Lindelof Exercise 2

The preceding post is an exercise showing that the product of countably many \sigma-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 \sigma-compact. This post gives several exercises surrounding the notion of \sigma-compactness.

Exercise 2.A

According to the preceding exercise, the product of countably many \sigma-compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many \sigma-compact spaces. More specifically, give an example of a product of uncountably many \sigma-compact spaces such that the product space is not Lindelof.

Exercise 2.B

Any \sigma-compact space is Lindelof. Since \mathbb{R}=\bigcup_{n=1}^\infty [-n,n], the real line with the usual Euclidean topology is \sigma-compact. This exercise is to find an example of “Lindelof does not imply \sigma-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 \sigma-compact. The example sought is a non-metric 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 \sigma-compact space.

Exercise 2.E

Prove that the product of finitely many \sigma-compact spaces is a \sigma-compact space. Give an example of a space showing that the product of countably and infinitely many \sigma-compact spaces does not have to be \sigma-compact. For example, show that \mathbb{R}^\omega, the product of countably many copies of the real line, is not \sigma-compact.

Comments

The Lindelof property and \sigma-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 \sigma-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.

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Hints for Exercise 2.A

Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if X is a Lindelof space, then every closed subspace of X is also Lindelof.

Prove that if X is a Lindelof space, then every closed and discrete subset of X 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 \mathbb{R}^c has a closed and discrete subspace of cardinality continuum where c is cardinality of continuum. Hence \mathbb{R}^c is not Lindelof.
  • Show that the product space \mathbb{R}^{\omega_1} has a closed and discrete subspace of cardinality \omega_1 where \omega_1 is the first uncountable ordinal. Hence \mathbb{R}^{\omega_1} is not Lindelof.

Hints for Exercise 2.B

Let \mathbb{P} be the set of all irrational numbers. Show that \mathbb{P} as a subspace of the real line is not \sigma-compact.

Hints for Exercise 2.C

Let S be the real line with the topology generated by the half open and half closed intervals of the form [a,b)=\{ x \in \mathbb{R}: a \le x < b \}. The real line with this topology is called the Sorgenfrey line. Show that S is Lindelof and is not \sigma-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 X be a space. Let Y be a compact space. Suppose that U is an open subset of X \times Y and suppose that \{ x \} \times Y \subset U where x \in X. Then there exists an open subset V of X such that \{ x \} \times Y \subset V \times Y \subset U.

Hints for Exercise 2.E

Since the real line \mathbb{R} is homeomorphic to the open interval (0,1), \mathbb{R}^\omega is homeomorphic to (0,1)^\omega. Show that (0,1)^\omega is not \sigma-compact.

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Further Hints for Exercise 2.A

The hints here focus on the example \mathbb{R}^c.

Let I=[0,1]. Let \omega be the first infinite ordinal. For convenience, consider \omega the set \{ 0,1,2,3,\cdots \}, the set of all non-negative integers. Since \omega^I is a closed subset of \mathbb{R}^I, any closed and discrete subset of \omega^I is a closed and discrete subset of \mathbb{R}^I. The task at hand is to find a closed and discrete subset of Y=\omega^I. To this end, we define W=\{W_x: x \in I  \} after setting up background information.

For each t \in I, choose a sequence O_{t,1},O_{t,2},O_{t,3},\cdots of open intervals (in the usual topology of I) such that

  • \{ t \}=\bigcap_{j=1}^\infty O_{t,j},
  • \overline{O_{t,j+1}} \subset O_{t,j} for each j (the closure is in the usual topology of I).

Note. For each t \in I-\{0,1 \}, the open intervals O_{t,j} are of the form (a,b). For t=0, the open intervals O_{t,j} are of the form [0,b). For t=1, the open intervals O_{t,j} are of the form (a,1].

For each t \in I, define the map f_t: I \rightarrow \omega as follows:

    f_t(x) = \begin{cases} 0 & \ \ \ \mbox{if } x=t \\ 1 & \ \ \ \mbox{if } x \in I-O_{t,1} \\ 2 & \ \ \ \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \ \ \ \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \\ j & \ \ \ \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \end{cases}

We are now ready to define W=\{W_x: x \in I  \}. For each x \in I, W_x is the mapping W_x:I \rightarrow \omega defined by W_x(t)=f_t(x) for each t \in I.

Show the following:

  • The set W=\{W_x: x \in I  \} has cardinality continuum.
  • The set W is a discrete space.
  • The set W is a closed subspace of Y.

Further Hints for Exercise 2.B

A subset A of the real line \mathbb{R} is nowhere dense in \mathbb{R} if for any nonempty open subset U of \mathbb{R}, there is a nonempty open subset V of U such that V \cap A=\varnothing. If we replace open sets by open intervals, we have the same notion.

Show that the real line \mathbb{R} 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 X and Y are \sigma-compact, then the product X \times Y is \sigma-compact, hence Lindelof.

Prove that S, the Sorgenfrey line, is Lindelof while its square S \times S is not Lindelof.

Further Hints for Exercise 2.D

As suggested in the hints given earlier, prove that X \times Y is Lindelof if X is Lindelof and Y is compact. As suggested, the Tube lemma is a useful tool.

Further Hints for Exercise 2.E

The product space (0,1)^\omega is a subspace of the product space [0,1]^\omega. Since [0,1]^\omega is compact, we can fall back on a Baire category theorem argument to show why (0,1)^\omega cannot be \sigma-compact. To this end, we consider the notion of Baire space. A space X is said to be a Baire space if for each countable family \{ U_1,U_2,U_3,\cdots \} of open and dense subsets of X, the intersection \bigcap_{i=1}^\infty U_i is a dense subset of X. Prove the following results.

Fact E.1
Let X be a compact Hausdorff space. Let O_1,O_2,O_3,\cdots be a sequence of non-empty open subsets of X such that \overline{O_{n+1}} \subset O_n for each n. Then the intersection \bigcap_{i=1}^\infty O_i is non-empty.

Fact E.2
Any compact Hausdorff space is Baire space.

Fact E.3
Let X be a Baire space. Let Y be a dense G_\delta-subset of X such that X-Y is a dense subset of X. Then Y is not a \sigma-compact space.

Since X=[0,1]^\omega is compact, it follows from Fact E.2 that the product space X=[0,1]^\omega is a Baire space.

Fact E.4
Let X=[0,1]^\omega and Y=(0,1)^\omega. The product space Y=(0,1)^\omega is a dense G_\delta-subset of X=[0,1]^\omega. Furthermore, X-Y is a dense subset of X.

It follows from the above facts that the product space (0,1)^\omega cannot be a \sigma-compact space.

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Proofs of Key Steps for Exercise 2.A

The proof here focuses on the example \mathbb{R}^c.

To see that W=\{W_x: x \in I  \} has the same cardinality as that of I, show that W_x \ne W_y for x \ne y. This follows from the definition of the mapping W_x.

To see that W is discrete, for each x \in I, consider the open set U_x=\{ b \in Y: b(x)=0 \}. Note that W_x \in U_x. Further note that W_y \notin U_x for all y \ne x.

To see that W is a closed subset of Y, let k: I \rightarrow \omega such that k \notin W. Consider two cases.

Case 1. k(r) \ne 0 for all r \in I.
Note that \{ O_{t,k(t)}: t \in I \} is an open cover of I (in the usual topology). There exists a finite H \subset I such that \{ O_{h,k(h)}: h \in H \} is a cover of I. Consider the open set G=\{ b \in Y: \forall \ h \in H, \ b(h)=k(h) \}. Define the set F as follows:

    F=\{ c \in I: W_c \in G \}

The set F can be further described as follows:

    \displaystyle \begin{aligned} F&=\{ c \in I: W_c \in G \} \\&=\{ c \in I: \forall \ h \in H, \ W_c(h)=f_h(c)=k(h) \ne 0 \} \\&=\{ c \in I: \forall \ h \in H, \ c \in I-O_{h,k(h)} \}  \\&=\bigcap_{h \in H} (I-O_{h,k(h)}) \\&=I-\bigcup_{h \in H} O_{h,k(h)}=I-I =\varnothing \end{aligned}

The last step is \varnothing because \{ O_{h,k(h)}: h \in H \} is a cover of I. The fact that F=\varnothing means that G is an open subset of Y containing the point k such that G contains no point of W.

Case 2. k(r) = 0 for some r \in I.
Since k \notin W, k \ne W_x for all x \in I. In particular, k \ne W_r. This means that k(t) \ne W_r(t) for some t \in I. Define the open set G as follows:

    G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}

Clearly k \in G. Observe that W_r \notin G since W_r(t) \ne k(t). For each p \in I-\{ r \}, W_p \notin G since W_p(r) \ne 0. Thus G is an open set containing k such that G \cap W=\varnothing.

Both cases show that W is a closed subset of Y=\omega^I.

Proofs of Key Steps for Exercise 2.B

Suppose that \mathbb{P}, the set of all irrational numbers, is \sigma-compact. That is, \mathbb{P}=A_1 \cup A_2 \cup A_3 \cup \cdots where each A_i is a compact space as a subspace of \mathbb{P}. Any compact subspace of \mathbb{P} is also a compact subspace of \mathbb{R}. As a result, each A_i is a closed subset of \mathbb{R}. Furthermore, prove the following:

    Each A_i is a nowhere dense subset of \mathbb{R}.

Each singleton set \{ r \} where r is any rational number is also a closed and nowhere dense subset of \mathbb{R}. This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus \mathbb{P} cannot be \sigma-compact.

Proofs of Key Steps for Exercise 2.C

The Sorgenfrey line S is a Lindelof space whose square S \times S 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 non-normality of S \times S uses the Baire category theorem and is found here.

If the Sorgenfrey line is \sigma-compact, then S \times S would be \sigma-compact and hence Lindelof. Thus S cannot be \sigma-compact.

Proofs of Key Steps for Exercise 2.D

Suppose that X is Lindelof and that Y is compact. Let \mathcal{U} be an open cover of X \times Y. For each x \in X, let \mathcal{U}_x \subset \mathcal{U} be finite such that \mathcal{U}_x is a cover of \{ x \} \times Y. Putting it another way, \{ x \} \times Y \subset \cup \mathcal{U}_x. By the Tube lemma, for each x \in X, there is an open O_x such that \{ x \} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x. Since X is Lindelof, there exists a countable set \{ x_1,x_2,x_3,\cdots \} \subset X such that \{ O_{x_1},O_{x_2},O_{x_3},\cdots \} is a cover of X. Then \mathcal{U}_{x_1} \cup \mathcal{U}_{x_2} \cup \mathcal{U}_{x_3} \cup \cdots is a countable subcover of \mathcal{U}. This completes the proof that X \times Y is Lindelof when X is Lindelof and Y is compact.

To complete the exercise, observe that if X is Lindelof and Y is \sigma-compact, then X \times Y is the union of countably many Lindelof subspaces.

Proofs of Key Steps for Exercise 2.E

Proof of Fact E.1
Let X be a compact Hausdorff space. Let O_1,O_2,O_3,\cdots be a sequence of non-empty open subsets of X such that $latex \overline{O_{n+1}} \subset O_n for each n. Show that the intersection \bigcap_{i=1}^\infty O_i is non-empty.

Suppose that \bigcap_{i=1}^\infty O_i=\varnothing. Choose x_1 \in O_1. There must exist some n_1 such that x_1 \notin O_{n_1}. Choose x_2 \in O_{n_1}. There must exist some n_2>n_1 such that x_2 \notin O_{n_2}. Continue in this manner we can choose inductively an infinite set A=\{ x_1,x_2,x_3,\cdots \} \subset X such that x_i \ne x_j for i \ne j. Since X is compact, the infinite set A has a limit point p. This means that every open set containing p contains some x_j (in fact for infinitely many j). The point p cannot be in the intersection \bigcap_{i=1}^\infty O_i. Thus for some n, p \notin O_n. Thus p \notin \overline{O_{n+1}}. We can choose an open set U such that p \in U and U \cap \overline{O_{n+1}}=\varnothing. However, U must contain some point x_j where j>n+1. This is a contradiction since O_j \subset \overline{O_{n+1}} for all j>n+1. Thus Fact E.1 is established.

Proof of Fact E.2
Let X be a compact space. Let U_1,U_2,U_3,\cdots be open subsets of X such that each U_i is also a dense subset of X. Let V a non-empty open subset of X. We wish to show that V contains a point that belongs to each U_i. Since U_1 is dense in X, O_1=V \cap U_1 is non-empty. Since U_2 is dense in X, choose non-empty open O_2 such that \overline{O_2} \subset O_1 and O_2 \subset U_2. Since U_3 is dense in X, choose non-empty open O_3 such that \overline{O_3} \subset O_2 and O_3 \subset U_3. Continue inductively in this manner and we have a sequence of open sets O_1,O_2,O_3,\cdots just like in Fact E.1. Then the intersection of the open sets O_n is non-empty. Points in the intersection are in V and in all the U_n. This completes the proof of Fact E.2.

Proof of Fact E.3
Let X be a Baire space. Let Y be a dense G_\delta-subset of X such that X-Y is a dense subset of X. Show that Y is not a \sigma-compact space.

Suppose Y is \sigma-compact. Let Y=\bigcup_{n=1}^\infty B_n where each B_n is compact. Each B_n is obviously a closed subset of X. We claim that each B_n is a closed nowhere dense subset of X. To see this, let U be a non-empty open subset of X. Since X-Y is dense in X, U contains a point p where p \notin Y. Since p \notin B_n, there exists a non-empty open V \subset U such that V \cap B_n=\varnothing. This shows that each B_n is a nowhere dense subset of X.

Since Y is a dense G_\delta-subset of X, Y=\bigcap_{n=1}^\infty O_n where each O_n is an open and dense subset of X. Then each A_n=X-O_n is a closed nowhere dense subset of X. This means that X is the union of countably many closed and nowhere dense subsets of X. More specifically, we have the following.

(1)………X= \biggl( \bigcup_{n=1}^\infty A_n \biggr) \cup \biggl( \bigcup_{n=1}^\infty B_n \biggr)

Statement (1) contradicts the fact that X is a Baire space. Note that all X-A_n and X-B_n are open and dense subsets of X. Further note that the intersection of all these countably many open and dense subsets of X is empty according to (1). Thus Y cannot not a \sigma-compact space.

Proof of Fact E.4
The space X=[0,1]^\omega is compact since it is a product of compact spaces. To see that Y=(0,1)^\omega is a dense G_\delta-subset of X, note that Y=\bigcap_{n=1}^\infty U_n where for each integer n \ge 1

(2)………U_n=(0,1) \times \cdots \times (0,1) \times [0,1] \times [0,1] \times \cdots

Note that the first n factors of U_n are the open interval (0,1) and the remaining factors are the closed interval [0,1]. It is also clear that X-Y is a dense subset of X. This completes the proof of Fact E.4.

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Concluding Remarks

Exercise 2.A
The exercise is to show that the product of uncountably many \sigma-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that \mathbb{R}^{c} has uncountable extent where c 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 \mathbb{R}^{\omega_1} is discussed in this post.

For \mathbb{R}^{c} and \mathbb{R}^{\omega_1}, there is another way to show non-Lindelof. 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 \omega^{\omega_1} as a closed subspace. The non-normality of \omega^{\omega_1} is discussed here.

Exercise 2.B
The hints given above is to show that the set of all irrational numbers, \mathbb{P}, is not \sigma-compact (as a subspace of the real line). The same argument showing that \mathbb{P} is not \sigma-compact can be generalized. Note that the complement of \mathbb{P} is \mathbb{Q}, the set of all rational numbers (a countable set). In this case, \mathbb{Q} 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 B, the complement of a set A, 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 A cannot be a \sigma-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 \sigma-compact. In this case, the suggested space is \mathbb{R}^{\omega}, 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 \mathbb{R}^{\omega} is embedded in the compact space [0,1]^{\omega}, 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 \omega^{\omega}, the product of countably many copies of the countably infinite discrete space, is not \sigma-compact. The space \omega of the non-negative integers, as a subspace of the real line, is certainly \sigma-compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not \sigma-compact. With the product space \omega^{\omega}, there is a connection with Exercise 2.B. The product \omega^{\omega} is homeomorphic to \mathbb{P}. The idea of the homeomorphism is discussed here. Thus the non-\sigma-compactness of \omega^{\omega} 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 non-\sigma-compactness of the Sorgenfrey line S can be achieved by a Baire category argument. The non-normality of the Sorgenfrey plane S \times S can be achieved by Jones’ lemma argument or by the fact that \mathbb{P} 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.

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