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.

Statement[]

The lemma uses the following terminology:

• If ${\displaystyle X}$ and ${\displaystyle Y}$ are topological spaces and ${\displaystyle X\times Y}$ is the product space, endowed with the product topology, a slice in ${\displaystyle X\times Y}$ is a set of the form ${\displaystyle \{x\}\times Y}$ for ${\displaystyle x\in X}$.
• A tube in ${\displaystyle X\times Y}$ is a subset of the form ${\displaystyle U\times Y}$ where ${\displaystyle U}$ is an open subset of ${\displaystyle X}$. It contains all the slices ${\displaystyle \{x\}\times Y}$ for ${\displaystyle x\in U}$.

Using the concept of closed maps, this can be rephrased concisely as follows: if ${\displaystyle X}$ is any topological space and ${\displaystyle Y}$ a compact space, then the projection map ${\displaystyle X\times Y\to X}$ is closed.

Examples and properties[]

1. Consider ${\displaystyle \mathbb {R} \times \mathbb {R} }$ in the product topology, that is the Euclidean plane, and the open set ${\displaystyle N=\{(x,y)\in \mathbb {R} \times \mathbb {R} ~:~|xy|<1\}.}$ The open set ${\displaystyle N}$ contains ${\displaystyle \{0\}\times \mathbb {R} ,,}$ but contains no tube, so in this case the tube lemma fails. Indeed, if ${\displaystyle W\times \mathbb {R} }$ is a tube containing ${\displaystyle \{0\}\times \mathbb {R} }$ and contained in ${\displaystyle N,}$ ${\displaystyle W}$ must be a subset of ${\displaystyle \left(-1/x,1/x\right)}$ for all ${\displaystyle x>0}$ which means ${\displaystyle W=\{0\}}$ contradicting the fact that ${\displaystyle W}$ is open in ${\displaystyle \mathbb {R} }$ (because ${\displaystyle W\times \mathbb {R} }$ is a tube). This shows that the compactness assumption is essential.

2. The tube lemma can be used to prove that if ${\displaystyle X}$ and ${\displaystyle Y}$ are compact spaces, then ${\displaystyle X\times Y}$ is compact as follows:

Let ${\displaystyle \{G_{a}\}}$ be an open cover of ${\displaystyle X\times Y}$. For each ${\displaystyle x\in X}$, cover the slice ${\displaystyle \{x\}\times Y}$ by finitely many elements of ${\displaystyle \{G_{a}\}}$ (this is possible since ${\displaystyle \{x\}\times Y}$ is compact, being homeomorphic to ${\displaystyle Y}$). Call the union of these finitely many elements ${\displaystyle N_{x}.}$ By the tube lemma, there is an open set of the form ${\displaystyle W_{x}\times Y}$ containing ${\displaystyle \{x\}\times Y}$ and contained in ${\displaystyle N_{x}.}$ The collection of all ${\displaystyle W_{x}}$ for ${\displaystyle x\in X}$ is an open cover of ${\displaystyle X}$ and hence has a finite subcover ${\displaystyle \{W_{x_{1}},\dots ,W_{x_{n}}\}}$. Thus the finite collection ${\displaystyle \{W_{x_{1}}\times Y,\dots ,W_{x_{n}}\times Y\}}$ covers ${\displaystyle X\times Y}$. Using the fact that each ${\displaystyle W_{x_{i}}\times Y}$ is contained in ${\displaystyle N_{x_{i}}}$ and each ${\displaystyle N_{x_{i}}}$ is the finite union of elements of ${\displaystyle \{G_{a}\}}$, one gets a finite subcollection of ${\displaystyle \{G_{a}\}}$ that covers ${\displaystyle X\times Y}$.

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 ${\displaystyle A=\{x\}}$ and ${\displaystyle B=Y.}$ It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each ${\displaystyle (a,b)\in A\times B}$ there are open sets ${\displaystyle U_{a,b}\subseteq X}$ and ${\displaystyle V_{a,b}\subseteq Y}$ such that ${\displaystyle (a,b)\in U_{a,b}\times V_{a,b}\subseteq N.}$ For any ${\displaystyle a\in A,}$ ${\displaystyle \left\{V_{a,b}~:~b\in B\right\}}$ is an open cover of the compact set ${\displaystyle B}$ so this cover has a finite subcover; namely, there is a finite set ${\displaystyle B_{0}(a)\subseteq B}$ such that ${\displaystyle V_{a}:=\bigcup _{b\in B_{0}(a)}V_{a,b}}$ contains ${\displaystyle B,}$ where observe that ${\displaystyle V_{a}}$ is open in ${\displaystyle Y.}$ For every ${\displaystyle a\in A,}$ let ${\displaystyle U_{a}:=\bigcap _{b\in B_{0}(a)}U_{a,b},}$ which is an open in ${\displaystyle X}$ set since ${\displaystyle B_{0}(a)}$ is finite. Moreover, the construction of ${\displaystyle U_{a}}$ and ${\displaystyle V_{a}}$ implies that ${\displaystyle \{a\}\times B\subseteq U_{a}\times V_{a}\subseteq N.}$ We now essentially repeat the argument to drop the dependence on ${\displaystyle a.}$ Let ${\displaystyle A_{0}\subseteq A}$ be a finite subset such that ${\displaystyle U:=\bigcup _{a\in A_{0}}U_{a}}$ contains ${\displaystyle A}$ and set ${\displaystyle V:=\bigcap _{a\in A_{0}}V_{a}.}$ It then follows by the above reasoning that ${\displaystyle A\times B\subseteq U\times V\subseteq N}$ and ${\displaystyle U\subseteq X}$ and ${\displaystyle V\subseteq Y}$ are open, which completes the proof.

See also[]

• Alexander's sub-base theorem
• Tubular neighborhood
• Tychonoff theorem

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)