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 .
(On the condition of the axiom of choice.)
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
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.
(On the condition of the axiom of choice.)
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 .
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.
- 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.