The preceding post is an exercise showing that the product of countably many compact spaces is a Lindelof space. The result is an example of a situation where the Lindelof property is countably productive if each factor is a “nice” Lindelof space. In this case, “nice” means compact. This post gives several exercises surrounding the notion of compactness.
Exercise 2.A
According to the preceding exercise, the product of countably many compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many compact spaces. More specifically, give an example of a product of uncountably many compact spaces such that the product space is not Lindelof.
Exercise 2.B
Any compact space is Lindelof. Since , the real line with the usual Euclidean topology is compact. This exercise is to find an example of “Lindelof does not imply compact.” Find one such example among the subspaces of the real line. Note that as a subspace of the real line, the example would be a separable metric space, hence would be a Lindelof space.
Exercise 2.C
This exercise is also to look for an example of a space that is Lindelof and not compact. The example sought is a nonmetric one, preferably a space whose underlying set is the real line and whose topology is finer than the Euclidean topology.
Exercise 2.D
Show that the product of two Lindelof spaces is a Lindelof space whenever one of the factors is a compact space.
Exercise 2.E
Prove that the product of finitely many compact spaces is a compact space. Give an example of a space showing that the product of countably and infinitely many compact spaces does not have to be compact. For example, show that , the product of countably many copies of the real line, is not compact.
Comments
The Lindelof property and compactness are basic topological notions. The above exercises are natural questions based on these two basic notions. One immediate purpose of these exercises is that they provide further interaction with the two basic notions. More importantly, working on these exercise give exposure to mathematics that is seemingly unrelated to the two basic notions. For example, finding compactness on subspaces of the real line and subspaces of compact spaces naturally uses a Baire category argument, which is a deep and rich topic that finds uses in multiple areas of mathematics. For this reason, these exercises present excellent learning opportunities not only in topology but also in other useful mathematical topics.
If preferred, the exercises can be attacked head on. The exercises are also intended to be a guided tour. Hints are also provided below. Two sets of hints are given – Hints (blue dividers) and Further Hints (maroon dividers). The proofs of certain key facts are also given (orange dividers). Concluding remarks are given at the end of the post.
Hints for Exercise 2.A
Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if is a Lindelof space, then every closed subspace of is also Lindelof.
Prove that if is a Lindelof space, then every closed and discrete subset of is countable (every space that has this property is said to have countable extent).
Show that the product of uncountably many copies of the real line does not have countable extent. Specifically, focus on either one of the following two examples.
 Show that the product space has a closed and discrete subspace of cardinality continuum where is cardinality of continuum. Hence is not Lindelof.
 Show that the product space has a closed and discrete subspace of cardinality where is the first uncountable ordinal. Hence is not Lindelof.
Hints for Exercise 2.B
Let be the set of all irrational numbers. Show that as a subspace of the real line is not compact.
Hints for Exercise 2.C
Let be the real line with the topology generated by the half open and half closed intervals of the form . The real line with this topology is called the Sorgenfrey line. Show that is Lindelof and is not compact.
Hints for Exercise 2.D
It is helpful to first prove: the product of two Lindelof space is Lindelof if one of the factors is a compact space. The Tube lemma is helpful.
Tube Lemma
Let be a space. Let be a compact space. Suppose that is an open subset of and suppose that where . Then there exists an open subset of such that .
Hints for Exercise 2.E
Since the real line is homeomorphic to the open interval , is homeomorphic to . Show that is not compact.
Further Hints for Exercise 2.A
The hints here focus on the example .
Let . Let be the first infinite ordinal. For convenience, consider the set , the set of all nonnegative integers. Since is a closed subset of , any closed and discrete subset of is a closed and discrete subset of . The task at hand is to find a closed and discrete subset of . To this end, we define after setting up background information.
For each , choose a sequence of open intervals (in the usual topology of ) such that
 ,
 for each (the closure is in the usual topology of ).
Note. For each , the open intervals are of the form . For , the open intervals are of the form . For , the open intervals are of the form .
For each , define the map as follows:
We are now ready to define . For each , is the mapping defined by for each .
Show the following:
 The set has cardinality continuum.
 The set is a discrete space.
 The set is a closed subspace of .
Further Hints for Exercise 2.B
A subset of the real line is nowhere dense in if for any nonempty open subset of , there is a nonempty open subset of such that . If we replace open sets by open intervals, we have the same notion.
Show that the real line with the usual Euclidean topology cannot be the union of countably many closed and nowhere dense sets.
Further Hints for Exercise 2.C
Prove that if and are compact, then the product is compact, hence Lindelof.
Prove that , the Sorgenfrey line, is Lindelof while its square is not Lindelof.
Further Hints for Exercise 2.D
As suggested in the hints given earlier, prove that is Lindelof if is Lindelof and is compact. As suggested, the Tube lemma is a useful tool.
Further Hints for Exercise 2.E
The product space is a subspace of the product space . Since is compact, we can fall back on a Baire category theorem argument to show why cannot be compact. To this end, we consider the notion of Baire space. A space is said to be a Baire space if for each countable family of open and dense subsets of , the intersection is a dense subset of . Prove the following results.
Fact E.1
Let be a compact Hausdorff space. Let be a sequence of nonempty open subsets of such that for each . Then the intersection is nonempty.
Fact E.2
Any compact Hausdorff space is Baire space.
Fact E.3
Let be a Baire space. Let be a dense subset of such that is a dense subset of . Then is not a compact space.
Since is compact, it follows from Fact E.2 that the product space is a Baire space.
Fact E.4
Let and . The product space is a dense subset of . Furthermore, is a dense subset of .
It follows from the above facts that the product space cannot be a compact space.
Proofs of Key Steps for Exercise 2.A
The proof here focuses on the example .
To see that has the same cardinality as that of , show that for . This follows from the definition of the mapping .
To see that is discrete, for each , consider the open set . Note that . Further note that for all .
To see that is a closed subset of , let such that . Consider two cases.
Case 1. for all .
Note that is an open cover of (in the usual topology). There exists a finite such that is a cover of . Consider the open set . Define the set as follows:
The set can be further described as follows:
The last step is because is a cover of . The fact that means that is an open subset of containing the point such that contains no point of .
Case 2. for some .
Since , for all . In particular, . This means that for some . Define the open set as follows:
Clearly . Observe that since . For each , since . Thus is an open set containing such that .
Both cases show that is a closed subset of .
Proofs of Key Steps for Exercise 2.B
Suppose that , the set of all irrational numbers, is compact. That is, where each is a compact space as a subspace of . Any compact subspace of is also a compact subspace of . As a result, each is a closed subset of . Furthermore, prove the following:

Each is a nowhere dense subset of .
Each singleton set where is any rational number is also a closed and nowhere dense subset of . This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus cannot be compact.
Proofs of Key Steps for Exercise 2.C
The Sorgenfrey line is a Lindelof space whose square is not normal. This is a famous example of a Lindelof space whose square is not Lindelof (not even normal). For reference, a proof is found here. An alternative proof of the nonnormality of uses the Baire category theorem and is found here.
If the Sorgenfrey line is compact, then would be compact and hence Lindelof. Thus cannot be compact.
Proofs of Key Steps for Exercise 2.D
Suppose that is Lindelof and that is compact. Let be an open cover of . For each , let be finite such that is a cover of . Putting it another way, . By the Tube lemma, for each , there is an open such that . Since is Lindelof, there exists a countable set such that is a cover of . Then is a countable subcover of . This completes the proof that is Lindelof when is Lindelof and is compact.
To complete the exercise, observe that if is Lindelof and is compact, then is the union of countably many Lindelof subspaces.
Proofs of Key Steps for Exercise 2.E
Proof of Fact E.1
Let be a compact Hausdorff space. Let be a sequence of nonempty open subsets of such that $latex for each . Show that the intersection is nonempty.
Suppose that . Choose . There must exist some such that . Choose . There must exist some such that . Continue in this manner we can choose inductively an infinite set such that for . Since is compact, the infinite set has a limit point . This means that every open set containing contains some (in fact for infinitely many ). The point cannot be in the intersection . Thus for some , . Thus . We can choose an open set such that and . However, must contain some point where . This is a contradiction since for all . Thus Fact E.1 is established.
Proof of Fact E.2
Let be a compact space. Let be open subsets of such that each is also a dense subset of . Let a nonempty open subset of . We wish to show that contains a point that belongs to each . Since is dense in , is nonempty. Since is dense in , choose nonempty open such that and . Since is dense in , choose nonempty open such that and . Continue inductively in this manner and we have a sequence of open sets just like in Fact E.1. Then the intersection of the open sets is nonempty. Points in the intersection are in and in all the . This completes the proof of Fact E.2.
Proof of Fact E.3
Let be a Baire space. Let be a dense subset of such that is a dense subset of . Show that is not a compact space.
Suppose is compact. Let where each is compact. Each is obviously a closed subset of . We claim that each is a closed nowhere dense subset of . To see this, let be a nonempty open subset of . Since is dense in , contains a point where . Since , there exists a nonempty open such that . This shows that each is a nowhere dense subset of .
Since is a dense subset of , where each is an open and dense subset of . Then each is a closed nowhere dense subset of . This means that is the union of countably many closed and nowhere dense subsets of . More specifically, we have the following.
(1)………
Statement (1) contradicts the fact that is a Baire space. Note that all and are open and dense subsets of . Further note that the intersection of all these countably many open and dense subsets of is empty according to (1). Thus cannot not a compact space.
Proof of Fact E.4
The space is compact since it is a product of compact spaces. To see that is a dense subset of , note that where for each integer
(2)………
Note that the first factors of are the open interval and the remaining factors are the closed interval . It is also clear that is a dense subset of . This completes the proof of Fact E.4.
Concluding Remarks
Exercise 2.A
The exercise is to show that the product of uncountably many compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that has uncountable extent where is continuum. Having uncountable extent (i.e. having an uncountable subset that is both closed and discrete) implies the space is not Lindelof. The uncountable extent of the product space is discussed in this post.
For and , there is another way to show nonLindelof. For example, both product spaces are not normal. As a result, both product spaces cannot be Lindelof. Note that every regular Lindelof space is normal. Both product spaces contain the product as a closed subspace. The nonnormality of is discussed here.
Exercise 2.B
The hints given above is to show that the set of all irrational numbers, , is not compact (as a subspace of the real line). The same argument showing that is not compact can be generalized. Note that the complement of is , the set of all rational numbers (a countable set). In this case, is a dense subset of the real line and is the union of countably many singleton sets. Each singleton set is a closed and nowhere dense subset of the real line. In general, we can let , the complement of a set , be dense in the real line and be the union of countably many closed nowhere dense subsets of the real line (not necessarily singleton sets). The same argument will show that cannot be a compact space. This argument is captured in Fact E.3 in Exercise 2.E. Thus both Exercise 2.B and Exercise 2.E use a Baire category argument.
Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not compact. In this case, the suggested space is , the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product space is embedded in the compact space , which is a Baire space. As mentioned earlier, Fact E.3 is essentially the same argument used for Exercise 2.B.
Using the same Baire category argument, it can be shown that , the product of countably many copies of the countably infinite discrete space, is not compact. The space of the nonnegative integers, as a subspace of the real line, is certainly compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not compact. With the product space , there is a connection with Exercise 2.B. The product is homeomorphic to . The idea of the homeomorphism is discussed here. Thus the noncompactness of can be achieved by mapping it to the irrationals. Of course, the same Baire category argument runs through both exercises.
Exercise 2.C
Even the noncompactness of the Sorgenfrey line can be achieved by a Baire category argument. The nonnormality of the Sorgenfrey plane can be achieved by Jones’ lemma argument or by the fact that is not a first category set. Links to both arguments are given in the Proof section above.
See here for another introduction to the Baire category theorem.
The Tube lemma is discussed here.
Dan Ma topology
Daniel Ma topology
Dan Ma math
Daniel Ma mathematics
2019 – Dan Ma