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

Statement

The lemma uses the following terminology: Using the concept of closed maps, this can be rephrased concisely as follows: if X is any topological space and Y 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 N contains but contains no tube, so in this case the tube lemma fails. Indeed, if is a tube containing and contained in N, W must be a subset of for all x>0 which means W = { 0 } contradicting the fact that W is open in \mathbb{R} (because is a tube). This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if X and Y are compact spaces, then X \times Y is compact as follows: Let { G_a } be an open cover of X \times Y. For each x \in X, cover the slice by finitely many elements of { G_a } (this is possible since is compact, being homeomorphic to Y). Call the union of these finitely many elements N_x. By the tube lemma, there is an open set of the form containing and contained in N_x. The collection of all W_x for x \in X is an open cover of X and hence has a finite subcover. Thus the finite collection covers X\times Y. Using the fact that each is contained in N_{x_i} and each N_{x_i} is the finite union of elements of {G_a}, one gets a finite subcollection of {G_a} that covers X \times Y.
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 A = { x } and B = Y. 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 a \in A, is an open cover of the compact set B so this cover has a finite subcover; namely, there is a finite set such that contains B, where observe that V_a is open in Y. For every a \in A, let which is an open in X set since B_0(a) is finite. Moreover, the construction of U_a and V_a implies that We now essentially repeat the argument to drop the dependence on a. Let be a finite subset such that contains A and set It then follows by the above reasoning that and and are open, which completes the proof.