Products of Metric Spaces

Shortly we’re going to need a construction that’s sort of interesting in its own right.

We know about products of topological spaces. We can take products of metric spaces, too, and one method comes down to us all the way from Pythagoras.

The famous Pythagorean theorem tells us that in a right triangle the length of the side opposite the right angle stands in a certain relation to the lengths and of the other two sides: . So let’s say we’ve got metric spaces and . For the moment we’ll think of them as being perpendicular and define a distance function on by

The quantity inside the radical here must be nonnegative, since it’s the sum of two nonnegative numbers. Since the result needs to be nonnegative, we take the unique nonnegative square root.

Oops, I don’t think I mentioned this before. Since the function has as its derivative, it’s always increasing where is positive. And since we can eventually a square above any real number we choose, its values run from zero all the way up to infinity. Now the same sort of argument as we used to construct the exponential function gives us an inverse sending any nonnegative number to a unique nonnegative square root.

Okay, that taken care of, we’ve got a distance function. It’s clearly nonnegative and symmetric. The only way for it to be zero is for the quantity in the radical to be zero, and this only happens if each of the terms and are zero. But since these are distance functions, that means and , so .

The last property we need is the triangle inequality. That is, for any three pairs , , we have the inequality

Substituting from the definition of we get the statement

The triangle inequalities for and tell us that and . So if we make these substitutions on the left, it increases the left side of the inequality we want. Thus if we can prove the stronger inequality

we’ll get the one we really want. Now since squaring preserves the order on the nonnegative reals, we can find this equivalent to

Some cancellations later:

We square and cancel some more:

Moving these terms around we find

So at the end of the day, our triangle inequality is equivalent to asking if a certain quantity squared is nonnegative, which it clearly is!

Now here’s the important thing at the end of all that calculation: this is just one way to get a metric on the product of two metric spaces. There are many other ones which give rise to different distance functions, but the same topology and the same uniform structure. And often it’s the topology that we’ll be most interested in.

In particular, this will give us a topology on any finite-dimensional vector space over the real numbers, but we don’t want to automatically equip that vector space with this norm unless we say so very explicitly. In fact, we don’t even want to make that same assumption about the two spaces being perpendicular to each other. The details of exactly why this is so I’ll leave until we get back to linear algebra, but I want to be clear right now that topology comes for free, but we may have good reason to use different “distances”.

Metric Spaces are Categories!

A guest post by Tom Leinster over at The n-Category Café reminded me of an interesting fact I haven’t mentioned yet: a metric space is actually an example of an enriched category!

First we’ll need to pick out our base category , in which we’ll find our hom-objects. Consider the set of nonnegative real numbers with their real-number order, and add in a point called that’s above all the other points. This is a totally ordered set, and orders are categories. Let’s take the opposite of this category. That is, the objects of our category are the points in the “interval” , and we have an arrow exactly when .

This turns out to be a monoidal category, and the monoidal structure is just addition. Clearly this gives a monoid on the set of objects, but we need to check it on morphisms to see it’s functorial. But if and then , and so we can see addition as a functor.

So we’ve got a monoidal category, and we can now use it to form enriched categories. Let’s keep out lives simple by considering a small -category . Here’s how the definition looks.

We have a set of objects that we’ll call “points” in a set . Between any two points and we need a hom-object . That is, we have a function .

For a triple of objects we need an arrow . In more quotidian terms, this means that .

Also, for each point there is an arrow from the identity object of to the hom-object . That is, , so .

These conditions are the first, fourth, and half of the second conditions in the definition of a metric space! In fact, there’s a weaker notion of a “pseudometric” space, wherein the second condition is simply that , and so we’re almost exactly giving the definition of a pseudometric space.

The only thing we’re missing is the requirement that . The case can be made (and has been, by Lawvere) that this requirement is actually extraneous, and that it’s in some sense more natural to work with “asymmetric” (pseudo)metric spaces that are exactly those given by this enriched categorical framework.

Some theorems about metric spaces

We need to get down a few facts about metric spaces before we can continue on our course. Firstly, as I alluded in an earlier comment, compact metric spaces are sequentially compact — every sequence has a convergent subsequence.

To see this fact, we’ll use the fact that compact spaces are the next best thing to finite. Specifically, in a finite set any infinite sequence would have to hit one point infinitely often. Here instead, we’ll have an accumulation point in our compact metric space so that for any and point in our sequence there is some with . That is, though the sequence may move away from , it always comes back within of it again. Once we have an accumulation point , we can find a subsequence converging to just as we found a subnet converging to any accumulation point of a net.

Let’s take our sequence and define — the closure of the sequence from onwards. Then these closed sets are nested , and the intersection of any finite number of them is the smallest one, which is clearly nonempty since it contains a tail of the sequence. Then by the compactness of we see that the intersection of all the is again nonempty. Since the points in this intersection are in the closure of any tail of the sequence, they must be accumulation points.

Okay, that doesn’t quite work. See the comments for more details. Michael asks where I use the fact that we’re in a metric space, which was very astute. It turns out on reflection that I did use it, but it was hidden.

We can still say we’re looking for an accumulation point first and foremost, because if the sequence has an accumulation point there must be some subsequence converging to that point. Why not a subnet in general? Because metric spaces must be normal Hausdorff (using metric neighborhoods to separate
closed sets) and first-countable! And as long as we’re first-countable (or, weaker, “sequential”) we can find a sequence converging to any limit point of a net.

What I didn’t say before is that once we find an accumulation point there will be a subsequence converging to that point. My counterexample is compact, and any sequence in it has accumulation points, but we will only be able to find subnets of our sequence converging to them, not subsequences. Unless we add something to assure that our space is sequential, and metric spaces do that.

We should note in passing that the special case where is a compact subspace of is referred to as the Bolzano-Weierstrass Theorem.

Next is the Heine-Cantor theorem, which says that any continuous function from a compact metric space to any metric space is uniformly continuous. In particular, we can use the interval as our compact metric space and the real numbers as our metric space to see that any continuous function on a closed interval is uniformly continuous.

So let’s assume that is continuous but not uniformly continuous. Then there is some so that for any there are points and in with but . In particular, we can pick as our and get two sequences and with but . By the above theorem we can find subsequences converging to and converging to .

Now , which converges to , and so . Therefore we must have also converging to by the continuity of . But this can’t happen, since each of these distances must be at least ! Thus must have been uniformly continuous to begin with.

The Heine-Borel Theorem

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 .

Tychonoff’s Theorem

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!

The Image of a Compact Space

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.

Some compact subspaces

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.

Compact Spaces

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.

Separation Axioms

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.

Separation Properties

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