was a bit long, but it wasn't wrong, so there's at least that. I'm re-writing how I think it should be written, so I may repeat some of your steps with a different notation.

  1. Open subsets of $X$ are also open subsets of $\hat X$, so $\varnothing$ is open in $\hat X$. But your definition is a tiny bit unclean because normally we only take the complement of a subset $A \subseteq X$ and then write $X \setminus A$ in those cases only, but you take $U \subseteq \hat X$ and consider $X \setminus U$. To make things clearer, I would write $X \cap (\hat X \setminus U)$. (Some may allow this abuse of notation, it's a matter of taste.) It is clear with this notation that $X \cap (\hat X \setminus \hat X) = X \cap \varnothing = \varnothing$ is closed and compact, thus $\hat X$ is open in $\hat X$.

  2. Given a family $\{A_i\}_{i \in I}$ of open subsets of $\hat X$, if $\bigcup_{i \in I} A_i$ is contained in $X$, then the same is true of each $A_i$, and therefore being open in $\hat X$ is equivalent to being open in $X$, so their union is open in $X$, and therefore also in $\hat X$. If this is not the case, let $j \in I$ be such that $* \in A_j$. Since that means $* \in A_j \subseteq \bigcup_{i \in I} A_i$, we have to check that $\hat X \setminus \bigcup_{i \in I} A_i = \bigcap_{i \in I} (\hat X \setminus A_i)$ is closed and compact. But $\bigcap_{i \in I} (\hat X \setminus A_i) \subseteq \hat X \setminus A_j$, and since the latter is closed and compact, and the former is an intersection of closed sets (and therefore closed), the intersection is a closed subset of a compact subset, and therefore compact (this is true in any topological space).

  3. We can use a little trick here: if $A$ is open in $\hat X$, then $A \cap X$ is open in $X$. That's because it's trivial if $A \subseteq X$ by definition, and if $* \in A$, then $\hat X \setminus A = X \setminus (A \cap X)$ is closed in $X$, and in particular $A \cap X$ is open in $X$. So if $A_1, A_2$ are open in $\hat X$ but $*$ is not in $A_2$, then $A_1 \cap A_2 = (A_1 \cap X) \cap A_2$ is an intersection of two open subsets of $X$, and therefore is open in $X$. To finish the proof that open sets in $\hat X$ are stable under taking finite intersections, we may assume $* \in A_1 \cap A_2$, which means we can assume $\hat X \setminus A_1$ and $\hat X \setminus A_2$ are both compact and closed. Now we have to show that $\hat X \setminus (A_1 \cap A_2) = (\hat X \setminus A_1) \cup (\hat X \setminus A_2)$ is compact and closed. Both of these are subsets of $X$, so the question is equivalent to asking if in a topological space $X$, the union of two compact and closed subsets is compact and closed. The 'closed' part is trivial by the axioms of a topology, and the 'compact' part is trivial because if we have an open cover of the union of two compact sets $F_1$ and $F_2$, then a finite open subcover covers $F_1$, and another finite open subcover covers $F_2$, so both together cover the union.

This is, however, very lengthy. You have to pass to complements every time. An easier way to define this topology is to work only with closed sets. So an equivalent way to define this topology is to say that $F \in \hat X$ is closed if and only if $* \in F$ and $F \cap X$ is closed in $X$, or $* \notin F$ and $F$ is closed and compact. Another way to phrase this is: $F$ is closed in $\hat X$ if $F \cap X$ is closed in $X$, and if $* \notin F$, we require that $F$ is also compact.

  1. $* \notin \varnothing$ and $\varnothing$ is closed in $X$ and compact, so $\varnothing$ is closed in $\hat X$. $* \in \hat X$ and $X \cap \hat X = X$ is closed in $X$, so $\hat X$ is closed in $\hat X$.

  2. If $\{F_i\}_{i \in I}$ is a family of closed sets in $\hat X$, then $$ X \cap \bigcap_{i \in I} F_i = \bigcap_{i \in I} (F_i \cap X) $$ is closed in $X$ because each $F_i$ is closed in $X$. If $* \notin F_j$ for a certain $j$, then the intersection doesn't contain $*$ and it is contained in $F_j \cap X$ which is compact, and therefore our intersection is a closed subset of a compact set, thus compact as well.

  3. If $F_1,F_2$ are closed in $\hat X$, then $(F_1 \cup F_2) \cap X = (F_1 \cap X) \cup (F_2 \cap X)$ is closed in $X$. If $*$ is in neither of $F_1$ or $F_2$, then both are compact, and therefore their union is also compact (by the same argument as above).

I find the second argument shorter because you avoid passing to complements. What do you think?

Hope that helps,

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  • $\begingroup$ Patrick, thanks for this awesome answer! The explanations regarding the unclean definition really helps, and your rewrite offers some nice explanations. It's still long though ;) I completely agree that the second version is much nicer. I would not have guessed that it gets so much cleaner when defining the topology in terms of closed sets. I'll keep that in mind... I'll not edit my version, so future viewers can see the difference clearly, but I'll make sure to write it down again by hand for practice. Thanks! $\endgroup$ Commented Jan 18, 2021 at 14:57
  • $\begingroup$ @noam.szyfer : The only reason why I re-wrote your answer is to highlight the usage of complements in your proof. When you just take that into account and realize that your arguments really work on the complement of open sets instead of the open sets themselves (like using the fact that the union of two compact subsets is compact, or that a closed subset of a compact subset is also compact), working on the right objects really simplifies the discourse. Otherwise you are constantly stuck translating between the language of open sets and closed sets, which is what makes your proof "lengthy"! $\endgroup$ Commented Jan 18, 2021 at 15:00
  • $\begingroup$ I thought the same about the complements, but it's true sometimes the definition already includes the intersection we were missing and I didn't want to be pedantic because it was understood what the OP meant, so I wrote nothing. $\endgroup$ Commented Jan 19, 2021 at 15:52
  • $\begingroup$ @Dog_69 I don't know what you're referring to! $\endgroup$ Commented Jan 19, 2021 at 16:49
  • $\begingroup$ It's this: en.wikipedia.org/wiki/… Usually we consider relatives with respect to the universe, but it's possible to do it with respect to another subset. $\endgroup$ Commented Jan 19, 2021 at 17:16

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