wherein is detailed Matt's experiences as he tries to figure out what to do with his life. Right now, that means lots of thinking about math.

Thursday, March 22, 2012

Last time we showed that any union of any collection of open sets in some metric space is itself any open set. We concluded from that that any intersection of closed sets is a closed set. We left off with the question of intersections of open sets and unions of closed sets, and the guess that an intersection of a collection of open sets should be an open set, while the union of closed sets should be a closed set.

This is true, but only if the collections are of a finite number of sets. Take some finite number of closed sets. Assume that x is a limit point of the union of the sets, but that x is not a member of any of the sets. Then each set has a closest point in the set to x. Since the number of sets is finite, one of those closest points is the closest point in the union of the sets, and is at some positive distance from x. Take a neighborhood of radius less than that distance around x, and it can't include any points from the union of the sets. Therefore, if x is not a member of the union, it can't be a limit point for the union, and the union of closed sets is itself closed.

However, the same argument does not hold true for an infinite number of closed sets. For any distance r from a point not in any set, there can always be sets with points within that radius. For a specific example, consider the collection of sets of real numbers, where each set is all numbers greater than or equal to 1/n for some natural number n. Each set is closed, no set contains the number 0, but 0 is a limit point of the union of these sets. Therefore, the union is not closed.

Again, we can show that intersections of open sets are open as long as we are taking an intersection of a finite number of sets, either through a direct argument or through complements and De Morgan's Law like we did before with intersections of closed sets.

This outcome is similar to what we got with the set functions. Some of the relationships between collections of open or closed sets that you might expect are true. Other relationships can be true, but are true only some of the time.

Now that we've looked at intersections and unions of open sets, we can go back to the question from the other day. If X is a metric space, and Y is a subset of X, then Y is also a metric space. Considering Y as a metric space, we can look at an open subset of Y. Call it E. However, if we consider E as a subset of X, then E may not be an open set in X. In this case, we can say that E is open relative to Y.

In our example from the other day, X is the metric space of complex numbers with distance function d(z,w)=|z−w|. Y is the metric space of real numbers with the same distance function, so Y is a subset of X. E is the open segment of real numbers greater than zero and less than one. E is open relative to Y, but E is not open in X, since if p is a point in E, every neighborhood of p in the space X includes complex numbers where the imaginary component is not zero.

Now, pick a point in E. Since E is open in Y, there exists a neighborhood of some radius r in the space Y, such that the neighborhood is a subset of E. Look at the neighborhood of the same radius, around the same point, in the space X. Since it's a neighborhood in X, it is an open set. For every point in E, we can therefore create an open set in X which does not contain any points in Y which are not also in E. (This is one of those times when drawing it is really helpful. Take a piece of paper. That's X. Draw a straight line across the page. That's Y. Mark off a start point and end point on Y. E is between the two points. Pick any point in E, and draw a circle around it, such that the circle doesn't extend beyond the start and end points. The interior of the circle is the open set in X.)

Now, create a neighborhood in X, the same way, for every point in E. Next, take the union of all of these neighborhoods. This union is a union of open sets, so it is itself an open set. Call this open set G. Therefore, any set E which is open relative to Y, where Y is any subset of X, is equal to the intersection of Y and some set G, which is an open set in X.

This logic also runs in the other direction, so start with the metric space X. Define some open set G in X. Define any set Y in X. Then the set E=G∩Y is always open relative to Y.

FAQ

What does "rolls a hoover" mean, anyway?

"Roll a hoover" was coined by Christopher Locke, aka RageBoy (not worksafe). He enumerated some Hooverian Principles, but that might not be too helpful. My interpretation is that rolling a hoover means doing something that you know is stupid without any clear sense of what the outcome will be, just to see what will happen. In my case, I quit my job in an uncertain economy to try to start a business. I'm still not sure how that will work out.