Tube lemma

In mathematics, particularly topology, the tube lemma is a useful tool in order to prove that the finite product of compact spaces is compact. It is in general a concept of point-set topology.

StatementEdit

The lemma uses the following terminology:

  • If   and   are topological spaces and   is the product space, which is endowed with the product topology, then a slice in   is a set of the form   for  
  • 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  

Examples and propertiesEdit

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.

ProofEdit

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.

See alsoEdit

ReferencesEdit

  • James Munkres (1999). Topology (2nd ed.). Prentice Hall. ISBN 0-13-181629-2.
  • Joseph J. Rotman (1988). An Introduction to Algebraic Topology. Springer. ISBN 0-387-96678-1. (See Chapter 8, Lemma 8.9)