A space is called a -compact space if it is the union of countably many compact subspaces. Clearly, any -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 -compact spaces is a Lindelof space. The discussion here centers on the following theorem.

*Theorem 1*

Let be -compact spaces. Then the product space 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 , let be compact spaces and let be the topological sum:

Then the product is Lindelof.

Note that in the topological sum , the spaces 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 is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many -compact spaces is Lindelof if each -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 be a compact space. Let be a subspace of . Suppose that there exist , closed subsets of , such that for all and where and , there exists such that and . Then is a Lindelof space.

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

*Theorem 4*

For each integer , let be the one-point compactification of in Theorem 2. Then the product is a compact space. Furthermore, is a subspace of . Prove that and satisfy Lemma 3.

Each in Theorem 2 is a locally compact space. To define the one-point compactifications, for each , choose . Make sure that for . Then is simply

with the topology defined as follows:

- Open subsets of continue to be open in .
- An open set containing is of the form where is open in and is contained in the union of finitely many .

For convenience, each point 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.

**Proofs of Key Steps**

*Proof of Lemma 3*

Let , and be as described in the statement for Lemma 3. Let be a collection of open subsets of such that covers . We would like to show that a countable subcollection of is also a cover of . Let . If , then is an open cover of and there is a finite subset of that is a cover of and thus a cover of . Thus we can assume that .

Let . Let , which is compact. We make the following claim.

**Claim**. Let be the union of all possible where is finite and . Then .

To establish the claim, let . For each , there exists such that and . This means that is an open cover of . By the compactness of , there are finitely many such that misses , or equivalently . Note that . Further note that . This establishes the claim that . The claim that is clear from the definition of .

Each set is compact since it is closed in . The intersection of finitely many is also compact. Thus the in the definition of in the above claim is compact. There can be only countably many in the definition of . Thus is a -compact space that is covered by the open cover . Choose a countable such that covers . Then is a cover of too. This completes the proof that is Lindelof.

*Proof of Theorem 4*

Recall that and that . Each is the one-point compactification of , which is the topological sum of the disjoint compact spaces .

For integers , define . For integers , define the product as follows:

Since is a product of compact spaces, is compact and thus closed in . There are only countably many .

We claim that the countably many have the property indicated in Lemma 3. To this end, let and . There exists an integer such that . This means that for all , i.e. (so must be the point at infinity). Choose large enough such that

for all . It follows that and . Thus the sequence of closed sets satisfies Lemma 3. By Lemma 3, is Lindelof.

**Reference**

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

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