Tube lemma

1. Consider R × R in the product topology, that is the Euclidean plane, and the open set N = { (x, y) : |x·y| < 1 }. The open set N contains {0} × R, but contains no tube, so in this case the tube lemma fails. Indeed, if W × R is a tube containing {0} × R and contained in N, W must be a subset of (−1/x, +1/x) for all positive integers x which means W = {0} contradicting the fact that W is open in R (because W × R 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 topological spaces, then X × Y is compact as follows:

Let {G_a} be an open cover of X × Y; for each x belonging to X, cover the slice {x} × Y by finitely many elements of {G_a} (this is possible since {x} × Y 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 W_x × Y containing {x} × Y and contained in N_x. The collection of all W_x for x belonging to X is an open cover of X and hence has a finite subcover W_x1  ∪ ... ∪ W_xn. Then for each x_i, W_xi × Y is contained in N_xi. Using the fact that each N_xi is the finite union of elements of G_a and that the finite collection (W_x1 × Y) ∪ ... ∪ (W_xn × Y) covers X × Y, the collection N_x1 ∪ ... ∪ N_xn is a finite subcover of X × Y.

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 ${\displaystyle A=\{x\}}$ 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 ${\displaystyle (a,b)\in U_{a,b}\times V_{a,b}\subseteq N}$. Fix some ${\displaystyle a\in A}$. Then ${\displaystyle \{V_{a,b}:b\in B\}}$ is an open cover of ${\displaystyle B}$. Since ${\displaystyle B}$ is compact, this cover has a finite subcover; namely, there is a finite ${\displaystyle B_{0}(a)\subseteq B}$ such that ${\displaystyle V'_{a}:=\bigcup _{b\in B_{0}(a)}V_{a,b}\supset B}$. Set ${\displaystyle U'_{a}:=\bigcap _{b\in B_{0}(a)}U_{a,b}}$. Since ${\displaystyle B_{0}(a)}$ is finite, ${\displaystyle U'_{a}}$ is open. Also ${\displaystyle V'_{a}}$ is open. 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 ${\displaystyle U'':=\bigcup _{a\in A_{0}}U'_{a}\supset A}$ 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.

• Tychonoff theorem
• Compact space
• Product topology

• 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)