We’ve talked about compact subspaces, particularly of compact spaces and Hausdorff spaces (and, of course, compact Hausdorff spaces). So how can we use this to understand the space of real numbers, or higher-dimensional versions like ?
First off, is Hausdorff, which should be straightforward to prove. Unfortunately, it’s not compact. To see this, consider the open sets of the form for all positive real numbers . Given any real number we can find an with , so . Therefore the collection of these open intervals covers . But if we take any finite number of them, one will be the biggest, and so we must miss some real numbers. This open cover does not have a finite subcover, and is not compact. We can similarly show that is Hausdorff, but not compact.
So, since is Hausdorff, any compact subset of must be closed. But not every closed subset is compact. What else does compactness imply? Well, we can take the proof that isn’t compact and adapt it to any subset . We take the collection of all open “cubes” consisting of -tuples of real numbers, each of which is between and , and we form open subsets of by the intersections . Now the only way for there to be a finite subcover of this open cover of is for there to be some so that . That is, every component of every point of has absolute value less than , and so we say that is “bounded”.
We see now that every compact subset of is closed and bounded. It turns out that being closed and bounded is not only necessary for compactness, but they’re also sufficient! To see this, we’ll show that the closed cube is compact. Then a bounded set is contained in some such cube, and a closed subset of a compact space is compact. This is the Heine-Borel theorem.
In the case, we just need to see that the interval is compact. Take an open cover of this interval, and define the set to consist of all so that a finite collection of the cover . Then define to be the least upper bound of . Basically, is as far along the interval as we can get with a finite number of sets, and we’re hoping to show that . Clearly it can’t go past , since . But can it be less than ?
In fact it can’t, because if it were, then we can find some open set from the cover that contains . As an open neighborhood of , the set contains some interval . Then must be in , and so there is some finite collection of the which covers . But then we can just add in to get a finite collection of the which covers , and this contradicts the fact that is the supremum of . Thus and there is a finite subcover of , making this closed interval compact!
Now Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube is compact. And since any closed and bounded set is contained in some such cube, it will be compact as a closed subspace of a compact space. Incidentally, since is finite, we don’t need to wave the Zorn talisman to get this invocation of the Tychonoff magic to work.
As a special case, we can look back at the one-dimensional case to see that a compact, connected space must be a closed interval . Then we know that the image of a connected space is connected, and that the image of a compact space is compact, so the image of a closed interval under a continuous function is another closed interval.
The fact that this image is an interval gave us the intermediate value theorem. The fact that it’s closed now gives us the extreme value theorem: a continuous, real-valued function on a closed interval attains a maximum and a minimum. That is, there is some so that for all , and similarly there is some so that for all .
One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family of compact spaces is again compact. Unsurprisingly, the really tough bit comes in when we look at an infinite product. Our approach will use the dual definition of compactness.
Let’s say that a collection of closed sets has the finite intersection hypothesis if all finite intersections of members of the collection are nonempty, so compactness says that any collection satisfying the finite intersection hypothesis has nonempty intersection. We can then form the collection of all collections of sets satisfying the finite intersection hypothesis. This can be partially ordered by containment — if every set in is also in .
Given any particular collection we can find a maximal collection containing it by finding the longest increasing chain in starting at . Then we simply take the union of all these collections to find the collection at its top. This is almost exactly the same thing as we did back when we showed that every vector space is a free module! And just like then, we need Zorn’s lemma to tell us that we can manage the trick in general, but if we look closely at how we’re going to use it we’ll see that we can get away without Zorn’s lemma for finite products.
Anyhow, this maximal collection has two nice properties: it contains all of its own finite intersections, and it contains any set which intersects each set in . These are both true because if didn’t contain one of these sets we could throw it in, make strictly larger, and still satisfy the finite intersection hypothesis.
Now let’s assume that is a collection of closed subsets of satisfying the finite intersection hypothesis. We can then get a maximal collection containing . Then given an index we can consider the collection of closed subsets of and see that it, too, satisfies the finite intersection hypothesis. Thus by compactness of the intersection of this collection is nonempty. Letting be a closed set containing one of these intersection points , we see that the preimage meets every , and so must itself be in .
Okay, so let’s take the point for each index and consider the point in with -th coordinate . Then pick some set containing from the base for the product topology. For all but a finite number of the , . For those finite number where it’s smaller, the closure of contains the point , and so is in . So their finite intersection must be nonempty, and so is itself!
Now, since is in , it must intersect each of the closed sets in the original collection . Since the only constraint on is that it contain , this point must be a limit point of each of the sets in . And because they’re closed, they must contain all of their limit points. Thus the intersection of all the sets in is nonempty, and the product space is compact!
One of the nice things about connectedness is that it’s preserved under continuous maps. It turns out that compactness is the same way — the image of a compact space under a continuous map is compact.
Let’s take an open cover of the image . Since is continuous, we can take the preimage of each of these open sets to get a bunch of open sets in . Clearly every point of is the preimage of some point of , so the form an open cover of . Then we can take a finite subcover by compactness of , picking out some finite collection of indices. Then looking back at the corresponding to these indices (instead of their preimages) we get a finite subcover of . Thus any open cover of the image has a finite subcover, and the image is compact.
Let’s say we have a compact space . A subset may not be itself compact, but there’s one useful case in which it will be. If is closed, then is compact.
Let’s take an open cover of . The sets are open subsets of , but they may not be open as subsets of . But by the definition of the subspace topology, each one must be the intersection of with an open subset of . Let’s just say that each is an open subset of to begin with.
Now, we have one more open set floating around. The complement of is open, since is closed! So between the collection and the extra set we’ve got an open cover of . By compactness of , this open cover has a finite subcover. We can throw out from the subcover if it’s in there, and we’re left with a finite open cover of , and so is compact.
In fact, if we restrict to Hausdorff spaces, must be closed to be compact. Indeed, we proved that if is compact and is Hausdorff then any point can be separated from by a neighborhood . Since there is such an open neighborhood, must be an interior point of . And since was arbitrary, every point of is an interior point, and so must be open.
Putting these two sides together, we can see that if is compact Hausdorff, then a subset is compact exactly when it’s closed.
An amazingly useful property for a space is that it be “compact”. We define this term by saying that if is any collection of open subsets of indexed by any (possibly infinite) set so that their union is the whole of — the sexy words are “open cover” — then there is some finite collection of the index set so that the union of this finite number of open sets still contains all of — the sexy words are “has a finite subcover”.
So why does this matter? Well, let’s consider a Hausdorff space , a point , and a finite collection of points . Given any point , we can separate and by open neighborhoods and , precisely because is Hausdorff. Then we can take the intersection and the union . The set is a neighborhood of , since it’s a finite intersection of neighborhoods, while the set is a neighborhood of . These two sets can’t intersect, and so we have separated and by neighborhoods.
But what if is an infinite set? Then the infinite intersection may not be a neighborhood of ! Infinite operations sometimes cause problems in topology, but compactness can make them finite. If is a compact subset of , then we can proceed as before. For each we have open neighborhoods and , and so — the open sets form a cover of . Then compactness tells us that we can pick a finite collection so that the union of that finite collection of sets still covers — we only need a finite number of the to cover . The finite intersection will then be a neighborhood of which doesn’t touch , and so we can separate any point and any compact set by neighborhoods.
As an exercise, do the exact same thing again to show that in a Hausdorff space we can separate any two compact sets and by neighborhoods.
In a sense, this shows that while compact spaces may be infinite, they sometimes behave as nicely as finite sets. This can make a lot of things simpler in the long run. And just like we saw for connectivity, we are often interested in things behaving nicely near a point. We thus define a space to be “locally compact” if every point has a neighborhood which is compact (in the subspace topology).
There’s an equivalent definition in terms of closed sets, which is dual to this one. Let’s say we have a collection of closed subsets of so that the intersection of any finite collection of the is nonempty. Then I assert that the intersection of all of the will be nonempty as well if is compact. To see this, assume that the intersection is empty:
Then the complement of this intersection is all of . We can rewrite this as the union of the complements of the :
Since we’re assuming to be compact, we can find some finite subcollection so that
which, taking complements again, implies that
but we assumed that all of the finite intersections were nonempty!
Now turn this around and show that if we assume this “finite intersection property” — that if all finite intersections of a collection of closed sets are nonempty, then the intersection of all the are nonempty — then we can derive the first definition of compactness from it.
Now that we have some vocabulary about separation properties down we can talk about properties of spaces as a whole, called the separation axioms.
First off, we say that a space is if every two distinct points can be topologically distinguished. This fails, for example, in the trivial topology on a set if has at least two points, because every point has the same collection of neighborhoods — for all points . As far as the topology is concerned, all the points are the same. This turns out to be particularly interesting in conjunction with other separation axioms, since we often will have one axiom saying that a property holds for all distinct points, and another saying that the property holds for all topologically distinguishable points. Adding turns the latter version into the former.
Next, we say that a space is if any two topologically distinguishable points are separated. That is, we never have a point in the closure of the singleton set without the point being in the closure of . Adding to this condition gives us . A space is one in which any two distinct points are not only topologically distinguishable, but separated. In particular, we can see that the singleton set is closed, since its closure can’t contain any other points than itself.
A space is if any two topologically distinguishable points are separated by neighborhoods. If this also holds for any pair of distinct points we say that the space is , or “Hausdorff”. This is where most topologists start to feel comfortable, though the topologies that arise in algebraic geometry are usually non-Hausdorff. To a certain extent (well, to me at least) Hausdorff spaces feel a lot more topologically natural and intuitive than non-Hausdorff spaces, and you almost have to try to construct pathological spaces to violate this property. Back in graduate school, some of us adapted the term to apply more generally, as in “That guy Steve is highly non-Hausdorff.”
One interesting and useful property of Hausdorff spaces is that the image of the diagonal map defined by is closed. To see this, notice that it means the complement of the image is open. That is, if is a pair of points of with then we can find an open neighborhood containing the point consisting only of pairs with . In fact, we have a base for the product topology on consisting of products two open sets in . That is, we can pick our open neighborhood of to be the set of all pairs with and , where is an open subset of containing and is an open subset containing . To say that this product doesn’t touch the diagonal means that , which is just what it means for and to be separated by neighborhoods!
We can strengthen this by asking that any two distinct points are separated by closed neighborhoods. If this holds we say the space is . There’s no standard name for the weaker version discussing topologically distinguishable points. Stronger still is saying that a space is “completely Hausdorff” or completely , which asks that any two distinct points be separated by a function.
A space is “regular” if given a point and a closed subset with we can separate and by neighborhoods. This is a bit stronger than being Hausdorff, where we only asked that this hold for two singletons. For regular spaces, we allow one of the two sets we’re separating to be any closed set. If we add on the condition we’re above , and so singletons are just special closed sets anyhow, but we’re strictly stronger than regularity now. We call this condition .
As for Hausdorff, we say that a space is completely regular if we can actually separate and by a function. If we take a completely regular space and add , we say it’s , or “completely regular Hausdorff”, or “Tychonoff”.
We say a space is “normal” if any two disjoint closed subsets are separated by neighborhoods. In fact, a theorem known as Urysohn’s Lemma tells us that we get for free that they’re separated by a function as well. If we add in (not this time) we say that it is “normal Hausdorff”, or .
A space is “completely normal” if any two separated sets are separated by neighborhoods. Adding in we say that the space is “completely normal Hausdorff”, or .
Finally, a space is “perfectly normal” if any two disjoint closed sets are precisely separated by a function. Adding makes the space “perfectly normal Hausdorff”, or .
The Wikipedia entry here is rather informative, and has a great schematic showing which of the axioms imply which others. Most of these axioms I won’t be using, but it’s good to have them out here in case I need them.
There’s a whole list of properties of topological spaces that we may want to refer to called the separation axioms. Even when two points are distinct elements of the underlying set of a topological space, we may not be able to tell them apart with topological techniques. Points are separated if we can tell them apart in some way using the topology. Today we’ll discuss various properties of separation, and tomorrow we’ll list some of the more useful separation axioms we can ask that a space satisfy.
First, and weakest, we say that points and in a topological space are “topologically distinguishable” if they don’t have the same collection of neighborhoods — if . Now maybe one of the collections of neighborhoods strictly contains the other: . In this case, every neighborhood of is a neighborhood of . a forteriori it contains a neighborhood of , and thus contains itself. Thus the point is in the closure of the set . This is really close. The points are topologically distinguishable, but still a bit too close for comfort. So we define points to be “separated” if each has a neighborhood the other one doesn’t, or equivalently if neither is in the closure of the other. We can extend this to subsets larger than just points. We say that two subsets and are separated if neither one touches the closure of the other. That is, and .
We can go on and give stronger conditions, saying that two sets are “separated by neighborhoods” if they have disjoint neighborhoods. That is, there are neighborhoods and of and , respectively, and . Being a neighborhood here means that contains some open set which contains and contains some open set which contains
, and so the closure of is contained in the open set, and thus in . Similarly, the closure of must be contained in .. We see that the closure of is contained in the complement of , and similarly the closure of is in the complement of , so neither nor can touch the other’s closure. Stronger still is being “separated by closed neighborhoods”, which asks that and be disjoint closed neighborhoods. These keep and even further apart, since these neighborhoods themselves can’t touch each other’s closures.
The next step up is that sets be “separated by a function” if there is a continuous function so that for every point we have , and for every point we have . In this case we can take the closed interval whose preimage must be a closed neighborhood of by continuity. Similarly we can take the closed interval whose preimage is a closed neighborhood of . Since these preimages can’t touch each other, we have separated and by closed neighborhoods. Stronger still is that and are “precisely separated by a function”, which adds the requirement that only points from go to and only points from go to .
This list of separation
I’m still bleary from a 17 hour drive, so I’m punting.
Over at Rigorous Trivialities, Charles has been working for a few weeks on a series of posts about algebraic geometry “from the beginning”. The expository style seems based on a “just-in-time” delivery model, introducing concepts right before they’re needed. Personally, I always worry that this will lead to way too much backtracking when I realize I forgot to mention a whole line of reasoning that I’ve internalized and don’t consciously think about anymore, but which isn’t trivial. But he seems to be pulling this off well.
I know a bit about algebraic geometry myself — my advisor was fond of saying he remembers it having something to do with solving polynomials — and I’ll get to what I know eventually. I’m sure I’ll be referring to Charles’ notes when I do.
I’ve been up all night so I can sleep early tonight, wake up really early tomorrow, and hit the road back to New Orleans. As a result, I’m really not up for thinking too hard right now, but I probably won’t even get a chance to post tomorrow, so I’ll take a cute from a commenter on my Intermediate Value Theorem post and mention the “ham sandwich” theorem.
The MathWorld entry is pretty good for references here. Basically, if we have three solids (whatever those are) floating out anywhere in three-dimensional space, like two slices of bread and a slice of ham, then you can cut them all in half with one plane. Or if you have four four-dimensional solids you can cut them all in half by a three-dimensional hyperplane. And so on.
The sketch of the proof on that page is pretty clear, I think. You pick a direction and consider planes perpendicular to it. The IVT gives you one plane for each of the three solids. Then you use these to construct a map from the sphere to the plane whose image you can show must contain the point , which corresponds to all three planes lining up in that direction.
Interestingly, the last step also boils down to something like saying that the image of a connected space under a continuous map is connected. But it’s not quite that. We’ll deal with it later, rest assured.
Here’s a nice, light application of the Intermediate Value Theorem we came up with yesterday. Let’s say you’ve got a square table with four legs at the corners. They’re all the same length, but the floor is warped and so the table wobbles. As it happens, you only need to rotate the table around its center to find a position with all four legs touching the ground.
Put three of the table’s legs on the floor so the fourth either hangs in the air or presses down into the ground. As we rotate the table, the distance this leg is from the floor is a real-valued function of the angle we’ve turned the table from its starting point. If the leg starts above the floor, then as we rotate the table a quarter-turn it must push down into the floor. To see this, notice that instead of rotating a quarter-turn we could just push that leg down to the floor. As we do, the next leg over has to get pushed into the floor. So our function will be positive for some angles and negative for others. Now since the height of the leg is a continuous function of the angle, and the set of angles is connected, the IVT tells us that there must be some angle where the height of the leg is zero — it exactly touches the floor along with the other three legs.
The details get a bit hairier than this rough sketch. For more about them, and about other shapes of tables, you can read this article on the arXiv. But the basic idea comes down to the IVT.