   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.   Friday, March 23, 2012

Friday Random Ten

1. Duran Duran - A View to a Kill
2. James Brown - Opening Fanfare
3. Anthony Rolfe Johnson/David Willison - Ralph Vaughn Williams: The House of Life No. 4, Heart's Haven
4. Howard Shore - Osgiliath Invaded
5. Yo-Yo Ma, Edgar Meyer, Mark O'Connor - Pickles
6. Amadeus Quartet/Cecil Aronowitz - Johannes Brahms: String Quintet No. 2, Third Movement
7. Eurythmics - Anything But Strong
8. Dresden Philharmonic Orchestra - Johann Sebastian Bach: Missa Brevis in G minor, BWV 235, Gratias
9. Netherlands Bach Collegium - Johann Sebastian Bach: Cantata #56, "Endlich, Endlich Wird Mein Joch"
10. Frank Zabel/Stefan Thomas - Ludwig van Beethoven: 8 Variations on a Theme by Count Waldstein, WoO 67, Theme

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.

Wednesday, March 21, 2012

After I introduced set functions, I looked at complements, unions, and intersections of sets as applied to set functions. Now that we are looking at open and closed sets, we can do the same thing. Are unions of closed sets necessarily closed?

Let's start with complements of sets. If X is a metric space, and E is an open set in that metric space, what can we say about about Ec? For example, take X as the real numbers with distance function d(p,q)=|p−q|, and take E as the set of all real numbers x such that x<1. It's easy to show that E is an open set. Take any point x in E. x must be less than one. Define a radius r = (1−x)/2. Then the neighborhood of x with radius r must lie entirely within E. Since this is true for any point in E, E is an open set.

Let's look at the complement of E. Ec is the set of all points in X which are not in E, so in our case it's all real numbers greater than or equal to 1. Look at any limit point in Ec. Any neighborhood of the limit point must contain a point in Ec. This means that no neighborhood of that point can be entirely contained in E, so that point can't be in E. Therefore, that point is in Ec. Since every limit point of Ec is a member of Ec, Ec is closed. Again, in our example, no point less than 1 can be a limit point of Ec, so every limit point of Ec is in Ec, and Ec is closed.

We can run this the other direction as well. Assume E is closed. Choose any point in Ec. This point is not in E, and can't be a limit point of E since E is closed. Since it's not a limit point of E, there exists a neighborhood of the point which contains no points in E, or which is a subset of Ec. Since this is true for every point in Ec, Ec is open.

In other words, if any set in a metric space is open, its complement is closed. If any set is closed, its complement is open. Note that the null set is both closed and open. Also, it's possible for a set to be neither closed nor open, in which case its complement is also neither closed nor open.

Let's look at unions of open sets. If we have a collection of open sets, with each set labeled Eλ for some index set Λ, then we can take the union of all of the sets ∪λ Eλ. If x is some point in the union, then x is a member of one of the individual sets. Since there exists a neighborhood of x in that individual set, and all of the points in that neighborhood are also in the union, every point in the union of sets has a neighborhood in the union, so the union is an open set.

We could probably prove that intersections of closed sets are closed directly. But let's be lazy. Last time I was looking at intersections, unions, and complements of sets, I stated that the complement of a union is equal to the intersection of the complements (De Morgan's Law). If we have a collection of closed sets, each of those is the complement of an open set. If Fα is closed, then Eα is open, where Fα=Eαc. Then ∩λ Fλ = ∩λ (Eλc). But by De Morgan's Law, this equals (∪λ Eλ)c. Since we've already shown that the union of open sets is an open set, and the complement of an open set is a closed set, it follows that the intersection of closed sets is a closed set.

This outcome isn't surprising. And you might expect that intersections of open sets are open, and unions of closed sets are closed. This is sometimes true, but there's a catch, which we'll look at next time.

Tuesday, March 20, 2012

We've now defined open sets and closed sets as subsets of metric spaces. Every point in an open set has a neighborhood of points that are also entirely within the open set. The neighborhood is defined relative to the space, so the neighborhood is a subset of the space. That neighborhood is then also a subset of the open set. There is no restriction on how big the neighborhood must be, but in order for the set to be open, every point in the set must have a neighborhood of some size that is a subset of the open set.

Every limit point of a closed set is a member of the closed set. A limit point of a set is a point such that every neighborhood of that point contains at least one point, other than the point itself, which is a member of the set. In general, a limit point of a set is not necessarily a member of the set itself. A closed set does not have to have any limit points, but if it does, all of the limit points must be members of the set.

"Open" and "closed" have an intuitive relationship which these definitions do not obviously have. Let's look at some examples to further clarify the relationship between open sets and closed sets.

Let's look at some metric space X. The set X is itself a subset of the space. It should be obvious that X is both open and closed according to the definitions. Since every point in the space is a member of X, every neighborhood of every point must be a subset of X, so X is open. Likewise, since every limit point of X (if it has any) is a member of X, X must be closed.

It's kind of weird to think of a set as both open and closed. There's one more important example. The null set, ∅, is a subset of every set. It has no points, which means that it is open. The definition of an open set doesn't require that there actually be any points in the set, just that every point that is in the set has a neighborhood in the set. Since the null set doesn't have any point, it doesn't have to have any neighborhoods. Likewise, since the null set has no points, it has no limit points. Any set with no limit points is closed. Therefore, the null set is both open and closed.

Whether a set is open or closed is dependent on both the set and the space. For example, consider the rational numbers as a set. If the space is the rational numbers, with the standard distance function d(p,q)=|p−q|, then the set is both open and closed. However, if the space is the real numbers, with the same distance function, then the set of rational numbers is neither open nor closed, as we've previously discussed. Every neighborhood of a rational number in the real space includes an irrational number, and every irrational number is a limit point of the rational numbers in the real space.

While we're here, we can introduce another definition. A set is dense in a space if every point in the space is either a member of the set or a limit point of the set (or both). The rational numbers are dense in the real space. (We may not actually use this definition much, but it shows a conceptual relationship between rational numbers and the real numbers.)

Let's look at another example in the real space. A segment is the set of real numbers such that a<x<b, for real numbers a and b where a<b. We can show that every segment is an open set in the space of real numbers. For any x in the segment, find x−a and b−x. Take the smaller of the two values, and call it r. Then a≤x−r<x<x+r≤b, so the neighborhood of x with radius r is a subset of the segment. Therefore, every x in the segment has a neighborhood in the segment, so the segment is an open set.

Since we've previously proved that every neighborhood is an open set, we also have a slightly more direct proof. Given a and b, find x=(a+b)/2 and r=(b−a)/2. Then the segment (a,b) is equal to the neighborhood of x with radius r, and since every neighborhood is an open set, the segment must be an open set.

We now have the fact that a segment is an open set in the space of real numbers. But what about the space of complex numbers (again, with the standard distance function)? Given a real number x and a radius r, there exists a real number y such that 0<y<r. Then the complex point x+iy is a member of the neighborhood of x with radius r. Since y≠0, x+iy is not a real number, and so isn't a member of the segment. Therefore, the segment is not an open set in the complex space. (Note that the segment is also not a closed set in the complex space, since a and b are both limit points of the segment and neither point is a member of the segment.)

Let's look at this in general terms. Suppose X is a metric space. Then suppose Y is some subset of X. Y is also a metric space with the same distance function as X. Finally, suppose E is an open subset of Y, or specifically that E is open relative to Y. Then E is not necessarily an open subset of X. In our specific case, X is the complex numbers, Y is the real numbers, and E is a real segment. We can say more about the relationships between X, Y, and E, but first we need to develop some more theory about intersections and unions of sets in metric spaces.

Monday, March 19, 2012

Last time we defined open sets in metric spaces. This time we're going to turn it around and define closed sets. Open sets are defined based on neighborhoods of points in the set. Unsurprisingly, closed sets are as well, but the neighborhoods are used differently.

A point, p, is a limit point of a set if every neighborhood of p contains a point other than p in the set. Importantly, p does not have to be a member of the set in order to be a limit point of the set. Consider the set of all real numbers less than 1. 1 is not itself a member of this set. Let's look at neighborhoods of 1. The neighborhood of radius r around 1 includes all real numbers x such that 1−r<x<1+r. Since this clearly includes points less than 1 for any positive value for r, 1 is a limit point of this set. Of course, if the set were the points less than or equal to 1, 1 would still be a limit point of the set, but it would also be a member of the set.

If p is a limit point of a set, every neighborhood of p must contain at least one point other than p in the set. This implies that every neighborhood of p in fact contains an infinite number of points in the set. Assume that the number of points in some neighborhood which are in the set is finite. (The neighborhood of course includes all the points in the space within some distance of p, but we are only interested in the points in the neighborhood which are also in the set.) Then there is a closest point to p in the set, and the distance from p to the closest point is some positive real number. Then the neighborhood of p with a radius smaller than that distance doesn't contain any points in the set other than p. Since by assumption every neighborhood of p must contain points in the set, the neighborhood must contain an infinite number of points from the set.

Moving on, let's start with the set of real numbers less than 1 again, and add to it the single point 2. It's obvious that any neighborhood of 2 with radius less than 1 doesn't include any other points in the set. Therefore, 2 is not a limit point of the set. Since 2 is member of the set but is not a limit point of the set, it is called an isolated point.

Let's look at the set of rational numbers of the form 1/n, where n is any natural number. This set has a limit point, 0. For any positive value of r, it's possible to pick an n>1/r, and the point 1/n is less than r, and so the neighborhood of 0 with radius r includes a point in the set. However, 0 is not itself a member of the set. Furthermore, every point in the set is an isolated point, because given two adjacent points, you can always find a radius less than the distance between them, so every point is in a little island neighborhood containing only itself.

If you were feeling poetic, you could imagine the set as an infinite set of lonely points, each reaching out to the one goal that could connect them all, the limit point 0, but no point ever connecting to any of the points around it. But it's probably better to just think of them as emotionless rational numbers.

Anyway, this gives us a definition of closed sets. A set is closed if every limit point of the set is also a member of the set. Notice the set can include, or entirely consist of, isolated points. There just can't be any limit points of the set which are outside the set itself. A finite set can't have any limit points, and therefore, if the set has any points (it's nonempty), every point in the set is an isolated point, and, the set is closed.         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.

Why is the HTML for this page not valid?

BlogSpot adds the advertisement that appears at the top of this page. That advertisement is not valid HTML and is outside of my control. I believe that aside from that ad, this page is valid HTML.  