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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Lindelof Exercise 1

A space X is called a \sigma-compact space if it is the union of countably many compact subspaces. Clearly, any \sigma-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. The most well known example is perhaps the square of the Sorgenfrey line. In certain cases, the Lindelof property can be productive. For example, the product of countably many \sigma-compact spaces is a Lindelof space. The discussion here centers on the following theorem.

Theorem 1
Let X_1,X_2,X_3,\cdots be \sigma-compact spaces. Then the product space \prod_{i=1}^\infty X_i is Lindelof.

Theorem 1 is Exercise 3.8G in page 195 of General Topology by Engelking [1]. The reference for Exercise 3.8G is [2]. But the theorem is not found in [2] (it is not stated directly and it does not seem to be an obvious corollary of a theorem discussed in that paper). However, a hint is provided in Engelking for Exercise 3.8G. In this post, we discuss Theorem 1 as an exercise by giving expanded hint. Solutions to some of the key steps in the expanded hint are given at the end of the post.

Expanded Hint

It is helpful to first prove the following theorem.

Theorem 2
For each integer i \ge 1, let C_{i,1},C_{i,2},\cdots be compact spaces and let C_i be the topological sum:

    C_i=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots=\oplus_{j=1}^\infty C_{i,j}

Then the product \prod_{i=1}^\infty C_i is Lindelof.

Note that in the topological sum C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots, the spaces C_{i,1},C_{i,2},C_{i,3},\cdots are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way to view this topology: each of the C_{i,j} is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many \sigma-compact spaces is Lindelof if each \sigma-compact space is the union of countably many disjoint compact spaces. The hint for Exercise 3.8G can be applied much more naturally on Theorem 2 than on Theorem 1. The following is Exercise 3.8F (a), which is the hint for Exercise 3.8G.

Lemma 3
Let Z be a compact space. Let X be a subspace of Z. Suppose that there exist F_1,F_2,F_3,\cdots, closed subsets of Z, such that for all x and y where x \in X and y \in Z-X, there exists F_i such that x \in F_i and y \notin F_i. Then X is a Lindelof space.

The following theorem connects the hint (Lemma 3) with Theorem 2.

Theorem 4
For each integer i \ge 1, let Z_i be the one-point compactification of C_i in Theorem 2. Then the product Z=\prod_{i=1}^\infty Z_i is a compact space. Furthermore, X=\prod_{i=1}^\infty C_i is a subspace of Z. Prove that Z and X satisfy Lemma 3.

Each C_i in Theorem 2 is a locally compact space. To define the one-point compactifications, for each i, choose p_i \notin C_i. Make sure that p_i \ne p_j for i \ne j. Then Z_i is simply

    Z_i=C_i \cup \{ p_i \}=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots \cup \{ p_i \}

with the topology defined as follows:

  • Open subsets of C_i continue to be open in Z_i.
  • An open set containing p_i is of the form \{ p_i \} \cup (C_i - \overline{D}) where D is open in C_i and D is contained in the union of finitely many C_{i,j}.

For convenience, each point p_i is called a point at infinity.

Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem 2 to the product space in Theorem 1.

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Proofs of Key Steps

Proof of Lemma 3
Let Z, X and F_1,F_2,F_3,\cdots be as described in the statement for Lemma 3. Let \mathcal{U} be a collection of open subsets of Z such that \mathcal{U} covers X. We would like to show that a countable subcollection of \mathcal{U} is also a cover of X. Let O=\cup \mathcal{U}. If Z-O=\varnothing, then \mathcal{U} is an open cover of Z and there is a finite subset of \mathcal{U} that is a cover of Z and thus a cover of X. Thus we can assume that Z-O \ne \varnothing.

Let F=\{ F_1,F_2,F_3,\cdots \}. Let K=Z-O, which is compact. We make the following claim.

Claim. Let Y be the union of all possible \cap G where G \subset F is finite and \cap G \subset O. Then X \subset Y \subset O.

To establish the claim, let x \in X. For each y \in K=Z-O, there exists F_{n(y)} such that x \in F_{n(y)} and y \notin F_{n(y)}. This means that \{ Z-F_{n(y)}: y \in K \} is an open cover of K. By the compactness of K, there are finitely many F_{n(y_1)}, \cdots, F_{n(y_k)} such that F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} misses K, or equivalently F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset O. Note that x \in F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}. Further note that F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset Y. This establishes the claim that X \subset Y. The claim that Y \subset O is clear from the definition of Y.

Each set F_i is compact since it is closed in Z. The intersection of finitely many F_i is also compact. Thus the \cap G in the definition of Y in the above claim is compact. There can be only countably many \cap G in the definition of Y. Thus Y is a \sigma-compact space that is covered by the open cover \mathcal{U}. Choose a countable \mathcal{V} \subset \mathcal{U} such that \mathcal{V} covers Y. Then \mathcal{V} is a cover of X too. This completes the proof that X is Lindelof.

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Proof of Theorem 4
Recall that Z=\prod_{i=1}^\infty Z_i and that X=\prod_{i=1}^\infty C_i. Each Z_i is the one-point compactification of C_i, which is the topological sum of the disjoint compact spaces C_{i,1},C_{i,2},\cdots.

For integers i,j \ge 1, define K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}. For integers n,j \ge 1, define the product F_{n,j} as follows:

    F_{n,j}=K_{1,j} \times \cdots \times K_{n,j} \times Z_{n+1} \times Z_{n+2} \times \cdots

Since F_{n,j} is a product of compact spaces, F_{n,j} is compact and thus closed in Z. There are only countably many F_{n,j}.

We claim that the countably many F_{n,j} have the property indicated in Lemma 3. To this end, let f \in X=\prod_{i=1}^\infty C_i and g \in Z-X. There exists an integer n \ge 1 such that g(n) \notin C_{n}. This means that g(n) \notin C_{n,j} for all j, i.e. g(n)=p_n (so g(n) must be the point at infinity). Choose j \ge 1 large enough such that

    f(i) \in K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}

for all i \le n. It follows that f \in F_{n,j} and g \notin F_{n,j}. Thus the sequence of closed sets F_{n,j} satisfies Lemma 3. By Lemma 3, X=\prod_{i=1}^\infty C_i is Lindelof.

Reference

  1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
  2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.

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A little corner in the world of set-theoretic topology

This post puts a spot light on a little corner in the world of set-theoretic topology. There lies in this corner a simple topological statement that opens a door to the esoteric world of independence results. In this post, we give a proof of this basic fact and discuss its ramifications. This basic result is an excellent entry point to the study of S and L spaces.

The following paragraph is found in the paper called Gently killing S-spaces by Todd Eisworth, Peter Nyikos and Saharon Shelah [1]. The basic fact in question is highlighted in blue.

A simultaneous generalization of hereditarily separable and hereditarily Lindelof spaces is the class of spaces of countable spread – those spaces in which every discrete subspace is countable. One of the basic facts in this little corner of set-theoretic topology is that if a regular space of countable spread is not hereditarily separable, it contains an L-space, and if it is not hereditarily Lindelof, it contains an S-space. [1]

The same basic fact is also mentioned in the paper called The spread of regular spaces by Judith Roitman [2].

It is also well known that a regular space of countable spread which is not hereditarily separable contains an L-space and a regular space of countable spread which is not hereditarily Lindelof contains an S-space. Thus an absolute example of a space satisfying (Statement) A would contain a proof of the existence of S and L space – a consummation which some may devoutly wish, but which this paper does not attempt. [2]

Statement A in [2] is: There exists a 0-dimensional Hausdorff space of countable spread that is not the union of a hereditarily separable and a hereditarily Lindelof space. Statement A would mean the existence of a regular space of countable spread that is not hereditarily separable and that is also not hereditarily Lindelof. By the well known fact just mentioned, statement A would imply the existence of a space that is simultaneously an S-space and an L-space!

Let’s unpack the preceding section. First some basic definitions. A space X is of countable spread (has countable spread) if every discrete subspace of X is countable. A space X is hereditarily separable if every subspace of X is separable. A space X is hereditarily Lindelof if every subspace of X is Lindelof. A space is an S-space if it is hereditarily separable but not Lindelof. A space is an L-space if it is hereditarily Lindelof but not separable. See [3] for a basic discussion of S and L spaces.

Hereditarily separable but not Lindelof spaces as well as hereditarily Lindelof but not separable spaces can be easily defined in ZFC [3]. However, such examples are not regular. For the notions of S and L-spaces to be interesting, the definitions must include regularity. Thus in the discussion that follows, all spaces are assumed to be Hausdorff and regular.

One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions that are not far from the definition of topological spaces. The same can be said about hereditarily separable and hereditarily Lindelof spaces. Out of these basic ingredients come the notion of S-spaces and L-spaces, the existence of which is one of the key motivating questions in set-theoretic topology in the twentieth century. The study of S and L-spaces is a body of mathematics that had been developed for nearly a century. It is a fruitful area of research at the boundary of topology and axiomatic set theory.

The existence of an S-space is independent of ZFC (as a result of the work by Todorcevic in early 1980s). This means that there is a model of set theory in which an S-space exists and there is also a model of set theory in which S-spaces cannot exist. One half of the basic result mentioned in the preceding section is intimately tied to the existence of S-spaces and thus has interesting set-theoretic implications. The other half of the basic result involves the existence of L-spaces, which are shown to exist without using extra set theory axioms beyond ZFC by Justin Moore in 2005, which went against the common expectation that the existence of L-spaces would be independent of ZFC as well.

Let’s examine the basic notions in a little more details. The following diagram shows the properties surrounding the notion of countable spread.

Diagram 1 – Properties surrounding countable spread

The implications (the arrows) in Diagram 1 can be verified easily. Central to the discussion at hand, both hereditarily separable and hereditarily Lindelof imply countable spread. The best way to see this is that if a space has an uncountable discrete subspace, that subspace is simultaneously a non-separable subspace and a non-Lindelof subspace. A natural question is whether these implications can be reversed. Another question is whether the properties in Diagram 1 can be related in other ways. The following diagram attempts to ask these questions.

Diagram 2 – Reverse implications surrounding countable spread

Not shown in Diagram 2 are these four facts: separable \not \rightarrow hereditarily separable, Lindelof \not \rightarrow hereditarily Lindelof, separable \not \rightarrow countable spread and Lindelof \not \rightarrow countable spread. The examples supporting these facts are not set-theoretic in nature and are not discussed here.

Let’s focus on each question mark in Diagram 2. The two horizontal arrows with question marks at the top are about S-space and L-space. If X is hereditarily separable, then is X hereditarily Lindelof? A “no” answer would mean there is an S-space. A “yes” answer would mean there exists no S-space. So the top arrow from left to right is independent of ZFC. Since an L-space can be constructed within ZFC, the question mark in the top arrow in Diagram 2 from right to left has a “no” answer.

Now focus on the arrows emanating from countable spread in Diagram 2. These arrows are about the basic fact discussed earlier. From Diagram 1, we know that hereditarily separable implies countable spread. Can the implication be reversed? Any L-space would be an example showing that the implication cannot be reversed. Note that any L-space is of countable spread and is not separable and hence not hereditarily separable. Since L-space exists in ZFC, the question mark in the arrow from countable spread to hereditarily separable has a “no” answer. The same is true for the question mark in the arrow from countable spread to separable

We know that hereditarily Lindelof implies countable spread. Can the implication be reversed? According to the basic fact mentioned earlier, if the implication cannot be reversed, there exists an S-space. Thus if S-space does not exist, the implication can be reversed. Any S-space is an example showing that the implication cannot be reversed. Thus the question in the arrow from countable spread to hereditarily Lindelof cannot be answered without assuming axioms beyond ZFC. The same is true for the question mark for the arrow from countable spread to Lindelf.

Diagram 2 is set-theoretic in nature. The diagram is remarkable in that the properties in the diagram are basic notions that are only brief steps away from the definition of a topological space. Thus the basic highlighted here is a quick route to the world of independence results.

We now give a proof of the basic result, which is stated in the following theorem.

Theorem 1
Let X is regular and Hausdorff space. Then the following is true.

  • If X is of countable spread and is not a hereditarily separable space, then X contains an L-space.
  • If X is of countable spread and is not a hereditarily Lindelof space, then X contains an S-space.

To that end, we use the concepts of right separated space and left separated space. Recall that an initial segment of a well-ordered set (X,<) is a set of the form \{y \in X: y<x \} where x \in X. A space X is a right separated space if X can be well-ordered in such a way that every initial segment is open. A right separated space is in type \kappa if the well-ordering is of type \kappa. A space X is a left separated space if X can be well-ordered in such a way that every initial segment is closed. A left separated space is in type \kappa if the well-ordering is of type \kappa. The following results are used in proving Theorem 1.

Theorem A
Let X is regular and Hausdorff space. Then the following is true.

  • The space X is hereditarily separable space if and only if X has no uncountable left separated subspace.
  • The space X is hereditarily Lindelof space if and only if X has no uncountable right separated subspace.

Proof of Theorem A
\Longrightarrow of the first bullet point.
Suppose Y \subset X is an uncountable left separated subspace. Suppose that the well-ordering of Y is of type \kappa where \kappa>\omega. Further suppose that Y=\{ x_\alpha: \alpha<\kappa \} such that for each \alpha<\kappa, C_\alpha=\{ x_\beta: \beta<\alpha \} is a closed subset of Y. Since \kappa is uncountable, the well-ordering has an initial segment of type \omega_1. So we might as well assume \kappa=\omega_1. Note that for any countable A \subset Y, A \subset C_\alpha for some \alpha<\omega_1. It follows that Y is not separable. This means that X is not hereditarily separable.

\Longleftarrow of the first bullet point.
Suppose that X is not hereditarily separable. Let Y \subset X be a subspace that is not separable. We now inductively derive an uncountable left separated subspace of Y. Choose y_0 \in Y. For each \alpha<\omega_1, let A_\alpha=\{ y_\beta \in Y: \beta <\alpha \}. The set A_\alpha is the set of all the points of Y chosen before the step at \alpha<\omega_1. Since A_\alpha is countable, its closure in Y is not the entire space Y. Choose y_\alpha \in Y-\overline{A_\alpha}=O_\alpha.

Let Y_L=\{ y_\alpha: \alpha<\omega_1 \}. We claim that Y_L is a left separated space. To this end, we need to show that each initial segment A_\alpha is a closed subset of Y_L. Note that for each \gamma \ge \alpha, O_\gamma=Y-\overline{A_\gamma} is an open subset of Y with y_\gamma \in O_\gamma such that O_\gamma \cap \overline{A_\gamma}=\varnothing and thus O_\gamma \cap \overline{A_\alpha}=\varnothing (closure in Y). Then U_\gamma=O_\gamma \cap Y_L is an open subset of Y_L containing y_\gamma such that U_\gamma \cap A_\alpha=\varnothing. It follows that Y-A_\alpha is open in Y_L and that A_\alpha is a closed subset of Y_L.

\Longrightarrow of the second bullet point.
Suppose Y \subset X is an uncountable right separated subspace. Suppose that the well-ordering of Y is of type \kappa where \kappa>\omega. Further suppose that Y=\{ x_\alpha: \alpha<\kappa \} such that for each \alpha<\kappa, U_\alpha=\{ x_\beta: \beta<\alpha \} is an open subset of Y.

Since \kappa is uncountable, the well-ordering has an initial segment of type \omega_1. So we might as well assume \kappa=\omega_1. Note that \{ U_\alpha: \alpha<\omega_1 \} is an open cover of Y that has no countable subcover. It follows that Y is not Lindelof. This means that X is not hereditarily Lindelof.

\Longleftarrow of the second bullet point.
Suppose that X is not hereditarily Lindelof. Let Y \subset X be a subspace that is not Lindelof. Let \mathcal{U} be an open cover of Y that has no countable subcover. We now inductively derive a right separated subspace of Y of type \omega_1.

Choose U_0 \in \mathcal{U} and choose y_0 \in U_0. Choose y_1 \in Y-U_0 and choose U_1 \in \mathcal{U} such that y_1 \in U_1. Let \alpha<\omega_1. Suppose that points y_\beta and open sets U_\beta, \beta<\alpha, have been chosen such that y_\beta \in Y-\bigcup_{\delta<\beta} U_\delta and y_\beta \in U_\beta. The countably many chosen open sets U_\beta, \beta<\alpha, cannot cover Y. Choose y_\alpha \in Y-\bigcup_{\beta<\alpha} U_\beta. Choose U_\alpha \in \mathcal{U} such that y_\alpha \in U_\alpha.

Let Y_R=\{ y_\alpha: \alpha<\omega_1 \}. It follows that Y_R is a right separated space. Note that for each \alpha<\omega_1, \{ y_\beta: \beta<\alpha \} \subset \bigcup_{\beta<\alpha} U_\beta and the open set \bigcup_{\beta<\alpha} U_\beta does not contain y_\gamma for any \gamma \ge \alpha. This means that the initial segment \{ y_\beta: \beta<\alpha \} is open in Y_L. \square

Lemma B
Let X be a space that is a right separated space and also a left separated space based on the same well ordering. Then X is a discrete space.

Proof of Lemma B
Let X=\{ w_\alpha: \alpha<\kappa \} such that the well-ordering is given by the ordinals in the subscripts, i.e. w_\beta<w_\gamma if and only if \beta<\gamma. Suppose that X with this well-ordering is both a right separated space and a left separated space. We claim that every point is a discrete point, i.e. \{ x_\alpha \} is open for any \alpha<\kappa.

To see this, fix \alpha<\kappa. The initial segment A_\alpha=\{ w_\beta: \beta<\alpha \} is closed in X since X is a left separated space. On the other hand, the initial segment \{ w_\beta: \beta < \alpha+1  \} is open in X since X is a right separated space. Then B_{\alpha}=\{ w_\beta: \beta \ge \alpha+1  \} is closed in X. It follows that \{ x_\alpha \} must be open since X=A_\alpha \cup B_\alpha \cup \{ w_\alpha \}. \square

Theorem C
Let X is regular and Hausdorff space. Then the following is true.

  • Suppose the space X is right separated space of type \omega_1. Then if X has no uncountable discrete subspaces, then X is an S-space or X contains an S-space.
  • Suppose the space X is left separated space of type \omega_1. Then if X has no uncountable discrete subspaces, then X is an L-space or X contains an L-space.

Proof of Theorem C
For the first bullet point, suppose the space X is right separated space of type \omega_1. Then by Theorem A, X is not hereditarily Lindelof. If X is hereditarily separable, then X is an S-space (if X is not Lindelof) or X contains an S-space (a non-Lindelof subspace of X). Suppose X is not hereditarily separable. By Theorem A, X has an uncountable left separated subspace of type \omega_1.

Let X=\{ x_\alpha: \alpha<\omega_1 \} such that the well-ordering represented by the ordinals in the subscripts is a right separated space. Let <_R be the symbol for the right separated well-ordering, i.e. x_\beta <_R \ x_\delta if and only if \beta<\delta. As indicated in the preceding paragraph, X has an uncountable left separated subspace. Let Y=\{ y_\alpha \in X: \alpha<\omega_1 \} be this left separated subspace. Let <_L be the symbol for the left separated well-ordering. The well-ordering <_R may be different from the well-ordering <_L. However, we can obtain an uncountable subset of Y such that the two well-orderings coincide on this subset.

To start, pick any y_\gamma in Y and relabel it t_0. The final segment \{y_\beta \in Y: t_0 <_L \ y_\beta \} must intersect the final segment \{x_\beta \in X: t_0 <_R \ x_\beta \} in uncountably many points. Choose the least such point (according to <_R) and call it t_1. It is clear how t_{\delta+1} is chosen if t_\delta has been chosen.

Suppose \alpha<\omega_1 is a limit ordinal and that t_\beta has been chosen for all \beta<\alpha. Then the set \{y_\tau: \forall \ \beta<\alpha, t_\beta <_L \ y_\tau \} and the set \{x_\tau: \forall \ \beta<\alpha, t_\beta <_R \ x_\tau \} must intersect in uncountably many points. Choose the least such point and call it t_\alpha (according to <_R). As a result, we have obtained T=\{ t_\alpha: \alpha<\omega_1 \}. It follows that T with the well-ordering represented by the ordinals in the subscript is a subset of (X,<_R) and a subset of (Y,<_L). Thus T is both right separated and left separated.

By Lemma B, T is a discrete subspace of X. However, X is assumed to have no uncountable discrete subspace. Thus if X has no uncountable discrete subspace, then X must be hereditarily separable and as a result, must be an S-space or must contain an S-space.

The proof for the second bullet point is analogous to that of the first bullet point. \square

We are now ready to prove Theorem 1.

Proof of Theorem 1
Suppose that X is of countable spread and that X is not hereditarily separable. By Theorem A, X has an uncountable left separated subspace Y (assume it is of type \omega_1). The property of countable spread is hereditary. So Y is of countable spread. By Theorem C, Y is an L-space or Y contains an L-space. In either way, X contains an L-space.

Suppose that X is of countable spread and that X is not hereditarily Lindelof. By Theorem A, X has an uncountable right separated subspace Y (assume it is of type \omega_1). By Theorem C, Y is an S-space or Y contains an S-space. In either way, X contains an S-space.

Reference

  1. Eisworth T., Nyikos P., Shelah S., Gently killing S-spaces, Israel Journal of Mathmatics, 136, 189-220, 2003.
  2. Roitman J., The spread of regular spaces, General Topology and Its Applications, 8, 85-91, 1978.
  3. Roitman, J., Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 295-326, 1984.
  4. Tatch-Moore J., A solution to the L space problem, Journal of the American Mathematical Society, 19, 717-736, 2006.

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Dan Ma math

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Every space is star discrete

The statement in the title is a folklore fact, though the term star discrete is usually not used whenever this well known fact is invoked in the literature. We present a proof to this well known fact. We also discuss some related concepts.

All spaces are assumed to be Hausdorff and regular.

First, let’s define the star notation. Let X be a space. Let \mathcal{U} be a collection of subsets of X. Let A \subset X. Define \text{St}(A,\mathcal{U}) to be the set \bigcup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}. In other words, the set \text{St}(A,\mathcal{U}) is simply the union of all elements of \mathcal{U} that contains points of the set A. The set \text{St}(A,\mathcal{U}) is also called the star of the set A with respect to the collection \mathcal{U}. If A=\{ x \}, we use the notation \text{St}(x,\mathcal{U}) instead of \text{St}( \{ x \},\mathcal{U}). The following is the well known result in question.

Lemma 1
Let X be a space. For any open cover \mathcal{U} of X, there exists a discrete subspace A of X such that X=\text{St}(A,\mathcal{U}). Furthermore, the set A can be chosen in such a way that it is also a closed subset of the space X.

Any space that satisfies the condition in Lemma 1 is said to be a star discrete space. The proof shown below will work for any topological space. Hence every space is star discrete. We come across three references in which the lemma is stated or is used – Lemma IV.2.20 in page 135 of [3], page 137 of [2] and [1]. The first two references do not use the term star discrete. Star discrete is mentioned in [1] since that paper focuses on star properties. This property that is present in every topological space is at heart a covering property. Here’s a rewording of the lemma that makes it look like a covering property.

Lemma 1a
Let X be a space. For any open cover \mathcal{U} of X, there exists a discrete subspace A of X such that \{ \text{St}(x,\mathcal{U}): x \in A \} is a cover of X. Furthermore, the set A can be chosen in such a way that it is also a closed subset of the space X.

Lemma 1a is clearly identical to Lemma 1. However, Lemma 1a makes it extra clear that this is a covering property. For every open cover of a space, instead of finding a sub cover or an open refinement, we find a discrete subspace so that the stars of the points of the discrete subspace with respect to the given open cover also cover the space.

Lemma 1a naturally leads to other star covering properties. For example, a space X is said to be a star countable space if for any open cover \mathcal{U} of X, there exists a countable subspace A of X such that \{ \text{St}(x,\mathcal{U}): x \in A \} is a cover of X. A space X is said to be a star Lindelof space if for any open cover \mathcal{U} of X, there exists a Lindelof subspace A of X such that \{ \text{St}(x,\mathcal{U}): x \in A \} is a cover of X. In general, for any topological property \mathcal{P}, a space X is a star \mathcal{P} space if for any open cover \mathcal{U} of X, there exists a subspace A of X with property \mathcal{P} such that \{ \text{St}(x,\mathcal{U}): x \in A \} is a cover of X.

It follows that every Lindelof space is a star countable space. It is also clear that every star countable space is a star Lindelof space.

Lemma 1 or Lemma 1a, at first glance, may seem like a surprising result. However, one can argue that it is not a strong result at all since the property is possessed by every space. Indeed, the lemma has nothing to say about the size of the discrete set. It only says that there exists a star cover based on a discrete set for a given open cover. To derive more information about the given space, we may need to work with more information on the space in question.

Consider spaces such that every discrete subspace is countable (such a space is said to have countable spread or a space of countable spread). Also consider spaces such that every closed and discrete subspace is countable (such a space is said to have countable extent or a space of countable extent). Any space that has countable spread is also a space that has countable extent for the simple reason that if every discrete subspace is countable, then every closed and discrete subspace is countable.

Then it follows from Lemma 1 that any space X that has countable extent is star countable. Any star countable space is obviously a star Lindelof space. The following diagram displays these relationships.

Countable spread and Lindelof property

According to the diagram, the star countable and star Lindelof are both downstream from the countable spread property and the Lindelof property. The star properties being downstream from the Lindelof property is not surprising. What is interesting is that if a space has countable spread, then it is star countable and hence star Lindelof.

Do “countable spread” and “Lindelof” relate to each other? Lindelof spaces do not have to have countable spread. The simplest example is the one-point compactification of an uncountable discrete space. More specifically, let X be an uncountable discrete space. Let p be a point not in X. Then Y=X \cup \{ p \} is a compact space (hence Lindelof) where X is discrete and an open neighborhood of p is of the form \{ p \} \cup U where X-U is a finite subset of X. The space Y is not of countable spread since X is an uncountable discrete subspace.

Does “countable spread” imply “Lindelof”? Is there a non-Lindelof space that has countable spread? It turns out that the answers are independent of ZFC. The next post has more details.

We now give a proof to Lemma 1. Suppose that X is an infinite space (if it is finite, the lemma is true since the space is Hausdorff). Let \kappa=\lvert X \lvert. Let \kappa^+ be the next cardinal greater than \kappa. Let \mathcal{U} be an open cover of the space X. Choose x_0 \in X. We choose a sequence of points x_0,x_1,\cdots,x_\alpha,\cdots inductively. If \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) \ne X, we can choose a point x_\alpha \in X such that x_\alpha \notin \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}).

We claim that the induction process must stop at some \alpha<\kappa^+. In other words, at some \alpha<\kappa^+, the star of the previous points must be the entire space and we run out of points to choose. Otherwise, we would have obtained a subset of X with cardinality \kappa^+, a contradiction. Choose the least \alpha<\kappa^+ such that \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) = X. Let A=\{x_\beta: \beta<\alpha \}.

Then it can be verified that the set A is a discrete subspace of X and that A is a closed subset of X. Note that x_\beta \in \text{St}(x_\beta, \mathcal{U}) while x_\gamma \notin \text{St}(x_\beta, \mathcal{U}) for all \gamma \ne \beta. This follows from the way the points are chosen in the induction process. On the other hand, for any x \in X-A, x \in \text{St}(x_\beta, \mathcal{U}) for some \beta<\alpha. As discussed, the open set \text{St}(x_\beta, \mathcal{U}) contains only one point of A, namely x_\beta.

Reference

  1. Alas O., Jumqueira L., van Mill J., Tkachuk V., Wilson R.On the extent of star countable spaces, Cent. Eur. J. Math., 9 (3), 603-615, 2011.
  2. Alster, K., Pol, R.,On function spaces of compact subspaces of \Sigma-products of the real line, Fund. Math., 107, 35-46, 1980.
  3. Arkhangelskii, A. V.,Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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Equivalent conditions for hereditarily Lindelof spaces

A topological space X is Lindelof if every open cover X has a countable subcollection that also is a cover of X. A topological space X is hereditarily Lindelof if every subspace of X, with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

Theorem 1
Let X be a topological space. The following conditions are equivalent.

  1. The space X is a hereditarily Lindelof space.
  2. Every open subspace of X is Lindelof.
  3. For every uncountable subspace Y of X, there exists a point y \in Y such that every open subset of X containing y contains uncountably many points of Y.

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence 1 \longleftrightarrow 3 is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

Proof of Theorem 1
The direction 1 \longrightarrow 2 is immediate. The direction 2 \longrightarrow 3 is straightforward.

3 \longrightarrow 1
We show \text{not } 1 \longrightarrow \text{not } 3. Suppose T is a non-Lindelof subspace of X. Let \mathcal{U} be an open cover of T such that no countable subcollection of \mathcal{U} can cover T. By a transfinite inductive process, choose a set of points \left\{t_\alpha \in T: \alpha < \omega_1 \right\} and a collection of open sets \left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\} such that for each \alpha < \omega_1, t_\alpha \in U_\alpha and t_\alpha \notin \cup \left\{U_\beta: \beta<\alpha \right\}. The inductive process is possible since no countable subcollection of \mathcal{U} can cover T. Now let Y=\left\{t_\alpha: \alpha<\omega_1 \right\}. Note that each U_\alpha can at most contain countably many points of Y, namely the points in \left\{t_\beta: \beta \le \alpha \right\}.

For each \alpha, let V_\alpha be an open subset of X such that U_\alpha=V_\alpha \cap Y. We can now conclude: for every point t_\alpha of Y, there exists an open set V_\alpha containing t_\alpha such that V_\alpha contains only countably many points of Y. This is the negation of condition 3. \blacksquare

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Remarks

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let p \in X. We say p is a limit point of the set Y \subset X if every open set containing p contains a point of Y different from p. Being a limit point of Y, we only know that each open set containing p contain infinitely many points of Y (assuming a T_1 space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if p is a limit point of Y satisfying condition 3, then p is said to be a condensation point of Y. According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set Y is a condensation point of Y.

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

Corollary 2
If the space X is hereditarily Lindelof, then every uncountable subspace Y of X contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

Theorem 3
If the space X is Lindelof, then every uncountable subspace Y of X has a limit point.

The condition “every uncountable subspace Y of X has a limit point” has another name. When a space satisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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Reference

  1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
  2. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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\copyright \ 2014 \text{ by Dan Ma}

An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space C_p(X) that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space X such that

  • C_p(X) is collectionwise normal and not paracompact,
  • C_p(X) is not Lindelof but contains a dense Lindelof subspace,
  • C_p(X) is not first countable but is a Frechet space,
  • As a corollary of the previous point, C_p(X) cannot contain a copy of the compact space \omega_1+1,
  • C_p(X) is homeomorphic to C_p(X)^\omega,
  • C_p(X) is not hereditarily normal,
  • C_p(X) is not metacompact.

A short and quick description of the space X is that X is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space C_p(X) is intimately related to a \Sigma-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the \Sigma-products of copies of real lines. Deep results about \Sigma-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space X=L_\tau. In the discussion that follows, the Greek letter \tau is always an uncountable cardinal number. Let D_\tau be a set with cardinality \tau. Let p be a point not in D_\tau. Let L_\tau=D_\tau \cup \left\{p \right\}. Consider the following topology on L_\tau:

  • Each point in D_\tau an isolated point, and
  • open neighborhoods at the point p are of the form L_\tau-K where K \subset D_\tau is countable.

It is clear that L_\tau is a Lindelof space. The Lindelof space L_\tau is sometimes called the one-point Lindelofication of the discrete space D_\tau since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space C_p(L_\tau). See this post for general information on the pointwise convergence topology of C_p(Y) for any completely regular space Y.

All the facts about C_p(X)=C_p(L_\tau) mentioned at the beginning follow from the fact that C_p(L_\tau) is homeomorphic to the \Sigma-product of \tau many copies of the real lines. Specifically, C_p(L_\tau) is homeomorphic to the following subspace of the product space \mathbb{R}^\tau.

    \Sigma_{\alpha<\tau}\mathbb{R}=\left\{ x \in \mathbb{R}^\tau: x_\alpha \ne 0 \text{ for at most countably many } \alpha<\tau \right\}

Thus understanding the function space C_p(L_\tau) is a matter of understanding a \Sigma-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of C_p(L_\tau) indicated above.

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The Homeomorphism

For each f \in C_p(L_\tau), it is easily seen that there is a countable set C \subset D_\tau such that f(p)=f(y) for all y \in D_\tau-C. Let W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}. Then each f \in W_0 has non-zero values only on a countable subset of D_\tau. Naturally, W_0 and \Sigma_{\alpha<\tau}\mathbb{R} are homeomorphic.

We claim that C_p(L_\tau) is homeomorphic to W_0 \times \mathbb{R}. For each f \in C_p(L_\tau), define h(f)=(f-f(p),f(p)). Here, f-f(p) is the function g \in C_p(L_\tau) such that g(x)=f(x)-f(p) for all x \in L_\tau. Clearly h(f) is well-defined and h(f) \in W_0 \times \mathbb{R}. It can be readily verified that h is a one-to-one map from C_p(L_\tau) onto W_0 \times \mathbb{R}. It is not difficult to verify that both h and h^{-1} are continuous.

We use the notation X_1 \cong X_2 to mean that the spaces X_1 and X_2 are homeomorphic. Then we have:

    C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R})  \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}

Thus C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}. This completes the proof that C_p(L_\tau) is topologically the \Sigma-product of \tau many copies of the real lines.

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Looking at the \Sigma-Product

Understanding the function space C_p(L_\tau) is now reduced to the problem of understanding a \Sigma-product of copies of the real lines. Most of the facts about \Sigma-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the \Sigma-product of separable metric spaces is collectionwise normal. Thus C_p(L_\tau) is collectionwise normal. The \Sigma-product of spaces, each of which has at least two points, always contains a closed copy of \omega_1 with the ordered topology (see the lemma in this previous post). Thus C_p(L_\tau) contains a closed copy of \omega_1 and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the \Sigma-product \Sigma_{\alpha<\tau}\mathbb{R}:

    \sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}

In this previous post, it is shown that \sigma_\tau is a Lindelof space. Though C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is not Lindelof, it has a dense Lindelof subspace, namely \sigma_\tau.

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A space Y is first countable if there exists a countable local base at each point y \in Y. A space Y is a Frechet space (or is Frechet-Urysohn) if for each y \in Y, if y \in \overline{A} where A \subset Y, then there exists a sequence \left\{y_n: n=1,2,3,\cdots \right\} of points of A such that the sequence converges to y. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number \tau, the product \mathbb{R}^\tau is not first countable. In fact, any dense subspace of \mathbb{R}^\tau is not first countable. In particular, the \Sigma-product \Sigma_{\alpha<\tau}\mathbb{R} is not first countable. In this previous post, it is shown that the \Sigma-product of first countable spaces is a Frechet space. Thus C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is a Frechet space.

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As a corollary of the previous point, C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space \omega_1+1 is not embeddable in C_p(L_\tau). The interest in compact subspaces of C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R} is that any compact space that is topologically embeddable in a \Sigma-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

    \Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}

Thus C_p(L_\tau) \cong C_p(L_\tau)^\omega. In particular, C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau) due to the following observation:

    C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)

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As a result of the peculiar fact that C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau), it can be concluded that C_p(L_\tau), though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if Y_1 \times Y_2 is hereditarily normal, then either Y_1 is perfectly normal or every countably infinite subset of Y_2 is closed and discrete (see this previous post). The function space C_p(L _\tau) is not perfectly normal since it contains a closed copy of \omega_1. On the other hand, there are plenty of countably infinite subsets of C_p(L _\tau) that are not closed and discrete. As a Frechet space, C_p(L _\tau) has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let \left\{x_1,x_2,x_3,\cdots \right\} be an infinite subset of D_\tau and consider the following:

    C=\left\{f_n: n=1,2,3,\cdots \right\}

where f_n is such that f_n(x_n)=n and f_n(x)=0 for each x \in L_\tau with x \ne x_n. Note that C is not closed and not discrete since the points in C converge to g \in \overline{C} where g is the zero-function. Thus C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau) is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since C_p(L_\tau) is collectionwise normal and not paracompact, C_p(L_\tau) can never be metacompact.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
  2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
  3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
  4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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\copyright \ 2014 \text{ by Dan Ma}

(Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space X=\prod_{\alpha \in A} X_\alpha. Fix a point b \in \prod_{\alpha \in A} X_\alpha, called the base point. The \Sigma-product of the spaces \left\{X_\alpha: \alpha \in A \right\} is the following subspace of the product space X:

    \Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}

In other words, the space \Sigma_{\alpha \in A} X_\alpha is the subspace of the product space X=\prod_{\alpha \in A} X_\alpha consisting of all points that deviate from the base point on at most countably many coordinates \alpha \in A. We also consider the following subspace of \Sigma_{\alpha \in A} X_\alpha.

    \sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}

For convenience , we call \Sigma_{\alpha \in A} X_\alpha the (upper case) Sigma-product (or \Sigma-product) of the spaces X_\alpha and we call the space \sigma the (lower case) sigma-product (or \sigma-product). Clearly, the space \sigma is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor X_\alpha is a separable metric space with at least two points, the \Sigma-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) \Sigma-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer n=0,1,2,3,\cdots, we define \sigma_n as follows:

    \sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}

Clearly, \sigma=\bigcup_{n=0}^\infty \sigma_n. We prove the following theorem. The fact that \sigma is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space X=\prod_{\alpha \in A} X_\alpha. We say V is a standard basic open subset of the product space X if V is of the form V=\prod_{\alpha \in A} V_\alpha such that each V_\alpha is an open subset of the factor space X_\alpha and V_\alpha=X_\alpha for all but finitely many \alpha \in A. The finite set F of all \alpha \in A such that V_\alpha \ne X_\alpha is called the support of the open set V.

Theorem 1
Let \sigma be the \sigma-product of the separable metrizable spaces \left\{X_\alpha: \alpha \in A \right\}. For each n, let \sigma_n be defined as above. The product space \sigma_n \times Y is Lindelof for each non-negative integer n and for all separable metric space Y.

Proof of Theorem 1
We prove by induction on n. Note that \sigma_0=\left\{b \right\}, the base point. Clearly \sigma_0 \times Y is Lindelof for all separable metric space Y. Suppose the theorem hold for the integer n. We show that \sigma_{n+1} \times Y for all separable metric space Y. To this end, let \mathcal{U} be an open cover of \sigma_{n+1} \times Y where Y is a separable metric space. Without loss of generality, we assume that each element of \mathcal{U} is of the form V \times W where V=\prod_{\alpha \in A} V_\alpha is a standard basic open subset of the product space X=\prod_{\alpha \in A} X_\alpha and W is an open subset of Y.

Let \mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\} be a countable subcollection of \mathcal{U} such that \mathcal{U}_0 covers \left\{b \right\} \times Y. For each j, let U_j=V_j \times W_j where V_j=\prod_{\alpha \in A} V_{j,\alpha} is a standard basic open subset of the product space X with b \in V_j and W_j is an open subset of Y. For each j, let F_j be the support of V_j. Note that \alpha \in F_j if and only if V_{j,\alpha} \ne X_\alpha. Also for each \alpha \in F_j, b_\alpha \in V_{j,\alpha}. Furthermore, for each \alpha \in F_j, let V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}. With all these notations in mind, we define the following open set for each \beta \in F_j:

    H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j

Observe that for each point y \in \sigma_{n+1} such that y \in V^c_{j,\beta} \times T_\beta, the point y already deviates from the base point b on one coordinate, namely \beta. Thus on the coordinates other than \beta, the point y can only deviates from b on at most n many coordinates. Thus \sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta) is homeomorphic to V^c_{j,\beta} \times \sigma_n. Note that V^c_{j,\beta} \times W_j is a separable metric space. By inductive hypothesis, V^c_{j,\beta} \times \sigma_n \times W_j is Lindelof. Thus there are countably many open sets in the open cover \mathcal{U} that covers points of H_{j,\beta} \cap (\sigma_{n+1} \times W_j).

Note that

    \sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)

To see that the left-side is a subset of the right-side, let t=(x,y) \in \sigma_{n+1} \times Y. If t \in U_j for some j, we are done. Suppose t \notin U_j for all j. Observe that y \in W_j for some j. Since t=(x,y) \notin U_j, x_\beta \notin V_{j,\beta} for some \beta \in F_j. Then t=(x,y) \in H_{j,\beta}. It is now clear that t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j). Thus the above set equality is established. Thus one part of \sigma_{n+1} \times Y is covered by countably many open sets in \mathcal{U} while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of \mathcal{U} covers \sigma_{n+1} \times Y. \blacksquare

Corollary 2
It follows from Theorem 1 that

  • If each factor space X_\alpha is a separable metric space, then each \sigma_n is a Lindelof space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a Lindelof space.
  • If each factor space X_\alpha is a compact separable metric space, then each \sigma_n is a compact space and that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that \Sigma-product of compact spaces is countably compact. Thus \Sigma_{\alpha \in A} X_\alpha is a countably compact space if each X_\alpha is compact. Note that each \sigma_n is a closed subset of \Sigma_{\alpha \in A} X_\alpha and is thus countably compact. Being a Lindelof space, each \sigma_n is compact. It follows that \sigma=\bigcup_{n=0}^\infty \sigma_n is a \sigma-compact space. \blacksquare

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each \alpha \in A, let X_\alpha be a separable metric space with at least two points. Then the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is collectionwise normal (see this previous post). According to the lemma in this previous post, the \Sigma-product \Sigma_{\alpha \in A} X_\alpha contains a closed copy of \omega_1. Thus the \Sigma-product \Sigma_{\alpha \in A} X_\alpha is not Lindelof. It is clear that the \sigma-product is a dense subspace of \Sigma_{\alpha \in A} X_\alpha. By Corollary 2, the \sigma-product is a Lindelof subspace of \Sigma_{\alpha \in A} X_\alpha.

Using specific factor spaces, if each X_\alpha=\mathbb{R} with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each X_\alpha=[0,1] with the usual topology, then \Sigma_{\alpha<\omega_1} X_\alpha is a non-Lindelof space with a dense \sigma-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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\copyright \ 2014 \text{ by Dan Ma}