Today I Learned

March 14th

Today I learned about one-point compactification, from the nlab page for example. Roughly speaking, the idea is to take topological space $X$ and add a point $\infty$ (along with lots of open sets) to make $X\cup\{\infty\}$ feel smaller and, in particular, compact. Anyways, we take the following definition.

We note that this topology is somewhat "natural'' because $X\setminus C$ is basically a large open set, so we're essentially endowing $X^*$ with the open sets endowed to it by $X,$ with some restriction on neighborhoods of $\infty.$

Already we can see that $X^*$ should be compact, for covering $\infty$ only leaves a compact subset $C$ of $X$ to cover, so covers of $X^*$ we can reduce to covers of $C,$ of which there exists a finite subcover. This is more or less our motivation; the fact that this makes a well-defined topology is remarkable.

Anyways, we should probably the claimed open sets actually a topology. In particular, we're not taking these open sets as a subbasis.

Because $C=\emp$ is compact, $X\cup\{\infty\}$ is open. Because $\emp$ is open in $X,$ $\emp$ is open in $X^*.$ It remains to deal with finite intersections and arbitrary unions.

We start with finite intersections. This is fairly quick and comes down to casework because our open sets only have two types.

• If our intersection looks like $U_1\cap U_2$ for open $U_1,U_2\subseteq X,$ then this is open because $X$ is a topological space.

• If our intersection looks like $U\cap[(X\setminus C)\cup\{\infty\}]$ for open $U\subseteq X$ and closed and compact $C\subseteq X,$ then we note that this is \[[U\cap(X\setminus C)]\cup[U\cap\{\infty\}]\] by distributing. The former set is open in $X$ because $X\setminus C$ is open in $X$; the latter set is open in $X$ because it is empty. So the union of these two is open.

• If our intersection looks like $[(X\setminus C_1)\cup\{\infty\}]\cap[(X\setminus C_2)\cup\{\infty\}]$ for closed and compact $C_1,C_2\subseteq X,$ then we note that this is \[[(X\setminus C_1)\cap(X\setminus C_2)]\cup\{\infty\}=[X\setminus(C_1\cup C_2)]\cup\{\infty\}\] by distributing and de Morgan's law. Thus, it remains to show that $C_1\cup C_2$ is closed and compact in $X.$ Well, it's closed in $X$ because $X$ is a topological space. It's compact because an open cover of $C_1\cup C_2$ can be restricted to covers of $C_1$ and $C_2$; extracting the finite subcovers of $C_1$ and $C_2$ and then joining the subcovers together gives a finite subcover of $C_1\cup C_2.$

This finishes the casework for finite intersections.

It remains to deal with arbitrary unions, which we do in a linear order for clarity. We begin by dealing with the an arbitrary union of (more than zero) open sets like $(X\setminus C)\cup\{\infty\}.$

We note that the arbitrary union has to contain $\infty,$ so it had better have this form. Now, we write\[\bigcup_{\alpha\in\lambda}(X\setminus C_\alpha)\cup\{\infty\}=\left(\bigcup_{\alpha\in\lambda}(X\setminus C_\alpha)\right)\cup\{\infty\}.\]Applying de Morgan's law, this is\[\bigcup_{\alpha\in\lambda}(X\setminus C_\alpha)\cup\{\infty\}=\left(X\setminus\bigcap_{\alpha\in\lambda}C_\alpha\right)\cup\{\infty\}.\]So it suffices to show that $C:=\bigcap_{\alpha\in\lambda}C_\alpha$ is closed and compact. Closed sets are closed under arbitrary intersection, so it is certainly closed.

As for compact, fixing any $C_\bullet\in\{C_\alpha\}_{\alpha\in\lambda},$ we note that an open cover of $C$ can be turned into an open cover of $C_\bullet$ by appending the open $X\setminus C.$ Then extracting our finite subcover from $C_\bullet,$ we can afterwards remove $X\setminus C$ from the finite subcover (if it is in the subcover) to get a finite subcover for $C$—$X\setminus C$ doesn't help cover $C.$ $\blacksquare$

Now take any open sets in $X^*,$ and we'll do the general-case arbitrary union. Our open sets can be sorted into being $\{U_\alpha\}_{\alpha\in\lambda}$ for $U_\alpha\subseteq X$ open or $(X\setminus C_\beta)\cup\{\infty\}$ for closed compact $\{C_\beta\}_{\beta\in\mu}.$ If the $(X\setminus C_\bullet)\cup\{\infty\}$ collection is empty, the union of the $U_\alpha$ is open because $X$ is a topological space. Otherwise, we are considering\[\left(\bigcup_{\alpha\in\lambda}U_\alpha\right)\cup\left(\bigcup_{\beta\in\mu}(X\setminus C_\beta)\cup\{\infty\}\right),\]where the right collection is nonempty. The left set here is some open $U\subseteq X$ again because $X$ is a topological space. The right set here takes the form $(X\setminus C)\cup\{\infty\}$ for some closed and compact $C$ because of lemma. So we have to show\[U\cup(X\setminus C)\cup\{\infty\}\]is open. This should contain $\infty,$ so we don't have many options. By de Moragn's law, this is $[X\setminus(C\cap(X\setminus U))]\cup\{\infty\},$ so we have to show that $C\cap(X\setminus U)$ is closed and compact in $X.$ Well, it is closed because both $C$ and $X\setminus U$ are.

As for compact, we repeat the trick from the lemma. Fix any cover of $C\cap(X\setminus U).$ If we add the set $U$ to this cover, we cover\[C\cap(X\setminus U)\cup U\supseteq C,\]so we get a finite subcover covering $C.$ We can turn this into a finite subcover of $C\cap(X\setminus U)$ by removing the open set $U$ (which is possibly in the finite subcover of $C$), which is legal because $U$ can't possibly help us cover $C\cap(X\setminus U).$ This completes the proof that we actually have a topology on $X^*.$ $\blacksquare$

While we're here, I guess I should actually show that $X^*$ is compact, justifying the name "compactification.''

This is done as outlined at the beginning. Fix any open cover $\{U_\alpha\}_{\alpha\in\lambda}$ of $X^*.$ One of these open sets needs to cover $\infty,$ but our only open sets covering $\infty$ take the form $(X\setminus C)\cup\{\infty\},$ so pick up some\[(X\setminus C)\cup\{\infty\}\in\{U_\alpha\}_{\alpha\in\lambda}.\]It more or less suffices to deal with open covers of $X$ now. Indeed, each $U_\bullet\in\{U_\alpha\}_{\alpha\in\lambda}$ is defined so that $U_\bullet\setminus\{\infty\}$ is open in $X$ (either it is $U_\bullet$ or $X\setminus C_\bullet$ for some closed $C_\bullet\subseteq X$), so we can view the collection\[\{U_\alpha\setminus\{\infty\}\}_{\alpha\in\lambda}\]as an open cover of $X.$ In particular, it is an open cover of $C,$ so we get to extract a finite subcover $\{U_k\setminus\{\infty\}\}_{k=1}^N$ from the $U_\bullet,$ and then $\{U_k\}_{k=1}^N$ is a finite subcover of $C$ directly from the $U_\bullet.$ So taking\[\{(X\setminus C)\cup\{\infty\}\}\cup\{U_k\}_{k=1}^N\]makes a finite subcover of $C\cup(X\setminus C)\cup\{\infty\}=X^*,$ which is what we wanted. $\blacksquare$

In other news, happy $\pi$ day, I guess.