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
. 