Tube lemma
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:
- 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 .
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 1 — 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
Generalized Tube Lemma 2 — Let be topological spaces and consider the product space For each , let be a compact subset of If is an open set containing then there exists open in with for all but finite amount of , such that
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 (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 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 . 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 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 also
- Alexander's sub-base theorem – Collection of subsets that generate a topologyPages displaying short descriptions of redirect targets
- Tubular neighborhood – neighborhood of a submanifold homeomorphic to that submanifold’s normal bundlePages displaying wikidata descriptions as a fallback
- Tychonoff theorem – Product of any collection of compact topological spaces is compactPages displaying short descriptions of redirect targets
References
- 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)