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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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.

\text{ }

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.

\text{ }

\text{ }

\text{ }

Dan Ma topology

Daniel Ma topology

Dan Ma math

Daniel Ma mathematics

\copyright 2019 – Dan Ma

1 thought on “Lindelof Exercise 1

  1. Pingback: Lindelof Exercise 2 | Dan Ma's Topology Blog

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s