Topology · Blog Entry
So, today, I actually didn't really do much, to be honest. I woke up bright and early, around 8 if my memory serves me correctly, and I served myself pancakes for breakfast from the RoMa dining hall.
After eating breakfast, I received a text from one of my friends, Lex, essentially asking me if I wanted to study some algebraic topology at Fine Hall today. I obliged, and headed over to Fine Hall. There, I met Professor Ozsvath, who teaches MAT 216 (Multivariable Analysis and Linear Algebra I), which I have skipped due to the permission of Professor McConnell, the placement officer.
To be honest, not much studying was going on then. But I did meet another friend, Petr, who I lent my Gunning analysis book. Then, I went to go get lunch with Lex at the Whitman dining hall.
That afternoon, I didn't do much. I called my parents, and then took a nap. I was woken up by my roommate, Kofi, so that I would be able to make it to the Step Sing!
"What's the Step Sing?" you might ask. Well, it is a Princeton tradition that the entire freshman class comes together at the steps of the Blair Arch and sings some traditional Princeton songs as well as songs chosen by the class, such as Viva La Vida and other songs which I cannot remember. I was the only one around me who knew Princeton's fight song, the Princeton Cannon.
I also met one of my other friends, Ioanna, who was looking to borrow copies of Needham's books (on both complex analysis and differential geometry). I ate dinner at the RoMa dining hall, and here I am, writing this. I intend on reviewing paracompactness now or perhaps later, and I will also be participating (as an audience member) in the Clash of the Colleges, in which all the students from each residential college (there are 7 at Princeton) compete in various events.
Now onto the real stuff. Let us review some topics from paracompactness. These are my informal notes; they will not be as polished as the ones in pdf format.
Here, we discuss the notion of paracompactness. The reader is hopefully familiar with the notion of compactness, in which every open covering admits a finite subcovering. Paracompactness is similar in some ways, but also different. Its main use is that a paracompact space may always admit a partition of unity, which is important in manifold theory. In fact, differentiable manifolds are often defined as a second-countable paracompact topological manifold. To be honest, I learned it the old-fashioned way from Do Carmo which involves coordinate charts and so on (I think this is a better way for those who do not know topology as well), but I will admit that the more topological definition of a differentiable manifold is much cleaner.
First, a little bit of preliminary definitions.
Suppose that we have open coverings $\textbf{U}$ and $\textbf{V}$ of some topological space. We say that $\textbf{U}$ is a
In other words, a refinement of an open covering is another open covering where all sets are "smaller," or perhaps finer (specifically subsets) than the first open covering.
Suppose that we have a collection $\textbf{U}$ of subsets of a topological space $X$. We say that $\textbf{U}$ is
This is a pretty complicated definition. What it means in plain language, is that, a collection of subsets is locally finite if every point in the space has some neighborhood around it that intersects only a finite amount of subsets in the aforementioned collection. This definition might seem rather contrived at first, but just wait. It is important to define paracompactness.
Now we come to what we have been waiting for: paracompactness.
A
Recall that a Hausdorff space is one in which any two points can have two disjoint open sets, with one set including one point, the other including the other. So, a paracompact space is one in which, if we have any open covering, we can find a covering where each set is a subset of one of the sets in the first collection such that if we take any point, a neighborhood of that point intersects only a finite number of those sets in the new covering. Does that sort of make sense?
A closed subspace of a paracompact space is paracompact.
Suppose $A$ is the closed subspace we are referring to, and $X$ is the paracompact space. Let us cover $A$ with sets that are open in $X$, and include the set $X-A$, which is open. We have thus created an open covering of $X$. Therefore, by paracompactness, we may take a locally finite refinement of the open covering with $X$. Now, take the intersection of each of these sets in the refinement with $A$. It is still locally finite, but now it is a refinement of the original covering of $A$. Let me convince you why this is true. Consider a set in our original covering of $A$. We refine it, and we consider a new set (if it exists) that is a subset of the old set. Intersecting this new set with $A$ gives us a newer set, which is still a subset of the old set. Therefore, our new covering must be a refinement of the original covering of $A$. Furthermore, since the refinement of $X$ is locally finite, simply intersecting each set with $A$ does not change the fact that the refinement is locally finite, as, colloquially speaking, the sets are only getting smaller, so each point in $A$ will have a neighborhood intersecting even less points, if anything. Hopefully this explanation sort of makes sense.
A paracompact space is normal.
First, we must show that a paracompact space $X$ is regular (recall what that means). So, suppose we have $x \in X$ and a closed $C \subset X$ such that $x \not\in C$. Consider that for all $y \in C$, since a paracompact space is Hausdorff, we may create disjoint open sets $U_y$ and $V_y$ such that $x \in U_y$ and $y \in V_y$. We can create a covering of $X$ by considering all $V_y$ and then $X-C$, as $C$ is surely contained in the union of $V_y$. By paracompactness, there is an open locally finite refinement, which we will denote by the sets $U_\alpha$. Now, we consider the set $$U = \bigcup \{U_\alpha \mid U_\alpha \subset \text{some }V_y\}$$ which is the set of all points in any $U_\alpha$ that is a subset of some $V_y$. This of course contains $C$, as any point in $C$ must be contained in a subset of $V_y$, not $X-C$, so there exists some set $U_\alpha$ containing that arbitrary point in $C$ that is a subset of some $V_y$. This is a locally finite collection, from the definition of paracompactness (this next part of the reasoning I actually don't understand right now), so its closure $\overline{U}$ is the union of the closure of all the $U_\alpha$ that are accounted for in the union. We know that $x$ is not in any of the $\overline{U}_\alpha$, as $x \not\in V_y$, and $U_\alpha \in V_y$, at least, for the $U_\alpha$ we are discussing and for some $V_y$. Therefore, $x \not \in \overline{U}$. Therefore, $U$ and $X-\overline{U}$ separate $x$ and $C$. Furthermore, by letting $C$ play the role of $x$ and the other closed set play the role of $C$, a similar argument will show that $X$ is normal as well. But that will be left for the reader to ponder.