Cute Proofs: Closure of the Union is the Union of the Closures.
March 24, 2011
Every so often I like to go back and read through some basic algebra or basic point-set topology and try to think of new ways to look at the easier, more fundamental problems. I’ve been going through Bredon’s Algebraic Topology to prepare for a topology topics class next semester and there are two wonderful things about this book:
- It is concise*. When you read Bredon and Hatcher back-to-back you’ll understand what I mean here: while I do love Hatcher’s long motivating paragraphs (and, in fact, I attempted to mimic his method of explaining things in depth when I first began this blog), I tend to get lost in his prose at times. (On the other hand, Bredon’s book had significantly less motivation. Significantly less.)
- The questions are few, but proud. No one can deny that Hatcher’s book has a ton of enlightening and challenging questions. On the other hand, Bredon’s questions are generally more focused toward the section immediately preceding it (helping the beginning student) and are generally quite interesting (which is not to say the Hatcher ones are not!). The nice part about Bredon’s book is that you could easily assign students a section and ALL of the associated questions and they would probably not complain too much since there’s usually only three or four associated questions. Not so with Hatcher.
(*Perhaps not as concise as May’s book, though!)
Closure of a Union, Union of Closure.
The question I picked to write about today was one that stuck out to me in one of the introductory sections of Bredon. At first glance, I didn’t actually believe it was true as stated. The statement is as follows:
Proposition. Given some topological space and . Then .
Why should I doubt such a nice claim? In the language of limit points, this means that if we take and all the limit points of as well as and all the limit points of and we union these together, then this union can’t have any limit points. This didn’t quite sit well with me — couldn’t we have so many limit points so densely packed that they’d make another limit point that we didn’t include at first? Well, let’s prove this proposition which will show that I was insane for thinking such things!
Proof. One direction is easy. It was told to me today (by a wise professor!) that a good way to see if someone understands a topic is to give him an if and only if proposition and see if he knows which is the "easy direction." In this case, if then or . Let’s suppose is true without loss of generality. Then or is a limit point of . Either way, (if you don’t see this, prove it straight from the definition of limit points).
Now suppose . Then either or it is a limit point of the union. If then or ; then clearly . Now suppose and . The best way I’ve found to do this "limit point" part is to do it by the contrapositive: suppose that is not a limit point of either or ; we’ll then show that it isn’t a limit point of . This is equivalent to showing that if it is a limit point of then it must be a limit point of either or .
If isn’t a limit point of or , then we have that there exists some open containing such that
but then we note that will also be open, and it will also contain . But this implies that
(Can you see this? Suppose there was some point in the intersection; then it’s in BOTH and . But that means that it’s in and and it’s in either or . If it’s in this is a contradiction, since it’s also in , but . Same for and . Cool.)
And that completes the proof.
Let me note here that I am a huge fan of limit point arguments but there are most likely alternate, more clever ways to do this proof. I simply think this one is nice and visual.