Tube lemma

In mathematics, particularly topology, the tube lemma, also called Wallace's theorem, is a useful tool in order to prove that the product of finitely many compact spaces is compact.

Statement

The lemma uses the following terminology:

If and are topological spaces and is the product space, endowed with the product topology, a slice in is a set of the form for.
A tube in is a subset of the form where is an open subset of. It contains all the slices for.

Using the concept of closed maps, this can be rephrased concisely as follows: if is any topological space and a compact space, then the projection map is closed.

Examples and properties

1. Consider in the product topology, that is the Euclidean plane, and the open set The open set contains but contains no tube, so in this case the tube lemma fails. Indeed, if is a tube containing and contained in must be a subset of for all which means contradicting the fact that is open in . This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if and are compact spaces, then is compact as follows:
Let be an open cover of. For each, cover the slice by finitely many elements of .
Call the union of these finitely many elements
By the tube lemma, there is an open set of the form containing and contained in
The collection of all for is an open cover of and hence has a finite subcover. Thus the finite collection covers.
Using the fact that each is contained in and each is the finite union of elements of, one gets a finite subcollection of that covers.
3. By part 2 and induction, one can show that the finite product of compact spaces is compact.
4. The tube lemma cannot be used to prove the Tychonoff theorem, which generalizes the above to infinite products.

Proof

The tube lemma follows from the generalized tube lemma by taking and
It therefore suffices to prove the generalized tube lemma.
By the definition of the product topology, for each there are open sets and such that
For any is an open cover of the compact set so this cover has a finite subcover; namely, there is a finite set such that contains where is open in
For every let which is an open set in since is finite.
Moreover, the construction of and implies that
We now essentially repeat the argument to drop the dependence on
Let be a finite subset such that contains and set
It then follows by the above reasoning that, and and are respectively open, which completes the proof.