Consider the product space . Fix a point , called the base point. The -product of the spaces is the following subspace of the product space :

In other words, the space is the subspace of the product space consisting of all points that deviate from the base point on at most countably many coordinates . We also consider the following subspace of .

For convenience , we call the (upper case) Sigma-product (or -product) of the spaces and we call the space the (lower case) sigma-product (or -product). Clearly, the space is a dense subspace of . 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 is a separable metric space with at least two points, the -product, though not Lindelof, has a dense Lindelof subspace. The (upper case) -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 , we define as follows:

Clearly, . We prove the following theorem. The fact that 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 . We say is a standard basic open subset of the product space if is of the form such that each is an open subset of the factor space and for all but finitely many . The finite set of all such that is called the support of the open set .

**Theorem 1**

Let be the -product of the separable metrizable spaces . For each , let be defined as above. The product space is Lindelof for each non-negative integer and for all separable metric space .

**Proof of Theorem 1**

We prove by induction on . Note that , the base point. Clearly is Lindelof for all separable metric space . Suppose the theorem hold for the integer . We show that for all separable metric space . To this end, let be an open cover of where is a separable metric space. Without loss of generality, we assume that each element of is of the form where is a standard basic open subset of the product space and is an open subset of .

Let be a countable subcollection of such that covers . For each , let where is a standard basic open subset of the product space with and is an open subset of . For each , let be the support of . Note that if and only if . Also for each , . Furthermore, for each , let . With all these notations in mind, we define the following open set for each :

Observe that for each point such that , the point already deviates from the base point on one coordinate, namely . Thus on the coordinates other than , the point can only deviates from on at most many coordinates. Thus is homeomorphic to . Note that is a separable metric space. By inductive hypothesis, is Lindelof. Thus there are countably many open sets in the open cover that covers points of .

Note that

To see that the left-side is a subset of the right-side, let . If for some , we are done. Suppose for all . Observe that for some . Since , for some . Then . It is now clear that . Thus the above set equality is established. Thus one part of is covered by countably many open sets in while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of covers .

**Corollary 2**

It follows from Theorem 1 that

- If each factor space is a separable metric space, then each is a Lindelof space and that is a Lindelof space.
- If each factor space is a compact separable metric space, then each is a compact space and that is a -compact space.

**Proof of Corollary 2**

The first bullet point is a clear corollary of Theorem 1. A previous post shows that -product of compact spaces is countably compact. Thus is a countably compact space if each is compact. Note that each is a closed subset of and is thus countably compact. Being a Lindelof space, each is compact. It follows that is a -compact space.

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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 , let be a separable metric space with at least two points. Then the -product is collectionwise normal (see this previous post). According to the lemma in this previous post, the -product contains a closed copy of . Thus the -product is not Lindelof. It is clear that the -product is a dense subspace of . By Corollary 2, the -product is a Lindelof subspace of .

Using specific factor spaces, if each with the usual topology, then is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each with the usual topology, then is a non-Lindelof space with a dense -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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