6:00 PM

Friday Random Ten

1. Ma, Meyer, O'Connor - Appalachia Waltz
2. Netherlands Bach Collegium - Johann Sebastian Bach: Cantata #87, "In Der Welt Habt Ihr Angst"
3. The Choir of Corpus Christi Church - Jacob Handl (Gallus): Aspiciens a longe
4. Burt Bacharach - Who Are These People?
5. Queen - Don't Stop Me Now
6. Erasure - Lay All Your Love On Me
7. Netherlands Bach Collegium: Johann Sebastian Bach: Cantata #47, "Wer Ein Wahrer Christ Will Heissen"
8. Banda, Jenez, Lukacz, Szendrey-Karper - Franz Schubert: Quartet in G Major, Theme and Variations
9. Arthur Grumiaux/Clara Haskil - Ludwig van Beethoven: Violin Sonata #5, Op. 24 "Spring", 2nd movement
10. Honest Bob and the Factory-To-Dealer Incentives - Coda

And one more for the road: Berlin Chamber Orchestra - Bach: Cantata #211, "Coffee Cantata", "Hat Man Nicht Mit Seinen Kindern". Yes, Bach actually wrote a cantata about coffee.

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7:26 PM

What do we know about compact sets, besides their kind of clunky and nonintuitive definition? So far, not much. We know that if a set is compact, it is compact in any space which contains it, which is a good start. Let's look at other properties of compact sets.

Assume E is a compact set in some space X. As usual, I'm picturing E as a square drawn on a piece of paper, which represents X. Choose some point p which is outside of E. For every point q in E, we are going to draw two neighborhoods. V_q is a neighborhood of p, and W_q is a neighborhood of q. Note that both subscripts are q, because we are taking any single fixed point p which is not in E, and then we are considering lots of different q's in E, and we are drawing neighborhoods around p and q depending on the particular q we are currently looking at.

Okay, that was a horrible abstract mess. Mark p on the page, somewhere not inside the square. Pick any point in the square, mark it, and call it q₁. Measure the distance between p and q₁ with a ruler. Call the distance d₁. Draw a circle with radius d₁/2 around p, and call the circle V_q1. Draw a circle with the same radius around q₁, and call that circle W_q1. Each circle represents an open neighborhood. The circles touch on the page, but they don't overlap. There are no points in both circles.

Next, mark a second point in the square and call it q₂. Measure the distance from q₂ to p, and call the distance d₂. Draw a second circle around p with radius d₂/2, and call it V_q2. Draw a circle around q₂ with the same radius, and call it W_q2.

Skip to the end. Assume we've marked every single point in E, measured the distance, and drawn lots of circles. p will be at the center of lots of circles. Each circle will have its own radius, but it is of course okay if two different circles have the same radius. Looking in E, every point will be the center of exactly one circle, but in general each point will be in the interior of lots of circles.

Now, every neighborhood of every q is an open set. Since every point in E is the center of one circle, the union of the collection of all the neighborhoods is an open cover of E. (Formally, E ⊂ ∪_q(W_q). Since each W_q is an open set, the collection {W_q} is an open cover of E.) Since E is compact, a finite subcollection of {W_q} is also an open cover.

Look at that finite subcollection. For each set in that subcollection, there is a corresponding neighborhood of p. (If W_q1 is in the subcollection, V_q1 is the corresponding neighborhood of p.) Since the collection is finite, there is a smallest neighborhood in the collection. (If the collection were infinite, it's possible that there would be no smallest neighborhood. We are dependent on the fact that E is compact to limit ourselves to a finite collection of sets.) Call the smallest neighborhood V_qsmall.

V_qsmall is a neighborhood of p. Furthermore, V_qsmall can't contain any points in E. In our finite subcollection of {W_q}, no W_qn can intersect its corresponding V_qn. Since we are looking at the smallest V_q, none of the W_q can intersect it. Go back to the piece of paper with the square and two points marked in the square. If each pair of circles for point q₁ has a smaller radius than the pair for point q₂, then W_q1 overlaps V_q2, but neither circle overlaps V_q1. Just extend that to a finite number of circles. (With an infinite number of circles, each circle could be smaller than the previous one, and it all breaks down.)

We've been talking about the set E. Let's turn things inside out and look at the complement of E, E^c. p is a member of E^c. V_qsmall is a neighborhood of p. V_qsmall does not intersect any set W_q in the finite open cover of E, and E is a subset of its open cover. Therefore, V_qsmall does not intersect E, so it is a subset of E^c. Therefore, p is an interior point of E^c. But p is just any arbitrary point in E^c, so every point in E^c is an interior point. Which means that E^c is an open set. (This chain of reasoning had a lot of individual steps, and if, like me, you're struggling to get up to speed on all these definitions, you may be wondering why each step follows from the last. Everything here comes from the definition of an open set.)

So if E is a compact set, then this means that E^c is an open set. Which means, in turn, that E is a closed set. Stated again, in conclusion, every compact set is closed. (The converse is not true. On the real number line, look at the set of all integers. This set is closed, but not compact.)

I keep thinking I'm going to get further talking about compact sets than I do, but I think I need to lay everything out in exhaustive detail to make sure that I understand it. I had hoped to cover more ground today, but I'm going to have to stop here and return to compact sets next week.

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8:10 PM

Last time we defined compact sets. Unfortunately, while open sets and closed sets have an intuitive meaning to me, and that intuitive meaning is backed up by the formal definition, with compact sets I only have the formal definition to work from. So, a set E in a metric space X is a compact set if for any collection of open sets, {G_α}, such that E is a subset of the union of G, there is a finite subcollection of the open sets such that E is a subset of the subcollection.

That feels really messy and nebulous to me. The important thing is that every open cover {G_α} of E is required to have a finite number of sets which is also an open cover of E. E is just a single set, and it's easy to think of different possibilities for E, and then ask which possibilities are compact. But "any open cover" is a really broad concept. I think the place to start is with lots of small neighborhoods of individual points. Working with my "metric space visualized as a sheet of paper" model from last time, it's straight forward to visualize E as some group of shapes, potentially including enclosed spaces, curved lines including all the points on the line, and individual points. Then any open cover is going to look like lots of little tiny overlapping circles which cover all of E. Start with one circle for each point in E, so if E contains an infinite number of points then there will be an infinite number of sets in the open cover. Then start removing redundant sets, and see if you can get down to a finite number of sets.

With that unwieldy definition, the goal now is to explore some properties of compact sets in the hope that we can get a better handle on them. Our first property is that a compact set is compact regardless of the space that contains it.

We looked a bit at open sets, and how openness is dependent on the underlying space. Our example was an open segment on the real number line. If the space is the real number line, with the standard definition for distance, then the segment is an open set. But if the space is the complex numbers, with the same definition for distance, then the segment is not an open set. Compact sets are better behaved.

Suppose E is a compact set in some space Y, and that Y is a subset of some larger space X. Since E is a subset of X, we can pick any open cover of E in X. In fact, we can consider every open cover of E in X. So {G_α} is a collection of open sets in X, and E is a subset of the union of all the sets in the collection. The collection can be infinite for now, but we will reduce it to a finite collection. For each set in the collection, take the intersection of that set with the set Y. We know that G_α∩Y is an open set in Y. Every point in E which is in G_α is also in G_α∩Y. Therefore the collection {G_α∩Y} is an open cover of E in Y. Since E is compact in Y, a finite subcollection of that open cover is also an open cover of E. If G_α∩Y (for some particular value of α) is in the finite open cover, then keep G_α as part of an open cover of E in X. Since the collection of sets in Y is finite, the collection of sets in X is the same size, and also finite, and therefore E is compact in X. If E⊂Y⊂X and E is compact in Y, then E is compact in X.

Suppose, instead, that E is a compact set in some space X. Furthermore, suppose E is a subset of Y, which is some subset of X. (In other words, we're considering the reverse of the previous situation. E⊂Y⊂X and E is compact in X.) Take some open cover of E in Y. (Again, this works with any possible open cover.) If V is one of the sets in the open cover, V is open relative to Y, but is not necessarily open relative to X. However, there exists a set G which is open relative to X, such that V=G∩Y. So, for every set in the open cover in Y, we can find an open set in X. This collection of open sets in X is an open cover of E in X. Since E is compact in X, a finite subcollection of these sets is also an open cover of E. Take the corresponding open sets in Y, and we now have a finite open cover of E in Y. Therefore, E is a compact set in Y.

This means, essentially, that once an compact set, always a compact set. If E is a compact set in any metric space which contains E, it is a compact set in every space which contains E. Since subsets of metric spaces are themselves metric spaces, and because E is compact regardless of what space may contain it, it makes sense to refer to E as a compact space. This makes it unlike open sets or closed sets, because they are only open or closed relative to a particular space. Every set is both open and closed relative to itself, so it doesn't really make sense to refer to an open or closed space.

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6:52 PM

Now that we've gotten a feel for open and closed sets, we can move on to the next topic, which is compact sets. I'm going to admit before I begin that this topic is really pushing the limits of my understanding. I'm not sure if that's because of lack of prior familiarity or because of the way it's presented in the textbook. Hopefully I won't get anything seriously wrong. One thing that I'm concerned about is that my examples are mostly in Euclidean spaces, and Euclidean spaces are sufficiently well behaved that I'm concerned that I might be overlooking important examples from other types of metric spaces.

Thinking about the process of studying math, one thing that is becoming clear is that you can't study just one topic. If you really want to understand one area of mathematics, you have to be willing to study lots of other, not necessarily related, subjects as well. In my case, one of my underlying interests is probability theory. However, at a certain level, to really understand what's going on in probability theory, you have to have a good understanding of certain topics in real analysis. So I'm studying real analysis, but now I'm finding that really understanding analysis requires an understanding of topology. So now I'm trying to make sense of compact sets. In another example, I studied a bit of abstract algebra last fall just for fun, and have since come to appreciate how that provides useful background for both probability theory and real analysis. I didn't realize at the time that it would be relevant, but now I appreciate having looked at the topic.

To study compact sets, let's start in the same place that we always start. We have a metric space, X. We have a set in the metric space, E. Right now we don't know anything about the set. For ease of visualization, we can picture the space as a piece of paper and the set as a square (or some other shape) marked out on the paper.

We are now going to define an open cover of E. An open cover is a collection of open sets in X such that every point in E is in at least one of the open sets. Going back to the piece of paper, drop a bunch of circles on the page. Each circle represents an open set. (The open sets don't have to be circles, so use other shapes if you want.) The open sets can partially overlap E, but they can also include points in X which are not in E. If every point in E is covered by a circle, then the collection of circles (open sets) is an open cover of E.

It should be obvious that lots of different collections of open sets can all be open covers of the same set E. In particular, there can potentially be an infinite number of sets in the open cover. One obvious way is if E extends to infinity in some direction. We can say that E is bounded if E is a subset of some neighborhood of some point. If E is not bounded, then if each set in the open cover of E is bounded, there will need to be an infinite number of sets in the open cover in order to cover every point. Sooner or later I'm going to extend the piece of paper example too far, but imagine that the paper extends infinitely in every direction, and that E is a column on the page that also extends infinitely. If the open sets are finite circles, then each additional set can cover another piece of the column, but the column will alway extend beyond the last circle. So the circles can only be an open cover of E if there is no last circle, in other words if the collection of circles is infinite.

But you can get an infinite open cover even if E is bounded. The trick is to go in the other direction, and start making the sets in the open cover smaller and smaller. There's no rule that the sets in the open cover can't be subsets of other sets in the open cover, so you can pick a point in E and consider all neighborhoods of that point as part of the open cover. As the radius gets smaller, each neighborhood is a subset of the larger neighborhoods, but they are all part of the open cover, so the number of sets in the open cover is infinite.

A set is compact if any open cover of the set always contains a finite subcollection which is also an open cover. Now that we've buried our square under an infinite pile of circles, start removing circles which entirely overlap other circles. If we can always reduce the pile to a finite set of circles, which still covers the square, regardless of what the collection of circles originally looked like, then the square is a compact set. Of course, when I say square here, I mean any set E in any metric space, and when I say circles, I really mean open sets. E could have several disconnected pieces. In three dimensions, E could be a box and the open sets could be balls. Don't get too attached to one visualization. In fact, it's better to always try to find new visualizations.

Let's look at an example. Define our metric space as the real number line, with our distance function being the usual absolute value of the difference between the two points. Define E as the set of all points that can be written as 1/n, where n is any natural number. We've looked at this set before. It's not open since the gaps between points in the set mean that every neighborhood of every point contains lots of points not in the set. It's not closed since 0 is a limit point of the set, but 0 is not a member of the set. Today's question is: is this a compact set?

Well, let's create an open cover for the set. Take the open segment (−1,2). Since every point in E is greater than −1 and less than 2, this set by itself is an open cover of the set. Since 1 set is a finite number of sets, this is a finite open cover. But E isn't compact unless every open cover includes a subcollection which is a finite open cover. So let's look at another open cover.

For the point 1/n, for any natural number n, the closest point in the set is the point 1/(n+1). So, for each point, take a neighborhood of radius of half the distance to the nearest point, (for any point 1/n, the algebra works out to a radius of 1/[2n(n+1)]). This neighborhood is an open set, so we can build an open cover of E by including one neighborhood for each point in E. Each neighborhood contains exactly one point in E, and there are an infinite number of points in E, so there are an infinite number of sets in the open cover. No sets can be removed while continuing to cover the entire set, so E is not compact.

Let's look at another example. In the same space, again look at the set of all points that can be written as 1/n, where n is a natural number, but also add 0. (Call this set F to distinguish it from the last example.) Now every open cover of F must contain an open set which contains 0. The open set which contains 0 must include all points in a neighborhood of some radius r around zero (by the definition of open sets). r is some positive real number. Therefore, there exists some number N such that 1/N < r. (This fact is the archimedean principle, which we have probably used before, but perhaps not by its name.) Therefore, every point of the form 1/n, where n is greater than or equal to N, is contained in the open set which contains 0. There are only a finite number of natural numbers less than N, so there only a finite number of points in F which are outside of the set containing 0. Therefore, even if every other open set contains a only a single point in F, since there must be an open set containing 0, every open cover of F has a finite subcollection which is also an open cover of F. Therefore, F is a compact set.

Now that we've looked at an example, next time we will start looking at properties of compact sets in general.

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6:08 PM

We like open sets and closed sets, since we know properties about all of the points in the set. For sets that are neither open nor closed, we can't say much, so it would be convenient if we turn those sets into closed sets. For that matter, it would be convenient to be able to turn open sets into closed sets.

As usual, take a metric space X and look at some set in that space, called E. We are not making any assumptions about E yet, so it could be open, closed, or neither. Find the set of all limit points of E, and call that E′. Then, the closure of E, written E, is defined as E ∪ E′. The closure of E has some useful properties.

First, if E is closed, then E=E. Since every limit point of E is in E, E′ is a subset of E, so E∪E′=E. In the other direction, if E=E, then E is closed. If E=E∪E′, E′ must be a subset of E, so every limit point of E is a member of E.

This prompts the more general conclusion that E is always closed, justifying the name "closure." The only way this could not be true is if there were some point which is a limit point of E′ but is not a limit point of E. Call this point x. We want to show that a neighborhood of radius r must include a point in E if x is a limit point of E′, for any positive value of r. Since x is a limit point of E′, there must be a point, call it p, in E′ at distance less than r/2 from x. Since p is in E′, and is therefore a limit point of E, there must be a point in E at distance less than r/2 from p. By the triangle inequality, the distance from this point in E to x is less than r/2 + r/2, so there is a point in E at distance less than r to x. This is true for any positive value of r, so x is a limit point of E, and is a member of E′. Therefore every limit point of E is a member of E′, so E is closed.

Finally, if E is a subset of F and F is closed, then E is a subset of F. Since E is a subset of F, every limit point of E is a limit point of F. Since F is closed, every limit point of F is a member of F. Therefore, E′ is a subset of F. Since E is a subset of F and E′ is a subset of F, E∪E′ is a subset of F.

The conclusion from all of this is that for any set E, the closure of E is the smallest closed set which contains E. It's like a handy package for E, closed up so you don't have to worry about its limit points leaking out all over the rest of the space.

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