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
One theorem turns out to be very important when we’re dealing with connected spaces, or even just with a connected component of a space. If is a continuous map from a connected space to any topological space , then the image is connected. Similarly, if is path-connected then its image is path-connected.
The path-connected version is actually more straightforward. Let’s say that we pick points and in . Then there must exist and with and . By path-connectedness there is a function with and , and so and . Thus the composite function is a path from to .
Now for the connected version. Let’s say that is disconnected. Then we can write it as the disjoint union of two nonempty closed sets and by putting some connected components in the one and some in the other. Taking complements we see that both of these sets are also open. Then we can consider their preimages and , whose union is since every point in lands in either or .
By the continuity of , each of these preimages is open. Seeing as each is the complement of the other, they must also both be closed. And neither one can be empty because some points in land in each of and . Thus we have a nontrivial clopen set in , contradicting the assumption that it’s connected. Thus the image must have been connected, as was to be shown.
From this theorem we see that the image of any connected component under a continuous map must land entirely within a connected component of the range of . For example, any map from a connected space to a totally disconnected space (one where each point is a connected component) must be constant.
When we specialize to real-valued functions, this theorem gets simple. Notice that a connected subset of is just an interval. It may contain one or both endpoints, and it may stretch off to infinity in one or both directions, but that’s about all the variation we’ve got. So if is a connected space then the image of a continuous function is an interval.
An immediate corollary to this fact is the intermediate value theorem. Given a connected space , a continuous real-valued function , and points with and (without loss of generality, ), then for any there is a so that . That is, a continuous function takes all the values between any two values it takes. In particular, if is itself an interval in we get back the old intermediate value theorem from calculus.
Tied in with the fundamental notion of continuity for studying topology is the notion of connectedness. In fact, once two parts of a space are disconnected, there’s almost no topological influence of one on the other, which should be clear from an intuitive idea of what it might mean for a space to be connected or disconnected. This intuitive notion can be illustrated by considering subspaces of the real plane .
First, just so we’re clear, a subset of the plane is closed if it contains its boundary and open if it contains no boundary points. Here there’s a lot more between open and closed than there is for intervals with just two boundary points. Anyhow, you should be able to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same answer, but it’s interesting to generalize this to find a metric on the product of two arbitrary metric spaces.
Anyhow, back to connectedness. Take a sheet of paper to be your plane, and draw a bunch of blobs on it. Use dotted lines sometimes to say you’re leaving out that section of the blob’s border. Have fun with it.
Now we’ll consider that collection of blobs as a subspace , and thus it inherits the subspace topology. We can take one blob and throw an open set around it that doesn’t hit any other blobs (draw a dotted curve around the blob that doesn’t touch any other). Thus the blob is an open subset because it’s the intersection of the open set we drew in the plane and the subspace . But we could also draw a solid curve instead of a dotted one and get a closed set in the plane whose intersection with is . Thus is also a closed subset of . Some people like to call such a subset in a topological space “clopen”.
In general, given any topological space we can break it into clopen sets. If the only clopen sets are the whole space and the empty subspace, then we’re done. Otherwise, given a nontrivial clopen subset, its complement must also be clopen (why?), and so we can break it apart into those pieces. We call a space with no nontrivial clopen sets “connected”, and a maximal connected subspace of a topological space we call a “connected component”. That is, if we add any other points from to , it will be disconnected.
An important property of connected spaces is that we cannot divide them into two disjoint nonempty closed subsets. Indeed, if we could then the complement of one closed subset would be the other. It would be open (as the complement of a closed subset) and closed (by assumption) and nontrivial since neither it nor its complement could be empty. Thus we would have a nontrivial clopen subset, contrary to our assumptions.
If we have a bunch of connected spaces, we can take their coproduct — their disjoint union — to get a disconnected space with the original family of spaces as its connected components.
Conversely, any space can be broken into its connected components and thus written as a coproduct of connected spaces. In general, morphisms from the coproduct of a collection of objects exactly correspond to collections of morphisms from the objects themselves. Here this tells us that a continuous function from any space is exactly determined by a collection of continuous functions, one for each connected component. So we don’t really lose much at all by just talking about connected spaces and trying to really understand them.
Sometimes we’re just looking near one point or another, like we’ve done for continuity or differentiability of functions on the real line. In this case we don’t really care whether the space is connected in general, but just that it looks like it’s connected near the point we care about. We say that a space is “locally connected” if every point has a connected neighborhood.
Sometimes just being connected isn’t quite strong enough. Take the plane again and mark axes. Then draw the graph of the function defined by on the interval , and by . We call this the “topologist’s sine curve”. It’s connected because any open set we draw containing the wiggly sine bit gets the point too. The problem is, we might want to draw paths between points in the space, and we can’t do that here. For two points in the sine part, we just follow the curve, but we can never quite get to or away from the origin. Incidentally, it’s also not locally connected because any small ball around the origin contains a bunch of arcs from the sine part that aren’t connected to each other.
So when we want to draw paths, we ask that a space be “path-connected”. That is, given points and in our space , there is a function with and . Slightly stronger, we might want to require that we can choose this function to be a homeomorphism from the closed interval onto its image in . In this case we say that the space is “arc-connected”.
Arc-connectedness clearly implies path-connectedness, and we’ll see in more detail later that path-connectedness implies connectedness. However, the converses do not hold. The topologist’s sine curve gives a counterexample where connectedness doesn’t imply path-connectedness, and I’ll let you try to find a counterexample for the other converse.
Just like we had local connectedness, we say that a space is locally path- or arc-connected if every point has a neighborhood which is path- or arc-connected. We also have path-components and arc-components defined as for connected components.
Unfortunately, we don’t have as nice a characterization as we did before for a space in terms of its path- or arc-components. In the topologist’s sine curve, for example, the wiggly bit and the point at the origin are the two path components, but they aren’t put together with anything so nice as a coproduct.
[UPDATE]: As discussed below in the comments, I made a mistake here, implicitly assuming the same thing edriv said explicitly. As I say below, point-set topology and analysis really live on the edge of validity, and there’s a cottage industry of crafting counterexamples to all sorts of theorems if you weaken the hypotheses just slightly.
We’ve defined the real numbers as a topological field by completing the rational numbers as a uniform space, and then extending the field operations to the new points by continuity. Now we extend the order on the rational numbers to make into an ordered field.
First off, we can simplify our work greatly by recognizing that we just need to determine the subset of positive real numbers — those with . Then we can say if . Now, each real number is represented by a Cauchy sequence of rational numbers, and so we say if has a representative sequence with each point .
What we need to check is that the positive numbers are closed under both addition and multiplication. But clearly if we pick and to be nonnegative Cauchy sequences representing and , respectively, then is represented by and is represented by , and these will be nonnegative since is an ordered field.
Now for each , , so . Also, if and , then and , so , and so . These show that defines a preorder on , since it is reflexive and transitive. Further, if and then and , so and thus . This shows that is a partial order. Clearly this order is total because any real number either has a nonnegative representative or it doesn’t.
One thing is a little hazy here. We asserted that if a number and its negative are both greater than or equal to zero, then it must be zero itself. Why is this? Well if is a nonnegative Cauchy sequence representing then represents . Now can we find a nonnegative Cauchy sequence equivalent to ? The lowest rational number that can be is, of course, zero, and so . But for and to be equivalent we must have for each positive rational an so that for . But this just says that converges to !
So is an ordered field, so what does this tell us? First off, we get an absolute value just like we did for the rationals. Secondly, we’ll get a uniform structure as we do for any ordered group. This uniform topology has a subbase consisting of all the half-infinite intervals and for all real . But this is also a subbase for the metric we got from completing the rationals, and so the two topologies coincide!
One more very important thing holds for all ordered fields. As a field is a kind of a ring with unit, and like any ring with unit there is a unique ring homomorphism . Now since in any ordered field, we have , and , and so on, to show that no nonzero integer can become zero under this map. Since we have an injective homomorphism of rings, the universal property of the field of fractions gives us a unique field homomorphism extending the ring homomorphism from the integers.
Now if is complete in the uniform structure defined by its order, this homomorphism will be uniformly complete. Therefore by the universal property of uniform completions, we will find a unique extension . That is, given any (uniformly) complete ordered field there is a unique uniformly continuous homomorphism of fields from the real numbers to the field in question. Thus is the universal such field, which characterizes it uniquely up to isomorphism!
So we can unambiguously speak of “the” real numbers, even if we use a different method of constructing them, or even no method at all. We can work out the rest of the theory of real numbers from these properties (though for the first few we might fall back on our construction) just as we could work out the theory of natural numbers from the Peano axioms.
We’ve defined the topological space we call the real number line as the completion of the rational numbers as a uniform space. But we want to be able to do things like arithmetic on it. That is, we want to put the structure of a field on this set. And because we’ve also got the structure of a topological space, we want the field operations to be continuous maps. Then we’ll have a topological field, or a “field object” (analogous to a group object) in the category of topological spaces.
Not only do we want the field operations to be continuous, we want them to agree with those on the rational numbers. And since is dense in (and similarly is dense in ), we will get unique continuous maps to extend our field operations. In fact the uniqueness is the easy part, due to the following general property of dense subsets.
Consider a topological space with a dense subset . Then every point has a sequence with . Now if and are two continuous functions which agree for every point in , then they agree for all points in . Indeed, picking a sequence in converging to we have
So if we can show the existence of a continuous extension of, say, addition of rational numbers to all real numbers, then the extension is unique. In fact, the continuity will be enough to tell us what the extension should look like. Let’s take real numbers and , and sequences of rational numbers and converging to and , respectively. We should have
but how do we know that the limit on the right exists? Well if we can show that the sequence is a Cauchy sequence of rational numbers, then it must converge because is complete.
Given a rational number we must show that there exists a natural number so that for all . But we know that there’s a number so that for , and a number so that for . Then we can choose to be the larger of and and find
So the sequence of sums is Cauchy, and thus converges.
What if we chose different sequences and converging to and ? Then we get another Cauchy sequence of rational numbers. To show that addition of real numbers is well-defined, we need to show that it’s equivalent to the sequence . So given a rational number does there exist an so that for all ? This is almost exactly the same as the above argument that each sequence is Cauchy! As such, I’ll leave it to you.
So we’ve got a continuous function taking two real numbers and giving back another one, and which agrees with addition of rational numbers. Does it define an Abelian group? The uniqueness property for functions defined on dense subspaces will come to our rescue! We can write down two functions from to defined by and . Since agrees with addition on rational numbers, and since triples of rational numbers are dense in the set of triples of real numbers, these two functions agree on a dense subset of their domains, and so must be equal. If we take the from as the additive identity we can also verify that it acts as an identity real number addition. We can also find the negative of a real number by negating each term of a Cauchy sequence converging to , and verify that this behaves as an additive inverse, and we can show this addition to be commutative, all using the same techniques as above. From here we’ll just write for the sum of real numbers and .
What about the multiplication? Again, we’ll want to choose rational sequences and converging to and , and define our function by
so it will be continuous and agree with rational number multiplication. Now we must show that for every rational number there is an so that for all . This will be a bit clearer if we start by noting that for each rational there is an so that for all . In particular, for sufficiently large we have , so the sequence is bounded above by some . Similarly, given we can pick so that for and get an upper bound for all . Then choosing to be the larger of and we will have
for . Now given a rational we can (with a little work) find and so that the expression on the right will be less than , and so the sequence is Cauchy, as desired.
Then, as for addition, it turns out that a similar proof will show that this definition doesn’t depend on the choice of sequences converging to and , so we get a multiplication. Again, we can use the density of the rational numbers to show that it’s associative and commutative, that serves as its unit, and that multiplication distributes over addition. We’ll just write for the product of real numbers and from here on.
To show that is a field we need a multiplicative inverse for each nonzero real number. That is, for each Cauchy sequence of rational numbers that doesn’t converge to , we would like to consider the sequence , but some of the might equal zero and thus throw us off. However, there can only be a finite number of zeroes in the sequence or else would be an accumulation point of the sequence and it would either converge to or fail to be Cauchy. So we can just change each of those to some nonzero rational number without breaking the Cauchy property or changing the real number it converges to. Then another argument similar to that for multiplication shows that this defines a function from the nonzero reals to themselves which acts as a multiplicative inverse.
Okay, in a uniform space we have these things called “Cauchy nets”, which are ones where the points of the net are getting closer and closer to each other. If our space is sequential — usually a result of assuming it to be first- or second-countable — then we can forget the more complicated nets and just consider Cauchy sequences. In fact, let’s talk as if we’re looking at a sequence to build up an intuition here.
Okay, so a sequence is Cauchy if no matter what entourage we pick to give a scale of closeness, there’s some point along our sequence where all of the remaining points are at least that close to each other. If we pick a smaller entourage we might have to walk further out the sequence, but eventually every point will be at least that close to all the points beyond it. So clearly they’re all getting pressed together towards a limit, right?
Unfortunately, no. And we have an example at hand of where it can go horribly, horribly wrong. The rational numbers are an ordered topological group, and so they have a uniform structure. We can give a base for this topology consisting of all the rays , the rays , and the intervals , which is clearly countable and thus makes second-countable, and thus sequential.
Okay, I’ll take part of that back. This is only “clear” if you know a few things about cardinalities which I’d thought I’d mentioned but it turns out I haven’t. It was also pointed out that I never said how to generate an equivalence relation from a simpler relation in a comment earlier. I’ll wrap up those loose ends shortly, probably tomorrow.
Back to the business at hand: we can now just consider Cauchy sequences, instead of more general Cauchy nets. Also we can explicitly give entourages that comprise a base for the uniform structure, which is all we really need to check the Cauchy condition: . I did do absolute values, didn’t I? So a sequence is Cauchy if for every rational number there is an index so that for all and we have .
We also have a neighborhood base for each rational number given by the basic entourages. For each rational number we have the neighborhood . These are all we need to check convergence. That is, a sequence of rational numbers converges to if for all rational there is an index so that for all we have .
And finally: for each natural number there are only finitely many square numbers less than . We’ll let be the largest such number, and consider the rational number . We can show that this sequence is Cauchy, but it cannot converge to any rational number. In fact, if we had such a thing this sequence would be trying to converge to the square root of two.
The uniform space is shot through with holes like this, making tons of examples of Cauchy sequences which “should” converge, but don’t. And this is all just in one little uniform space! Clearly Cauchy nets don’t converge in general. But we dearly want them to. If we have a uniform space in which every Cauchy sequence does converge, we call it “complete”.
Categorically, a complete uniform space is sort of alike an abelian group. The additional assumption is an extra property which we may forget when convenient. That is, we have a category of uniform spaces and a full subcategory of complete uniform spaces. The inclusion functor of the subcategory is our forgetful functor, and we’d like an adjoint to this functor which assigns to each uniform space its “completion” . This will contain as a dense subspace — the closure in is the whole of — and will satisfy the universal property that if is any other complete uniform space and is a uniformly continuous map, then there is a unique uniformly continuous extending .
To construct such a completion, we’ll throw in the additional assumption that is second-countable so that we only have to consider Cauchy sequences. This isn’t strictly necessary, but it’s convenient and gets the major ideas across. I’ll leave you to extend the construction to more general uniform spaces if you’re interested.
What we want to do is identify Cauchy sequences in — those which should converge to something in the completion — with their limit points in the completion. But more than one sequence might be trying to converge to the same point, so we can’t just take all Cauchy sequences as points. So how do we pick out which Cauchy sequences should correspond to the same point? We’ll get at this by defining what the uniform structure (and thus the topology) should be, and then see which points have the same neighborhoods.
Given an entourage of we can define an entourage as the set of those pairs of sequences where there exists some so that for all and we have . That is, the sequences which get eventually -close to each other are considered -close.
Now two sequences will be equivalent if they are -close for all entourages of . We can identify these sequences and define the points of to be these equivalence classes of Cauchy sequences. The entourages descend to define entourages on , thus defining it as a uniform space. It contains as a uniform subspace if we identify with (the equivalence class of) the constant sequence . It’s straightforward to show that this inclusion map is uniformly continuous. We can also verify that the second-countability of lifts up to .
Now it also turns out that is complete. Let’s consider a sequence of Cauchy sequences . This will be Cauchy if for all entourages there is an so that if and the pair is in . That is, there is an so that for and we have . We can’t take the limits in of the individual Cauchy sequences — the limits along — but we can take the limits along ! This will give us another Cauchy sequence, which will then give a limit point in .
As for the universal property, consider a uniformly continuous map to a complete uniform space . Then every point in comes from a Cauchy sequence in . Being uniformly continuous, will send this to a Cauchy sequence in , which must then converge to some limit since is complete. On the other hand, if is another representative of then the uniform continuity of will force , so is well-defined. It is unique because there can be only one continuous function on which agrees with on the dense subspace .
So what happens when we apply this construction to the rational numbers in an attempt to patch up all those holes and make all the Cauchy sequences converge? At long last we have the real numbers ! Or, at least, we have the underlying complete uniform space. What we don’t have is any of the field properties we’ll want for the real numbers, but we’re getting close to what every freshman in calculus thinks they understand.
Now I want to toss out a few assumptions that, if they happen to hold for a topological space, will often simplify our work. There are a lot of these, and the ones that I’ll mention I’ll dole out in small, related collections. Often we will impose one of these assumptions and then just work in the subcategory of of spaces satisfying them, so I’ll also say a few things about how these subcategories behave. Often this restriction to “nice” spaces will end up breaking some “nice” properties about , and Grothendieck tells us that it’s often better to have a nice category with some bad objects than to have a bad category with only nice objects. Still, the restrictions can come in handy.
First I have to toss out the concept of a neighborhood base, which is for a neighborhood filter like a base for a topology. That is, a collection of neighborhoods of a point is a base for the neighborhood filter if for every neighborhood there is some neighborhood with . Just like we saw for a base of a topology, we only need to check the definition of continuity at a point on a neighborhood base at .
Now we’ll say that a topological space is “first-countable” if each neighborhood filter has a countable base. That is, the sets in can be put into one-to-one correspondence with some subset of the natural numbers . We can take this collection of sets in the order given by the natural numbers: . Then we can define , , and in general . This collection will also be a countable base for the neighborhood filter, and it satisfies the extra property that implies that . From this point we will assume that our countable base is ordered like this.
Why does it simplify our lives to only have a countable neighborhood base at each point? One great fact is that a function from a first-countable space will be continuous at if each neighborhood contains the image of some neighborhood . But must contain a set from our countable base, so we can just ask if there is an with .
We also picked the to nest inside of each other. Why? Well we know that if isn’t continuous at then we can construct a net that converges to but whose image doesn’t converge to . But if we examine our proof of this fact, we can look only at the base and construct a sequence that converges to and whose image fails to converge to . That is, a function from a first-countable space is continuous if and only if for all sequences , and sequences are a lot more intuitive than general nets. When this happens we say that a space is “sequential”, and so we have shown that every first-countable space is sequential.
Every subspace of a first-countable space is first-countable, as is every countable product. Thus the subcategory of consisting of first-countable spaces has all countable limits, or is “countably complete”. Disjoint unions of first-countable spaces are also first-countable, so we still have coproducts, but quotients of first-countable spaces may only be sequential. On the other hand, there are sequential spaces which are not first-countable whose subspaces are not even sequential, so we can’t just pass to the subcategory of sequential spaces to recover colimits.
A stronger condition than first-countability is second-countability. This says that not only does every neighborhood filter have a countable base, but that there is a countable base for the topology as a whole. Clearly given any point we can take the sets in our base which contain and thus get a countable neighborhood base at that point, so any second-countable space is also first-countable, and thus sequential.
Another nice thing about second-countable spaces is that they are “separable”. That is, in a second-countable space there will be a countable subset whose closure is all of . That is, given any point there is a sequence — we don’t need nets because is sequential — so that converges to . That is, in some sense we can “approximate” points of by sequences of points in , and itself has only countably many points.
The subcategory of all second-countable spaces is again countably complete, since subspaces and countable products of second-countable spaces are again second-countable. Again, we have coproducts, but not coequalizers since a quotient of a second-countable space may not be second-countable. However, if the map sends open sets in to open sets in the quotient, then the quotient space is second-countable, so that’s not quite as bad as first-countability.
Second-countability (and sometimes first-countability) is a property that makes a number of constructions work out a lot more easily, and which doesn’t really break too much. It’s a very common assumption since pretty much every space an algebraic topologist or a differential geometer will think of is second-countable. However, as is usually the case with such things, “most” spaces are not second-countable. Still, it’s a common enough assumption that we will usually take it as read, making explicit those times when we don’t assume that a space is second-countable.