The Tube Lemma is a useful tool in working with Cartesian products of finitely many compact spaces. A general discussion is followed by three applications of the lemma.

Let and be topological spaces. A slice in the Cartesian product is a subspace of the form or where and . A tube is an open subset of the Cartesian product that is of the form or where is open in and is open in . In the Euclidean plane, a slice would be either a vertical line or a horizontal line and open strips (vertical or horizontal) are examples of tubes.

Tubes are one type of open subsets of the Cartesian product . The Tube Lemma is applicable when one of the factors is compact. Let be the factor that is compact. A good way of thinking about the lemma is that when you consider the slices as “points”, the tubes , where , behave like a base. The following is a statement of the lemma.

**The Tube Lemma**

Let be a space and let be a compact space. For each , and for each open set of such that , there is an open set such that .

* Proof.* Let and let be open in the product space such that . For each , choose open sets and such that . Since is compact, we can find finitely many whose union is , say, . Let . It follows that .

**Remarks**

The lemma is not true when none of the factors is compact. Let and with the usual topology. Let be defined by:

The open set contains the slice . But no tube can be situated between this slice and .

The Tube Lemma can be used in proving that the product of two compact spaces is compact. By induction, it follows that the product of finitely many compact spaces is compact. However, the lemma cannot be used in proving the compactness of product space with infinitely many compact factors (the Tychonoff Theorem).

The Tube Lemma also shows that both the Lindelof property and paracompactness are preserved in taking two-factor Cartesian product as long as one of the factors is compact. As a corollary, the product of two Lindelof spaces is Lindelof if one of the factors is -compact. We have the following theorems.

**Theorem 1**

Let and be compact spaces. Then is compact.

* Proof*. Let be an open cover of . For each , let be a finite subcollection of such that is a cover of . By the Tube Lemma, there is an open set such that . Since is compact, there are finitely many such that . It follows that covers .

**Theorem 2**

Let be a Lindelof space and be any compact space. Then is Lindelof.

* Proof*. The same proof in Theorem 1 applies except that there are countably many that cover , leading to a countable subcover of the original open cover.

**Corollary 3**

Let be a Lindelof space and be any -compact space. Then is Lindelof.

**Theorem 4**

Let be a paracompact space and be any compact space. Then is paracompact.

* Proof*. The proof begins just as in Theorem 1. Let be an open cover of . For each , let be a finite subcollection of such that is a cover of . By the Tube Lemma, there is an open set such that . Since is paracompact, let be a locally finite open refinement of such that for each .

Let . It can be shown that is a cover of , is a refinement of , and is locally finite. The first two points are clear. To show that it is a locally finite collection of sets, let . There is some open such that and meets only finitely many . Then meets only finitely many sets in .

Excellent! Thank’s a lot!