Consider the open disk in the plane, x2 + y2 < 1, and the closed disk, x2 + y2 ≤ 1. Are these spaces different? They feel different. One has a border you can stand on and the other does not. Still, there might be a homeomorphism that equates the two, wherein an open set at p, on the border, somehow becomes an open ball centered at f(p) inside the open disk. How do you know this can't happen?
A year of algebraic topology won't help you, because both spaces have no homotopy groups, no homology groups, and no cohomology groups. However, by the end of this chapter you will be able to say, "The closed disk is compact and the open disk is not." That's it in one sentence.
Cool - but have we lost sight of the title of this book, Discrete Math? Why all the continuous functions and continuous spaces? Because groups and rings will be given various topologies, and homomorphisms will turn into continuous functions from one topology to another. In some cases the topology of a ring or algebraic space is compact, and that is helpful in certain theorems. As mentioned in the prolog, branches of mathematics are not really independent. So let's stroll through a couple more chapters of topology, and then groups and rings will return.
An open cover for a set S, inside a topological space T, is a collection of open sets whose union contains S. A finite subcover is any finite subcollection of these open sets whose union also contains S. S is compact if every open cover of S admits a finite subcover. If T is a compact set then the space itself is compact.
In the above, is S a compact set or a compact space? The words really mean the same thing. S is a set that inherits a topology from T, and as such S becomes a subspace of T, and a space unto itself. The open cover might come from T, but it is still an open cover of S, a collection of open sets in S that covers S. So it doesn't really matter if you call S a compact set or a compact space.
A finite set is always compact.
Here is an alternate definition of compact. A collection of sets possesses the finite intersection property if every finite subcollection of these sets has a nonempty intersection. A set S is compact if every collection of closed subsets of S with the finite intersection property has a nonempty intersection. Stated another way, every collection of closed subsets of S with no intersection includes a finite subcollection that also has no intersection. Let's tie these definitions together.
A closed set in S is the complement of an open set in S. The open sets form a cover iff the closed sets in S have no intersection. The finite subcover of open sets corresponds to a finite subcollection of closed sets with no intersection. Hence the definitions are equivalent.
Any closed subset Y of a compact set S is compact. Select any open cover for Y, and throw in the complement of Y to cover S. The finite subcover of S, sans the complement of Y, covers Y.
Most compact sets are closed, but this isn't required. In the indiscrete topology, where only the empty set and the entire space are open, every subset of the space is compact.
The finite union of compact sets is compact. Given an open cover for all of them, select a finite subcover for each compact set, and the union of these subcovers remains finite.
Let f be a continuous function on the domain T, and let Y be the image of S, a compact set. An open cover of Y pulls back to an open cover of S. The resulting finite subcover maps forward, through f, to a finite subcover of Y. Thus the continuous image of a compact set is compact.
If a set is unbounded in a metric space, choose an open cover consisting of concentric spherical shells. One shell might be the points that are between 8 and 10 units away from the origin. The next shell might contain points between 9 and 11 units from the origin. And so on. This builds an open cover for the set; in fact it is an open cover for the entire space. Since S is unbounded, infinitely many shells are required. Therefore a compact set in a metric space is always bounded.
So - where do we go from here? You might think that we should begin by proving that the closed interval [0,1] is compact. Then we can extend this to paths, the continuous image of [0,1]. Next, the product of unit intervals makes unit cubes in n dimensions. We ought to be able to prove those are compact, eh? Then any closed set embedded in a hypercube is compact, and we finally have the long sought theorem: a closed bounded set in Rn is compact.
This is the approach I would take, but as it turns out, it's darn hard to prove [0,1] is compact from first principles. It's actually easier to solve a more general problem, then apply the result to a specific case such as the unit interval. It's a counterintuitive approach. If S is compact in a hausdorff space, it is closed. Consider any point x not in S. For each y in S, embed y in an open set disjoint from an open set containing x. Do this for each y in S, and we have an open cover. Take a finite subcover, intersect the corresponding open sets about x, and find an open set that contains x, and misses S. This holds for all x outside of S, hence the complement of S is open, and S is closed.
An important corollary is that a continuous map of a compact space onto a hausdorff space is bicontinuous. Any closed set in the domain is also compact, its continuous image is compact, and its image is closed. Taking complements, the image of open sets is open. Thus f is bicontinuous. If f happens to be injective then it is a homeomorphism, and the two spaces are topologically equivalent.
The word onto is important here. Embed the closed interval [0,1] into the real line, and the result is not bicontinuous. The interval [0,½) is open in the domain, but its image is not open in the range.
Let f map X onto Y and let g map Y into Z. Let X be compact and let Y be Hausdorff. Let the functions f and fg be continuous. Thus f is bicontinuous. Start with a closed set C in Z and take its preimage under fg. This is closed in X, and its image is closed in Y. The preimage of C is closed in Y, and g is continuous.
Let S be hausdorff and let U be a compact subspace of S. If U is also dense in S then U = S. Remember that a compact set in a hausdorff space is closed, so U is closed. If U misses the point x then U must miss some open set containing x, which contradicts the fact that U is dense. Thus U contains every x, and U = S.
Making a topology weaker keeps a space compact, for any open cover is an open cover in the original compact space. In contrast, making a topology stronger, i.e. adding more open sets, keeps a space hausdorff.
Let S be a space that is compact and hausdorff. Watch what happens to S with a stronger or weaker topology. Let f map S onto S, but let the domain have a stronger topology. Open sets still have open sets as preimage, so f is continuous. Yet some open sets in the domain are new, and don't map to open sets in S, hence f is not bicontinuous. We already showed that a map from a compact space onto a hausdorff space is a homeomorphism, and that isn't the case here; hence the domain of f is not compact. Similarly, when f maps S onto a copy of itself with a weaker topology, f is not bicontinuous, and the range is not hausdorff. A compact hausdorff space walks a fine line. Add more open sets and lose compactness. Take away open sets and the space is no longer hausdorff.
Assume a space is both compact and hausdorff. Let S be a closed set in this space, and let x be any point not in S. Separate every point in S from x, using disjoint open sets. A finite cover will do, so we can separate S from x, and the space is regular.
Let T be another closed set, disjoint from S. For every x in T, separate x and S in disjoint open sets. Again, a finite cover will do. A finite intersection of open sets contains S, while a finite union of open sets contains T, and the space is normal.
Again let S be compact and hausdorff. A homeomorphism embeds S in a hypercube. For all practical purposes, S is a subspace of the cube.
Consider all possible continuous functions from S into [0,1]. These form a composite function f that maps S into the product space P, a hypercube with the weak product topology. Select any pair of points x and y in S. Because S is normal, Urysohn's lemma provides a function fj that maps x to 0 and y to 1. Since x and y attain different values on at least one coordinate, f(x) ≠ f(y), and f is injective. In other words, f embeds S into P, and the map is continuous. Since the domain is compact and the range hausdorff, f is a homeomorphism. A space is countably compact if every countable open cover has a finite subcover.
Assume a space S is second countable, and let U be countably compact in S. Then U is compact (in the traditional sense). Given an open cover for U, perhaps an uncountable cover, replace each open set with a union of base sets. Thus a collection of base sets covers U, and there are countably many of these. A finite subcover will do. Let U be covered by n base open sets. Call them B1 through Bn for convenience. Find an open set, from the original cover, that contains B1 and bring it in. Then find an open set from the original cover that includes B2. Repeat this process for B3B4 etc, up to Bn, and build a finite subcover for U. Thus U is compact. If a space is second countable, compact and countably compact are the same thing.
Review the section at the top of this chapter, and extend those concepts to countably compact spaces. Verify the following four assertions.
Assume S has the bolzano weierstrass property, i.e. every sequence has a cluster point. Take a countable family of closed sets with the finite intersection property and let xn be a point in the intersection of the first n sets. Let p be a cluster point of this sequence. Let O be any open set containing p. Since p is cluster, O contains infinitely many points in the sequence. Thus O intersects every closed set in the collection. This holds for every open set O containing p. Thus p is in the closure of every closed set in our collection. This means p is in every closed set. The closed sets intersect, and S is countably compact.
Conversely, let S be countably compact. Start with a sequence x1 x2 x3 etc. Let B1 B2 B3 etc be a sequence of sets where Bn contains xk for k ≥ n. Thus B3 is the set containing x3 x4 x5 etc.
Let Cn be the closure of Bn. Note that every finite subcollection of these sets intersects. In other words, C has the finite intersection property. Since S is countably compact, there is a point p in the intersection. This point z will become a cluster point for the sequence. Let O be an arbitrary open set containing z, and suppose O has only a finite number of points from the x sequence. Beyond xn, O has none. This puts O, and z, outside of Cn+1, which is a contradiction. Therefore z is a cluster point of x.
In summary, countably compact is equivalent to bolzano weierstrass. A space is sequentially compact if every sequence has a convergent subsequence.
If the subsequence converges to p, then p is a cluster point for the original sequence. Therefore sequentially compact implies countably compact.
Next assume S is countably compact and first countable. Let a sequence x1 x2 x3 etc have a cluster point p. Let B1 B2 B3 etc be the base open sets assigned to p. Start with B1, which contains an infinite subsequence of x. Perhaps B1 contains x3 x7 x35 x2279 and so on. The intersection of B1 and B2 extracts an infinite subsequence from this subsequence. B2 may include some additional points from x, outside of the intersection, but we're not going to worry about that. Similarly, B3∩B2∩B1 pulls out another infinite subset. Continue this process, building a descending chain of infinite subsets from the sequence x.
Build the subsequence v as follows. Let v1 be any point of x that is in B1. In the above example, we might set v1 = x7. Let v2 be any point from x in B1∩ B2, that has an index beyond 7. Let v3 be any point from x in B1∩B2∩B3, beyond the previous two points. At each step, select a point farther out in the x sequence. Continue this process, building a subsequence of the original sequence x. Then consider any open set containing p. This includes a base open set about p, say Bj. Bj includes vj, and vj+1, and vj+2, and so on - the tail of the sequence v. This holds for every open set about p, hence v is a convergent subsequence of x, converging to p. Since x was an arbitrary sequence, countably compact and first countable implies sequentially compact.
The first uncountable ordinal, ℵ1, with the linear topology, is an example of a space that is sequentially compact, but not compact. Like any limit ordinal, it is the union of all the ordinals that came before, while a finite union of any of those ordinals is the maximum of those ordinals, which is something less. Thus ℵ1 is not compact. The fact that it is sequentially compact involves set theory that is beyond the scope of this book. Here's an adjective you may not have seen before. Let f be a function from a topological space S into the reals, or any other space with a linear topology. f is upper semicontinuous if the preimage of a ray extending downward is always an open set. Thus the preimage of the set (x<65) is an open set.
Lower semicontinuous is defined similarly, with the inequality reversed. Verify that f is continuous iff it is lower and upper semicontinuous. Hint, an open interval is the intersection of two rays.
Let f and g be upper semicontinuous, with f+g = h. If h(x) is less than u, let the difference be ε. Now the reals below f(x)+½ε have, as their preimage under f, an open set V in S. Similarly, W is an open set with g(W) below g(x)+½ε. The intersection is an open set, containing x, whose image under h is below u. The preimage, under h, is covered with open sets, and h is upper semicontinuous.
If f is upper semicontinuous and S is countably compact, then the image f(S) has an upper bound. Cover S with open sets On, where On is the preimage of the reals less than n. Choose a finite subcover, and the image of S is bounded.
Let u be the least upper bound of f(S). Build a family of closed sets in R, such that Cn is the set of reals ≥ u-1/n. Let Dn be the preimage of Cn under f, and note that Dn is closed. Also, the collection D1 D2 D3 etc has the finite intersection property. There is always some x in S whose image is arbitrarily close to u. Since S is countably compact, this collection of closed sets has a nonempty intersection. Call this point z. Since f(z) lies in every closed set Cn, and is bounded by u, f(z) = u. Therefore f(S) attains its upper bound.
If f is lower semicontinuous, f(S) attains its lower bound.
Let S be a metric space, and let V and W be countably compact sets in S. It is possible to measure the distance from V to W, as though they were closed sets. For a fixed y in S, and for any x in V, let f(x,y) be the distance |x,y|. Show that f(x) is continuous into the reals, hence it is upper and lower semicontinuous. This function attains its minimum/maximum. There is a point x in V that is closest, or farthest, from y.
Next, verify that the closest or farthest distance from y to V is a continuous function of y. This is a consequence of the triangular inequality.
Let y lie in W and let g(y) be the minimum or maximum distance from y to V. This is continuous into the reals. Reasoning as above, g attains its minimum and maximum. Thus V and W exhibit two points that are nearest each other, and two points that are farthest apart.
If V = W, there are two points x and y that lie on opposite sides of V. In other words, |x,y| is the diameter of V. If V and W are disjoint then they are a certain nonzero distance apart.
Let S be countably compact. Assume a sequence of upper semicontinuous functions fn(S) converges pointwise to 0. For any x in S, fn(x) approaches 0. Furthermore, let this progression be monotonic. The functions pull the image of x steadily towards 0. We will show that the functions converge uniformly.
Select an ε > 0. The reals below ε form an open ray; let On be the preimage of this ray under fn. Since images steadily drop to 0, On is an ascending chain of open sets. Every x gets below ε eventually, so S is covered by the infinite chain On. Choose a finite subcover. If n is the largest index in the finite subcover, fn(S), and all subsequent images of S, lie below ε. Thus the sequence of functions converges uniformly to 0.
Instead of 0, let the upper semicontinuous functions fn(S) decrease monotonically towards a limit function g(S). We will show g(S) is upper semicontinuous. (The domain S need not be countably compact for this step.)
Select an upper bound u, and a point x in S, with g(x) < u. The images of x approach g(x), so there is some n with fn(x) between g(x) and u. The preimage of the reals below u, under fn, is an open set W in S. All functions beyond n map W to values that are even lower. Thus the limit function g maps W to values below u. Therefore x is contained in an open set W, which is also in the preimage under g. The preimage is covered by open sets, and is open. This holds for all u, hence g is upper semicontinuous.
The classic example is xn on [0,1]. The limit function g is 0 on [0,1) and g(1) = 1. That is clearly not continuous, but it is upper semicontinuous. The product of topological spaces is taken across some indexing set J. Typically, J runs through the numbers 1 to n for a finite product, or J runs through the nonnegative integers, or perhaps the members of an uncountable ordinal. In any case, the product P can be given the weak or strong product topology. The weak topology crosses open sets in finitely many components for the base of P. The strong topology crosses open sets in all the components for the base of P. But if J is itself a topological space, there is another topology, intermediate between weak and strong, called the compact product topology. A base open set in P is the cross product of open sets, or base open sets, from the components of P corresponding to a compact set in J. Let's prove this is a base.
Let z be a point in the product space P, such that z is in the intersection of two base open sets. One base set comes from V in J, and the other comes from W. The first specifies open sets Bi in Si for each i in V, the second specifies Ci in Si for each i in W. When i is in V∩W, zi belongs to Bi and Ci simultaneously. Intersect Bi and Ci to find another open set containing zi. These are the open sets associated with V∩W. The other open sets in V and W are unchanged, and the components outside of V and W are unconstrained. Since V∪W is compact, there is no trouble here. The base defines a topology.
Any set J can be given the discrete topology, whence a set is compact iff it is finite. This reproduces the weak topology. On the other hand, J could be itself compact, or J could be given the indiscrete topology, which makes J compact, and in that case the compact product topology is the strong topology. If S is a metric space, or a set within a larger metric space, the concepts of compact, countably compact, and sequentially compact are equivalent. Let's see why this is so.
Every metric space is first countable, and that makes countably compact and sequentially compact equivalent. We only need show compact = countably compact.
Assume S is countably compact. Equivalently, every sequence has a cluster point.
Suppose, for a given ε, every finite set of points remains farther than ε from some point in S. As long as the set is finite, it is bounded away from some point in S. Build a sequence p1 p2 p3 etc, such that each point is at least ε away from all the earlier points. If x is a cluster point of this sequence, than there are infinitely many points within ½ε of x, which is impossible. We have built a sequence with no cluster point, and that contradicts the fact that S is countably compact. Therefore every ε has a finite set of points, such that all of S is within ε of the points in that finite set. And this can be done for each ε.
Set ε = 1/n, and find a finite set Wn for ε. Let U be the union of these finite sets Wn. Note that U is a countable set. Furthermore, the center of an open ball is always close to something in U. In other words, every open ball intersects U, and U is a dense set. Thus S is separable, and second countable. In a second countable space, compact and countably compact are equivalent. That completes the proof. In a metric space, compact = countably compact = sequentially compact.
Still within a metric space, compact implies totally bounded and complete. Suppose S is not totally bounded. For some ε, no finite set of ε balls will cover S. For each x in S, build a ball of radius ε, centered at x. S is covered by these open sets, yet there is no finite subcover. This is a contradiction, hence S is totally bounded after all.
Let c1 c2 c3 etc be a cauchy sequence in S. Since S is compact, this sequence has a cluster point x in S. In fact x has to be the limit of the sequence. This makes S a complete metric space.
Conversely, let S be complete and totally bounded. Demonstrate sequentially compact, which is the same as compact. Start with a sequence p1 p2 p3 etc, and cover S with finitely many balls of radius 1. Let B1 be a ball containing infinitely many terms of p.
Next, cover S with finitely many balls of radius ½. This means there are finitely many intersections between balls of radius 1 and balls of radius ½, and these intersections cover all of S. In particular, the intersections of the balls of radius ½ with B1 cover B1. Choose B2 so that B1∩B2 contains infinitely many points of p.
Cover S with finitely many balls of radius ¼, and choose B3 such that B1∩B2∩B3 contains infinitely many points of p. Continue this process, as the radius approaches 0.
Build a subsequence of p as follows. Start with any point of p that is contained in B1. Then select a later point of p in both B1 and B2. Then a third point of p from B1∩B2∩B3, and so on. Show that this subsequence is cauchy. Since S is complete, let the subsequence converge to x. This proves S is sequentially compact, hence compact. S in a metric space is compact iff it is complete and totally bounded.
If all this is true, then the rationals in [0,1] are not compact. Can we find a pathological open cover?
Select an irrational point q in the interval. Let On be the open set [0,b)∪(c,1], where b and c are carefully chosen rational numbers bracketing q, such that c-b < 1/n. This is an open cover with no finite subcover. In euclidean n space, S is closed and bounded iff S is compact. Take the forward direction first.
If S is closed in Rn it has all its limit points, and is complete. If S is bounded it is totally bounded. This makes S compact.
Conversely, let S be compact in Rn. If S is not bounded, ever expanding balls cover S, and there is no finite subcover. This is a contradiction, thus S is bounded.
If S does not contain one of its limit points x, build a sequence that approaches x, and S fails to be complete. Since compact implies complete, S has to be closed.
In Rn, compact is equivalent to closed and bounded. A line segment, a square, a circle, and an n dimensional hypercube; these are all compact sets.
The unit ball in Ej is closed and bounded, but not compact. Let the point xi be 1 in the ith component and 0 elsewhere. This sequence has no cluster point. Let S be a compact metric space and let T be a metric space. If f is a continuous function from S into T then f is uniform. This is a more general version of a better known theorem in real analysis: a continuous function on a closed bounded set in Rn is uniform.
Given ε, and a point x in S, there is some δ satisfying continuity at x, at the level of ½ε. Build a ball of radius ½δ about x. Do this for every x, and build an open cover for S. Select a finite subcover and set γ to the smallest radius. This is the value that will satisfy uniformity.
Let y and z be points in S, with |y,z| < γ. Refer to the open cover; y is contained in a ball of radius ½δ, centered at a point I will call x. (The value of δ may depend on x, but it is at least as large as γ.) By the triangular inequality, |x,z| < δ. That means |f(x),f(z)| < ½ε. The same is true of |f(x),f(y)|, hence f(y) and f(z) are within ε of each other. The value of γ satisfies continuity for every x, and f is uniform.
Continuing the above, the image of S is bounded in T. Fix an origin in T and consider the distance between f(S) and the origin. By composition, this is a continuous map from S into the reals. Since S is compact the image is compact, thus closed and bounded. Let b be the least upper bound. Being closed, the image attains its upper bound. Hence there is some x with |f(x)| = b. Once again this is a generalization of a theorem in real analysis: the image of a closed bounded set in Rn is bounded, and attains its upper and lower bounds. Let f and g be continuous functions from S into T, where S is compact and T is a metric space, and let h(x) be the distance from f(x) to g(x). Choose an open interval in the reals, and let h(x) lie inside this interval. In other words, x is in the preimage of an open set. Choose an open set in S that keeps f(y) close to f(x), and g(y) close to g(x). Within this open set, h(y) is close to h(x), and lies inside our interval. (This is the triangular inequality.) The preimage of every open interval is covered by open sets, and h is continuous.
As per the previous section, h is bounded. Every two functions differ by an upper bound b, and b = 0 iff the functions are equal. Let S be a compact region in a metric space. Let C be an open cover for S. There is then a distance d, such that every open ball in S, of diameter d or less, lies inside at least one of the open sets of C.
Suppose this is not the case. Build a sequence of balls, with decreasing diameters, such that none of these balls is contained in an open set in C. Since S is compact, the centers have a cluster point p. Restrict attention to a subsequence of balls whose centers approach p. At each step, choose a ball with a smaller diameter, so that the balls are shrinking as they approach p.
Since p is contained in an open set U in C, place p at the center of an open ball of radius e, contained in U. Beyond some index, the centers of our balls are within e/2 of p. Go farther, if necessary, so that the radius of each ball is less than e/2. This places the entire ball inside the set U, which is a contradiction. Therefore a distance d exists, such that every ball of diameter d or less lies in some open set U in our open cover C.
Two disjoint closed sets in a metric space can be 0 distance apart, as shown by the graphs of xy = 1 and xy = -1 in the plane. However, this cannot happen if one of the sets is compact.
Let S be compact and let T be closed in a metric space. Map any x in S to the distance between x and T. This is a continuous map from a compact set into the reals. Thus the image is compact, and closed. It cannot approach 0 without becoming 0.
If x is 0 distance from a closed set T, then x is a limit point of T, x lies in T, and the two sets are not disjoint. This is a contradiction, hence S and T are a positive distance apart. The Tikhonov (also spelled Tychonoff) product theorem states that a product space is compact iff all its component spaces are compact.
Compact is usually described in terms of open sets and open covers, but there is a complementary definition that refers to closed sets and the finite intersection property, which was presented at the top of this chapter. I'm going to use that here.
A collection of sets is larger than another if it includes more sets. This is a partial ordering on families of sets.
Restrict attention to collections of sets that have the finite intersection property. Once again, collection X is larger than collection Y if X brings in additional sets, and both X and Y have the finite intersection property. Verify that the union of an ascending chain yields an even larger collection with the finite intersection property. Use zorn's lemma to select a maximal collection of sets with the finite intersection property.
Let M be such a collection. Let U and V be two sets in this collection. Clearly U and V intersect; let W = U∩V. Let F be any finite subcollection of M, and note that the intersection of F and U and V is the intersection of F with W. The former is nonempty; so is the latter. If we bring in W, M still has the finite intersection property. Since M is maximal, it already contains W. In other words, the finite intersection of any of the sets in M produces another set in M.
Let C be a set that happens to intersect everything in M. Let F be a finite subcollection, and let G be their nonempty intersection. Since G is in M, C and G intersect. This means C intersects all the members of F. We can bring C into M, and it still has the finite intersection property. Since M is maximal, it already contains C. M contains any set that intersects all the members of M.
Let Si, for i in some indexing set, be an indexed family of topological spaces, with P as a topological product. Let ei be the ith projection from P onto Si. If P is compact, then each Si is the continuous image of P under ei, and is compact. That was the easy direction.
Conversely, assume each space Si is compact. Let T be a collection of closed sets in P with the finite intersection property. Embed T in a collection M of (not necessarily closed) sets, that is maximal with respect to the finite intersection property. Project M down to Si to give a collection of sets in Si, which I will call Mi. Show that Mi has the finite intersection property.
It follows that the closures of the sets in Mi, relative to Si, also have the finite intersection property. Since Si is compact, there is some qi in Si that lies in the closure of each set in Mi. Let the point q in P have projections qi.
What does this mean? Let W be any set in M and project W down to Wi in Si. The closure of Wi includes qi. Any open set in Si containing qi intersects Wi.
Select an arbitrary set W from M, and keep it on standby. Let U be a base open set in P that contains q, but not just any base open set. Let U restrict one component, say Si, to a base open set V, and leave the other components unconstrained. Now V intersects Wi in a point y. Let x be a point in W, whose ith component is y. There certainly is such a point, since the projection of W is Wi. Since y lies in V, x lies in U. Thus x is in W∩U.
Since U intersects every W in M, U belongs to M. This holds for every U of this form, i.e. containing q and restricting one component. The intersection of two such sets constrains two components. The intersection of finitely many such sets builds a base open set in P. Since M is closed under finite intersection, M includes every base open set in P that contains q.
Let C be a closed set taken from our original collection T, which is part of M. Since C intersects every base open set containing q, C contains q. Therefore T has a nonempty intersection, namely q, and P is compact. Let Si, for i in some indexing set, be an indexed family of topological spaces, with P as product. Assume P is sequentially compact. Let ei be the ith projection from P onto Si. Let y1 y2 y3 etc be a sequence in S, one of the component spaces of P. Assign constants to the other components. This produces a sequence x1 x2 x3 etc in P. Only the component of S changes, the other components are fixed as the sequence proceeds. Select a subsequence that approaches q. In fact, retain only the subsequence, and throw away the other terms of x. Throw away the corresponding terms of y. The remaining sequence x converges to q, while its projection y, (intuition suggests), converges to r, where r is the projection of q in S. Let V be an open set in S that contains r, and let U be the preimage of V under the projection. Since U, an open set in P, holds the tail of the convergent sequence x, V holds the tail of the sequence y. This is valid for every V containing r, hence y, a subsequence of the original, converges to r. If P is sequentially compact then every component space is sequentially compact.
Conversely, assume each component space is sequentially compact, and assume P is the product of countably many such spaces. Let x1 x2 x3 etc be a sequence in P. Let t1 be a subsequence of x that converges in the first component S1 of P. Thus t1,1 might be x7, and t1,2 might be x23, and so on.
Let t2 be a subsequence of t1 that converges in the second component S2 of P. Let t3 be a subsequence of t2 that converges in the third component S3 of P. Continue this process, building an infinite sequence of sequences. The nth sequence converges in the first n components.
Let tn converge to qn in Sn. Let q be the point in P with projections qn in Sn. Build a subsequence of x that converges to q as follows. Run down the diagonal, and set zn = tn,n, the nth term of the nth sequence. Let U be a base open set about q, which constrains finitely many components. Let Bj be the projection of U into Sj, where j designates one of the constrained components of U. Remember that Bj contains qj. Go out far enough, and the terms of tj are all inside Bj, so beyond some index dj, the terms of z all live in Bj. This holds across finitely many components, so let d be the maximum value over dj. Beyond d, z lives in U. This holds for all base open sets U containing q, hence the subsequence z converges to q, and P is sequentially compact. A space is locally compact if every point x has some open neighborhood Q about it whose closure is compact. If you want Q to lie inside a given open neighborhood U containing x, intersect Q with U. The result is open, and its closure is in the original closure, which was compact. Closed in a compact space is compact, so Q∩U works just as well.
Here is an equivalent definition. The space S is locally compact iff the open sets with compact closures form a base. Let's take the forward direction first.
Intersect two open sets with compact closures, and the closure of the intersection lies in the closure of the first open set. A closed subspace of a compact set is compact, so the intersection of two base sets is another base set.
Now consider a point x in an arbitrary open set W. Surround x with an open set in W whose closure is compact. Thus all of W is covered with base sets, and we really do have a base.
Conversely, let the open sets with compact closures form a base for the topology. Any point p is part of the entire space, which is open, hence p is covered by a base set. This is the definition of locally compact.
Compact implies locally compact, since you can always use the entire space as the open set. In contrast, Rn is locally compact, but not compact.
Ej is a complete metric space that is not locally compact. Let p be the origin and imagine an open set about p with compact closure. Inside this set is a ball of radius ε, which also has compact closure. Rescale everything, so the ball has radius 3. This ball is not totally bounded, which is a requirement for compact in a metric space, hence Ej is not locally compact. If S is locally compact and hausdorff, we can turn S into a compact space by adding the point at infinity. This process, developed by Aleksandrov, (also spelled Alexandroff), is called compactification. The classic example is the plane, which is not compact. Curl it up into a sphere with a hole at the top, then add the point at infinity, which becomes the north pole, and closes up the sphere. The sphere is compact, hence it is the "compactification" of the plane.
In this case we have a precise map. Place the sphere on the xy plane, its south pole at the origin. Draw a line from the north pole, through the surface of the sphere, and out to a point in the plane. These rays implement the bijection between the punctured sphere and the plane. You can derive the formula and prove it is bicontinuous. Thus the plane and the punctured sphere are homeomorphic.
The north pole does not map to any point in the plane; it is the point at infinity. It is also the key to making the sphere compact.
Let S be a space that is hausdorff and locally compact. Add the point ω to S. Open sets in S remain open. Also, the complement of a compact set in S, which necessarily contains ω, is open. We need to show this is a valid base for the topology, and there are no new open sets in S.
The intersection of two open sets in S yields an open set in S, so that's not a problem.
Intersect an open set in S and the complement of a compact set in S. Since S is hausdorff, compact sets are closed. Thus we are intersecting two open sets in S, and the result is open in S.
The intersection of the complements of two compact sets is the complement of their union, which is another compact set. In every case, the intersection of open sets is open, so we have a base.
If even one open set in a union of open sets contains ω, the union contains ω, and isn't going to make any unwanted open sets in S.
Let C be the new space, S∪ω, with its topology. If x and y are in S they can be placed in disjoint open sets in S. Given x and ω, find an open set about x whose closure is compact, and let the complement of the closure contain ω. Thus any two points in C can be separated in disjoint open sets, and C is hausdorff.
Now for compactness. Start with an open cover for C. Each open set is a union of base sets, thus C is covered by base sets. If a finite subcollection of base sets covers C, then bring in one open set for each of these base sets, and find a finite subcover from the original open cover. This is a general technique; a space is compact iff it is compact relative to its base open sets.
Base sets are open sets in S, and complements of compact sets in S adjoin ω. The first base set that contains ω has a complement Q that is compact in S, and closed in S. Other base sets containing ω are open in S after ω has been removed. Still other base sets are open in S from the get-go. These open sets cover Q, and a finite subcover will do. Bring the first base set back in, and C is covered, hence C is compact.
Embed S in C, and open sets remain open. Also, a new base open set is the complement of a compact set in S, but compact sets are closed, so this complement is already open. Embedding S into C is a homeomorphism.
If S and T are homeomorphic, verify that their compactifications are homeomorphic. Apply the preexisting homeomorphism between S and T, and map ωS to ωT. The new open sets correspond, and the compactifications are homeomorphic.
This has a surprising application. The open disk in the plane and the open ball in 3 space are different spaces, but how do you know that? They feel different; one is 2 dimensional and one is 3 dimensional. But maybe there is some magical homeomorphism h that makes them the same. There are no homotopy or homology groups to prove these spaces are distinct, so we have to think of something else. If they are homeomorphic then so are their compactifications. These are the 2 sphere and the 3 sphere respectively. These are indeed distinct spaces, as will be demonstrated later in this book. In fact the spheres of all dimensions are distinct, and so, the open balls of all dimensions are distinct. Furthermore, each ball in turn is homeomorphic to its containing space, as shown by the radial map r → r/(1-r). Thus each n dimensional space is a different topological space. However, we need the homology of the spheres to complete this proof.
In another application of compactification, a closed set crunches down to a point, while the rest of space stretches to accommodate, like squeezing everything above the Arctic Circle up to the north pole.
Let S be a compact hausdorff space and O an open subset of S. Let R be the complement of O, a closed set that will soon crunch down to a point. Since S is compact, so is R.
Select any x in O, and since R is compact, we can separate R and x in disjoint open sets. Let U be the set about x. The closure of U is entirely in O, and since it is closed in S, it is compact. This makes O a locally compact space. And of course, O remains hausdorff.
Let C be the compactification of O. Now O in S is homeomorphic to O in C. Map S onto C by mapping O to O, and R to ω. This crunches R down to a single point ω. Verify that this map is continuous. Open in C, and containing ω, comes from the complement of a compact set in O, which is the complement of a closed set in S, which is open. However an open set in R maps to the single point ω, thus the map is probably not bicontinuous. A continuous map between topologycal spaces is proper if the preimage of every compact set is compact.
Assume the domain S and range T of f are locally compact and hausdorff. Extend domain and range by taking their compactifications, giving S′ and T′, and extend f to a function g that agrees with f, and maps ω in S′ to ω in T′. f is proper iff g is continuous.
Let g be continuous and consider a compact set in T. This is closed in T′, with closed preimage in S′, which is compact in S′. An open cover that happens to come from S still has a finite subcover, thus the preimage of our compact set is compact, and f is proper.
If f is proper, a base open set in T′ containing ω pulls back to a base open set in S′ containing ω. This establishes continuity at ω. For any other point p in T, the continuity of f at p implies continuity of g at p. Thus g is continuous everywhere. Let S be a complete metric space, and let O1 O2 O3 etc be a countable collection of dense open sets in S. Let O be the intersection of all these open sets. O is nonempty and dense in S. This is the Baire category theorem.
Let W be an arbitrary nonempty open set in S, and intersect W with O1, having x1 in the intersection. Let S1 be an open ball in the intersection, with radius r1 centered at x1. Remember that the open ball S1 is the set of points less than r1 distance from x1. Let C1 be the closure of S1, and remember that C1 cannot include any points farther than r1 from x1. Cut r1 in half, if need be, so that C1 lies entirely inside the intersection of O1 and W.
Since O2 is dense, O2 intersects S1. As before, the intersection is a nonempty open set. Let x2 lie in the intersection, and build an open ball S2 about x2 with radius r2, subject to the following criteria.
Proceed by induction on n. At the nth step, Cn is contained in On and in Sn-1, and the radii shrink geometrically.
The sequence of centers x1 x2 x3 etc is cauchy, thanks to the shrinking radii, so let x be the limit of this cauchy sequence. Since the metric space is complete, x exists.
If x is not in Cn, then x is at least rn distance from xn. However, Sn+1 is bounded away from the edge of Sn, bounded away from x. The rest of the sequence cannot approach x, and that is a contradiction. Hence x lies in each Cn.
Since Cn lies in On, x lies in each On. Remember that O is the intersection O1∩O2∩O3∩O4∩… It follows that x lies in O, and O is nonempty.
Since x lies in C1 lies in W, x is in W. This means O and W intersect. Since W is arbitrary, O is a dense set.
Take a breath, that's the first half of the theorem. The same result holds if S is a locally compact hausdorff space. This includes the common metric spaces such as Rn, but it brings in some other spaces as well, so this is not a generalization of the above. We are really proving the theorem again for a new class of spaces.
As before, you are given O1 O2 O3 etc, a countable collection of dense open sets. Let W be an arbitrary open set. Let x1 be a point in W∩O1, and let C1 be the closure of an open set Q1, containing x1, such that C1 is compact. We can do this because S is locally compact.
Intersect Q1 with W and with O1 to give a possibly smaller open set S1 containing x1. The closure of S1 is a closed subspace inside C1, and is compact. Shrink C1 down to the closure of S1.
We're still not where we want to be. We want C1 to lie entirely inside W and O1. So far we only know this about S1, not its closure. We need to use the fact that the space is hausdorff. Let B1 be the boundary of S1, i.e. the points in C1 that are not in S1. This is a closed set C1, intersect the complement of an open set S1, hence B1 is closed. And it is part of C1, which is compact. Thus B1 is compact. If B1 is empty, then C1 is S1, and is entirely inside W and O1, and there is no trouble; so assume B1 is nonempty. Remember that x1 is in S1, and is not part of B1. Also x1 is closed, as is every point in a hausdorff space. These are two distinct closed sets in C1. Since C1 is both compact and hausdorff, it is normal. This means x1 and B1 can be separated in disjoint open sets. Let U contain x1 and let V contain B1. The closure of U misses V, and B1, hence the closure of U lies entirely inside S1, which is in O1 and W. Also, the closure of U is a closed set in C1, and is compact. Shrink S1 down to U, and let C1 be its closure. Now C1 is in W and O1, and contains x1, and is compact.
Since O2 is dense, O2 intersects S1. Let x2 lie in the intersection, and find an open set Q2 containing x2 whose closure is compact. Intersect Q2 with O2 and S1 to find S2. Let C2 be the closure of S2, which is compact. Shrink S2 down, if necessary, so that its closure C2 is entirely contained in O2 and S1. This uses the boundary hausdorff technique described above.
Continue this process forever, building a descending chain of closed compact sets Cn. At the nth step, Cn is contained in On and in Cn-1. Suppose this chain has no intersection. The complements form an ascending chain of open sets that cover C1. Since C1 is compact, a finite subcover will do. This means one of the closed sets Cn, and all its descendants, are empty, which is a contradiction. Therefore the chain intersects in at least one point; call it x.
As before, x is in every On, and in W, hence O intersects W, and O is a dense set in S.
This theorem fails when the collection of dense open sets is uncountable. The interval [0,1] is compact, hausdorff, a complete metric space, the nicest space you could think of, and the complement of any point is a dense open set, yet the intersection of these sets is empty. Let a set U be nowhere dense if the complement of its closure is dense. Equivalently, U closure contains no open sets.
If an open set B is in U it is in U closure. Let C be the complement of U, and assume B is in U closure but not in U, so that some part of B lives in C. C∩B cannot contain an open set, else B is not in U closure after all. Conversely, if U closure misses any part of B, then U misses the open set B intersect the complement of U closure. U closure does not contain B iff U misses some open set inside B.
In summary, U is nowhere dense iff each open set misses U, or includes a smaller open set that misses U. A subspace of a nowhere dense set misses even more sets, and remains nowhere dense.
If the topology has a base, it is sufficient to refer to base sets in the above characterization. For instance, if the topology is a metric space, then U is nowhere dense iff every open ball contains another open ball that misses U. This has nothing to do with first and second countable.
A set U is first category, or meager, if it is a countable union of nowhere dense sets. The space S is first category if it is a first category set within itself.
A subset of a first category set remains first category. Each nowhere dense set in the list is still nowhere dense.
Since the countable union of countable sets is countable, the countable union of first category sets is first category.
The complement of a first category set is called residual.
If U is not first category it is second category. That's the definition.
A corollary of the baire category theorem, and one reason it is called the baire `category' theorem, states that a complete metric space, or a locally compact hausdorff space, is second category. Suppose S is first category, and let O1 O2 O3 etc be the complements of the closures of the nowhere dense sets U1 U2 U3 etc. Thus O1 O2 O3 etc are open dense sets. By the baire category theorem, their intersection is nonempty. Let x lie in the intersection, hence x is not in the closure of U1, nor the closure of U2, nor the closure of U3, etc. This means x cannot lie in any of our nowhere dense sets, and their union does not cover all of S. Thus S is not first category, and must be second category.
If S is a complete metric space, as above, and S has no isolated points, then S, as a set, is uncountable. If S is countable then it is the countable union of its points. Remember that each point is closed in a metric space. If every point acts as a nowhere dense set then S becomes first category, which is a contradiction. Thus some point x fails to be a nowhere dense set. It's complement misses an open set, and that open set has to be x. If x is an open set then it is bounded away from the rest of S, and x is an isolated point.
The reals have no isolated points, and sure enough, they are uncountable.
Let S be a complete metric space, and assume the subspace U is first category within itself. We will show that its complement in S is dense.
Suppose the complement of U is not dense in S, hence U contains a nonempty open set O0. Replace O0 with an open ball inside O0. In other words, O0 has become a base set.
Let U1 be the first nowhere dense set inside U. Since U1 is nowhere dense, O0 contains a nonempty open ball O1 missing U1. Similarly O1 contains a nonempty open ball O2 missing U2, and so on. Choose ever smaller radii, so diameters approach 0, and make sure the closure of each open ball is entirely contained in the previous open ball. The centers approach some point x in the complete metric space S. Since x is in the closure of O1, x is in O0. Yet x is not in U1, or U2, or U3, etc, hence x is not in U. O0 is not entirely in U after all. The complement of U intersects the arbitrary open set O0, and the complement of U is dense. Let f be a family of continuous functions from a complete metric space S into a metric space T with a designated origin. Assume for every x in S, some real number m(x) is an upper bound for |fi(x)| over all the functions fi. There is a nonempty open subset of S, whose images in T are all bounded by some constant m.
For each positive integer m, let Um,i be the set of x in S with |fi(x)| ≤ m. Let Um be the intersection over i of Um,i. Since each fi is continuous, each Um,i is closed, hence each Um is closed.
By assumption, each x has an m(x) bounding all the images fi(x), which means each x belongs to some Um, and S is the countable union over Um.
Since S is not first category, some Um fails to be nowhere dense. It contains an open set, and the image of this open set is bounded by m.
Perhaps S is not a complete metric space, but the functions in f are all uniformly continuous. These functions can be extended to the completion of S. The uniform boundedness principle applies, And since each fi is the same on S and on the completion of S, this result pulls back to S. Some open set in S has an image bounded in T.