A tube in is just a basis element, in containing a slice in where is an open subset of
Tube Lemma — Let and be topological spaces with compact, and consider the product space If is an open set containing a slice in then there exists a tube in containing this slice and contained in
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.
Generalized Tube Lemma — Let and be topological spaces and consider the product space Let be a compact subset of and be a compact subset of If is an open set containing then there exists open in and open in such that
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 positive integers which means contradicting the fact that is open in (because is a tube). This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if and are compact topological spaces, then is compact as follows:
Let be an open cover of ;
for each cover the slice by finitely many elements of (this is possible since is compact being homeomorphic to ).
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
Then for each is contained in
Using the fact that each is the finite union of elements of and that the finite collection covers the collection is a finite subcover of
3. By example 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.
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 observe that is open in
For every let which is an open in set 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 open, which completes the proof.