General Topology/Compact spaces
Definition (compact space):
Let be a topological space. is called compact if and only if for every open cover of there exists a finite subcover, that is, indices so that .
Definition (compact subset):
Let be a topological space and be a subset. is called compact iff it is compact with respect to the subspace topology induced on by the topology of .
Proposition (the image of a compact set via a continuous map is compact):
Let be topological spaces, a compact subset and a continuous function. Then is a compact subset of .
Proof: Define to ease notation. Let be an open cover of , that is, by definition of the subspace topology, for suitable . Since
- ,
we obtain upon noting that , that the sets
form an open cover of with respect to its subspace topology; indeed, the continuity of insures that each set is open. Since is compact, a finite subcover, indexed by , may be chosen. Let be arbitrary. By the definition of , pick such that , and then such that . Then , so that .
Definition (proper map):
A function between topological spaces is called proper if and only if for each compact subset , the preimage is a compact subset of .
Note that the composition of proper maps is proper.
Proposition (closed subsets of a compact space are compact):
Let be a compact space, and let be closed. Then is compact.
Proof: Let be an open cover of . By definition of the subspace topology of , we take where is open in ; in order to avoid the axiom of choice, we may replace by the union of all for which . Then an open cover of is given by
- ,
and by compactness of , we may extract a finite subcover. Suppose that are the sets of that made it into the subcover. Then
- ,
since is contained in the whole subcover, but the only additional set in this subcover may be , which doesn't change whether or not is covered. Thus, is an open subcover of .
Theorem (Cantor's intersection theorem):
Let be a topological space, let be a directed set and let be a family of nonempty sets which are simultaneously compact and closed, such that . Then
- .
Proof: Suppose that
- .
Note that is compact, and since each is closed, its complement is open. Further, by definition of the subspace topology, the sets are open in , and by de Morgan () and distributivity of intersection over union, we get that the form an open cover of . By compactness of , we may extract a finite subcover , and upon choosing , we get that since and thus, since , also , a contradiction.
Proposition (compact nonempty Kolmogorov spaces contain a closed point):
Let be a nonempty, compact T0 space. Then contains a point such that is closed in .
Proof: The set of nonempty, closed subsets of , ordered by inverse inclusion, satisfies the hypotheses of Zorn's lemma, since the arbitrary intersection of closed sets is closed and also nonempty. Hence, there exists a minimal closed set . Suppose that contains two distinct points . Then by the hypothesis, select an open set that contains one point, but not the other, eg. . Then
Proposition (compact subsets of Hausdorff spaces are closed):
Let be a Hausdorff space, and let be compact. Then is closed.
Proof: Let be given. For each , there exist open sets and such that , and . Since is compact, select among the a finite subcover ; note that this step does not use the axiom of choice, since the cover in their totality; that is, we include in the cover not only one specific for each , but all sets of this form. Then set and obtain that is disjoint from ; indeed, it can't contain for any . Therefore,
- , ie. open.
Conversely, we have:
Proposition (compact sets being closed implies T1):
Let be a topological space where all compact sets are closed. Then is T1.
Proof: Any finite subset of is compact, so that we may apply the characterisation of T1 spaces.
Proposition (R1 space is Hausdorff iff all compact sets are closed):
Let be an R1 space. Then is Hausdorff iff all compact sets are closed.
Proof: One direction is clear since compact subsets of Hausdorff spaces are closed. For the other direction, we may apply the R-axiom characterisation of Hausdorff spaces, using the fact that is T1.
Proposition (intersection of compact sets in Hausdorff spaces is compact):
Let be compact subsets of a Hausdorff space . Then
is compact.
Proof: Since is Hausdorff, all the s are closed. Hence, the given set is a closed subset of the compact set , where is arbitrary.
Proposition (compact Hausdorff spaces are normal):
Let be a compact Hausdorff space. Then is normal.
Proof: Let be two closed subsets of which are disjoint. First, we note that are compact, since closed subsets of a compact space are compact. Then let be arbitrary. Since is Hausdorff, we may choose and disjoint open so that and . Since is compact, choose a finite subcover of . Then set and observe (as in the proof of the last proposition) that is an open subset of disjoint from . Then note that since is compact, we may choose a finite subcover of . Then define
- ,
and observe that , and that are open and disjoint.
Definition (finite intersection property):
Let be a topological space. is said to possess the finite intersection property if and only if for all families of closed subsets of such that
- ,
there exists a finite set of indices such that
- .
Proposition (compactness is equivalent to the finite intersection property):
Let be a topological space. is compact if and only if it satisfies the finite intersection property.
Proof: being compact is equivalent to the assertion that for all families that cover , there exists a finite subcover. Such covers are in bijective correspondence to families with empty intersection via
- , with inverse
Note further that under this correspondence, families of open sets cover if and only if the corresponding family of closed sets has empty intersection; this is a consequence of de Morgan. Hence, whenever we have the finite intersection property, we may translate an open cover into a family of closed sets with empty intersection, extract a finite subfamily with empty intersection, and revert back to see that the resulting open cover (which is a subcover of the original family) is finite and covers , and if we have compactness, an analogous argument works.
Proposition (continuous bijection from compact to Hausdorff is homeomorphism):
Let be a compact space and be a Hausdorff space. Suppose that is a bijective function which is continuous. Then is a homeomorphism.
Proof: Let be given; we prove that is continuous at . Set and suppose that where is open. Note that for each , we may choose open neighbourhoods and such that , and . Consider the family of sets ; it forms an open cover of , so that we may extract a finite subcover (note that is needed, since it's the only set of the cover that contains ). Then set so that is an open neighbourhood of , and observe that , because if , then for a suitable , a contradiction since then .
Proposition (a finite union of compact sets is compact):
Let be a topological space and let be compact subsets of . Then is a compact subset of .
Proof: Let be an open cover of . By definition of the subspace topology, this means that , where is open in , for all . Note that upon defining for , we obtain that forms an open cover of , so that we may extract a finite subcover . Then observe that
is an open cover of , since each is covered.
Definition (locally compact):
Let be a topological space. Then is said to be locally compact if and only if for each and each open neighbourhood of , there exists a compact neighbourhood of so that .
Proposition (proper continuous maps to a locally compact Hausdorff space are closed):
Let be topological spaces, where is locally compact. Let be a continuous and proper function. Then is in fact closed.
Proof: Suppose that is closed, and set . Let , we are then to show that . Since is locally compact, pick a compact neighbourhood of . Since is proper and continuous, will be a compact set. Suppose that does not contain a point which is mapped via to . Since is Hausdorff, whenever , we find open neighbourhoods of and of such that . Therefore, the sets and are disjoint. Now the cover , where runs through all points of . Therefore, by compactness of , we may pick a finite subcover , and then , which contradicts the fact that .
Definition (compactification):
Let be a topological space. A compactification of is a pair , where is a compact topological space and is continuous, such that is an embedding and is dense in .
Often, and is the inclusion.
Definition (Alexandroff compactification):
Let be a topological space. The Alexandroff compactification of is defined by adjoining to a formal symbol , ie. , and by defining the topology on as the union of the following sets:
- The topology of
- All -complements of closed, compact sets of
Proposition (Alexandroff compactification is well-defined):
Let be a topological space, and let be its Alexandroff compactification. Then is a compact topological space, and if is not already compact, together with the inclusion it gives a compactification of .
Proof: First, we prove that the given topology is indeed a topology. Clearly, is compact and closed. Hence, is in the topology, as is , since is a topological space. Then, let be open. If either or are open subsets of , then so is . If both are -complements of closed, compact sets of , then
- ,
and again is open because the union of two closed sets is closed, and the union of two compact subsets is compact.
Now suppose we are given a family of open sets of open sets of , and a family of complements of compact and closed sets. Then
- ,
and if , we conclude since closed subsets of compact sets are compact; if , we conclude since is a topological space.
Note now that the inclusion is a homeomorphism onto its image by definition of the topology on ; it is continuous, open and bijective. Then suppose that is not compact; we claim that is dense in . Indeed, let be any open set in . Since is not compact, must intersect . Hence, is dense in .
Proposition (Alexandroff compactification of locally compact Hausdorff space is Hausdorff):
Let be a locally compact Hausdorff space. Then the Alexandroff compactification is Hausdorff.
Proof: Let , ; as usual, we denote by the point that was added to in forming . Suppose first that neither nor are . Then and are separated by disjoint neighbourhoods because is Hausdorff. Suppose now wlog. that , then , so . Since is locally compact, pick a compact neighbourhood of . Since is a neighbourhood of , pick an open neighbourhood of . Set . Since is Hausdorff, is closed, so that are open neighbourhoods of that satisfy the requirements of the definition of a Hausdorff space.
Exercises
[edit | edit source]- Let be a set, a topological space, and a function. Then is a compact subset of if and only if there exists a topology on which makes into a compact topological space.
- Let be a set with two topologies and , with respect to which is compact. Prove that also, is compact with respect to the topologies and , where the latter shall denote the least upper bound topology of and , borrowing notation from lattice theory.
-
- Let be topological spaces and let and be compact sets. Prove that is a compact subset of , where the latter is given the product topology.
- Let be Hausdorff spaces and suppose that and are proper, continuous functions. On the condition of the axiom of choice, prove that is proper. Hint: Prove first that it suffices to show that preimages of products of compact sets of are compact.
- Use Alexander's subbasis theorem to prove Tychonoff's theorem.
- Prove that if is a compact space and is discrete with respect to the subspace topology, then is a finite set.
- Let be compact spaces, a set and for , let be a function. Suppose that carries the final topology by the (). Prove that is compact if and only if is cofinite in .
- Let be topological spaces, where is compact and is Hausdorff, and let be a continuous bijection. On the condition of the axiom of choice, prove that is Hausdorff and is compact.
- Let be a noncompact connected topological space. Prove that its Alexandroff compactification is connected.