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Let

  • $(E,\tau)$ be a topological space and $\Delta\not\in E$;
  • $E_\Delta:=E\cup\{\Delta\}$ and $$\tau_\Delta:=\tau\cup\underbrace{\{E_\Delta\setminus B:B\subseteq E\text{ is closed and compact}\}}_{=:\:\sigma_\Delta}.$$

We can easily show that $\tau_\Delta$ is a topology on $E_\Delta$.

How do we show that $\tau_\Delta$ is compact and why is it necessary to integrate the "closedness" in the definition of $\sigma_\Delta$?

Let $\mathcal B_\Delta\subseteq2^{E_\Delta}$ be an $\tau_\Delta$-open cover of $E_\Delta$.

Obviously, there must be a $B_\Delta\in\mathcal B_\Delta$ with $\{\Delta\}\subseteq B_\Delta$ and, by construction of $\tau_\Delta$, $$B_\Delta=E_\Delta\setminus B\tag1$$ for some $\tau$-closed $\tau$-compact $B\subseteq E$.

Now $\mathcal B_\Delta$ is clearly an $\tau_\Delta$-open cover of $B$ as well. So, there is a $\mathcal B\subseteq\mathcal B_\Delta$ with $$B\subseteq\bigcup\mathcal B\tag2.$$

However, why can we assume that $\mathcal B\subseteq\tau$?

Assuming $\mathcal B\subseteq\tau$, since $B$ is $\tau$-compact, there is a finite $\mathcal F\subseteq\mathcal B$.

How do we now conclude that $\mathcal F\cup\{B_\Delta\}$ is a finite subcover of $\mathcal B_\Delta$? And where did we need the closedness?

Tyrone
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0xbadf00d
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  • In my memory the one-point compactification is genuine, i.e. the resulting space is compact, if and only if the original space is locally compact – Richard Chen Apr 25 '22 at 15:42
  • Closedness is assumed to ensure that the collection $\sigma_\Delta$ is a topology. Otherwise it may fail to close under binary unions (recall that without Hausdroff separation, the intersection of two compact sets may fail to be compact.) – Tyrone Apr 25 '22 at 15:53
  • @RichardChen This is true only for Hausdorff spaces. – Paul Frost Apr 25 '22 at 16:57
  • @Tyrone One could take the topology generated by $\tau_\Delta$. But if we allow non-closed $B$ (if there exist compact non-closed $B$), then $E$ is no longer a subspace of $E_\Delta$. – Paul Frost Apr 25 '22 at 17:02
  • In your definition of $\tau_\Delta$, you don’t need $B$ to be closed. “$\tau_\Delta$ is compact“ is not a logical statement. We have essentially the same question https://math.stackexchange.com/q/4426018/861687 (third paragraph), don’t we? – user264745 Apr 26 '22 at 15:30

1 Answers1

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  1. Where do we need the closedness of the $B \subset E$ occuring in the definition of $\sigma_\Delta$?

We want that $E_\Delta$ contains $E$ (with its original topology!) as a subspace. This requires that all $(E_\Delta \setminus B) \cap E = E \setminus B$ belong to $\tau$, i.e. that all $B$ are closed in $E$.

  1. Why can we assume that $\mathcal B\subset \tau$?

We cannot. But we know that $B\subset \bigcup\mathcal B$. Since $B$ is a compact subspace of $E$ and $E$ is a subspace of $E_\Delta$, we see that $B$ is a compact subspace of $E_\Delta$. Therefore there exist finitely many $U_i \in \mathcal B$ such that $B \subset \bigcup U_i$. This completes the proof.

Paul Frost
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  • I'm sorry, but could you elaborate on how we show that "$B\subseteq E$ is $\tau$-compact implies $B$ is $\tau_\Delta$-compact"? – 0xbadf00d Apr 25 '22 at 19:16
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    @0xbadf00d Compactness is an internal property of a space $A$. But if $A$ is embedded as a subspace of a bigger space $A'$, we can say that $A$ is compact iff each cover of $A$ with open subsets of $A'$ has a finite subcover. We know that $B$ is compact and is embedded as a subspace of $E_\Delta$, thus we get a finite subcover of $\mathcal B$. – Paul Frost Apr 25 '22 at 23:11
  • Thank you; you're right. – 0xbadf00d Apr 26 '22 at 09:29
  • Am I missing something or doesn't it hold $\tau_\Delta=\tau\cup{\Delta}$? In fact, if $B_\Delta\in\tau_\Delta\setminus\tau$, then $$B_\Delta=\underbrace{E\setminus B}{\in:\tau}\cup{\Delta}$$ for some $\tau$-closed $B\subseteq E$. On the other hand, $E\in\tau\subseteq\tau\Delta$ and $E_\Delta\in\tau_\Delta$ so that $${\Delta}=E_\Delta\setminus E\in\tau_\Delta.$$ – 0xbadf00d Apr 26 '22 at 09:29
  • @0xbadf00d In fact, each $B_\Delta \in \tau_\Delta \setminus \tau$ has the form $B_\Delta = U \cup {\Delta}$ with some $U \in \tau$. But this does not mean that each $U \cup {\Delta}$ with $U \in \tau$ belongs to $\tau_\Delta$. This is only true if $E \setminus U$ is compact. In particular ${\Delta}$ is in general not open. This happens if and only if $E$ is compact. – Paul Frost Apr 26 '22 at 10:49
  • Hm, why is ${\Delta}$ not open? As I wrote above, $E\in\tau\subseteq\tau_\Delta$ and $E_\Delta\in\tau_\Delta$. So, it should hold ${\Delta}=E_\Delta\setminus E\in\tau_\Delta$. – 0xbadf00d Apr 26 '22 at 10:57
  • I'm sorry. Topologies and $\sigma$-algebras have a lot in common and that's why I've made the stupid mistake to thinkg $A,B\in\mathcal B$ implies $A\setminus B\in\mathcal B$ which is true for every $\sigma$-algebra $\mathcal B$, but not for a topology $\mathcal B$. Please let me know if I'm wrong, but I think that's what I was missing. – 0xbadf00d Apr 26 '22 at 14:48
  • @PaulFrost I’m not familiar with this(OP is using) notation. I have essentially same question as OP(showing compactness of $Y$), link: https://math.stackexchange.com/q/4426018/861687, third paragraph. Can you please help me with that? – user264745 Apr 26 '22 at 15:35
  • @0xbadf00d You are right, in a topology the difference of open sets is in general not open. In an "open difference space" all open sets are also closed (if $U$ is open, then also $X \setminus U$ is open, thus $U = X \setminus (X \setminus U)$ is closed). Conversely if all open sets are also closed, then $X$ is an "open difference space". You see that "open difference spaces" are exotic special cases. – Paul Frost Apr 26 '22 at 23:16