Tube lemma

The lemma uses the following terminology: • If X and Y are topological spaces and X \times Y is the product space, endowed with the product topology, a slice in X \times Y is a set of the form \{ x \} \times Y for x \in X. • A tube in X \times Y is a subset of the form U\times Y where U is an open subset of X. It contains all the slices \{x\} \times Y for x \in U. Using the concept of closed maps, this can be rephrased concisely as follows: if X is any topological space and Y a compact space, then the projection map X \times Y \to X is closed. {{Math theorem|name=Generalized Tube Lemma 2|math_statement= Let X_i, i\in I be topological spaces and consider the product space \prod_{i\in I} X_i. For each i\in I, let A_i be a compact subset of X_i. If N is an open set containing \prod_{i\in I} A_i, then there exists U_i open in X_i with U_i = X_i for all but finite amount of i\in I, such that \prod_{i\in I} A_i \subseteq \prod_{i\in I} U_i \subseteq N. }} == Examples and properties ==

1. Consider \mathbb{R} \times \mathbb{R} in the product topology, that is the Euclidean plane, and the open set N = \{ (x, y) \in \mathbb{R} \times \mathbb{R} ~:~ |xy| The open set N contains \{ 0 \} \times \mathbb{R}, but contains no tube, so in this case the tube lemma fails. Indeed, if W \times \mathbb{R} is a tube containing \{ 0 \} \times \mathbb{R} and contained in N, W must be a subset of \left( - 1/x, 1/x \right) for all x>0 which means W = \{ 0 \} contradicting the fact that W is open in \mathbb{R} (because 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 X and Y are compact spaces, then X \times Y is compact as follows: Let \{ G_a \} be an open cover of X \times Y. For each x \in X, cover the slice \{ x \} \times Y by finitely many elements of \{ G_a \} (this is possible since \{ x \} \times 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 \times Y containing \{ x \} \times Y and contained in N_x. The collection of all W_x for x \in X is an open cover of X and hence has a finite subcover \{W_{x_1},\dots,W_{x_n}\}. Thus the finite collection \{W_{x_1}\times Y,\dots,W_{x_n}\times Y\} covers X\times Y. Using the fact that each W_{x_i} \times Y is contained in N_{x_i} and each N_{x_i} is the finite union of elements of \{G_a\}, one gets a finite subcollection of \{G_a\} that covers 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 A = \{ x \} and B = Y. It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each (a, b) \in A \times B there are open sets U_{a, b} \subseteq X and V_{a, b} \subseteq Y such that (a, b) \in U_{a, b} \times V_{a, b} \subseteq N. For any a \in A, \left\{ V_{a,b} ~:~ b \in B \right\} is an open cover of the compact set B so this cover has a finite subcover; namely, there is a finite set B_0(a) \subseteq B such that V_{a} := \bigcup_{b \in B_0(a)} V_{a,b} contains B, where V_a is open in Y. For every a \in A, let U_a := \bigcap_{b \in B_0(a)} U_{a,b}, which is an open set in X since B_0(a) is finite. Moreover, the construction of U_a and V_a implies that \{ a \} \times B \subseteq U_a \times V_a \subseteq N. We now essentially repeat the argument to drop the dependence on a. Let A_0 \subseteq A be a finite subset such that U := \bigcup_{a \in A_0} U_a contains A and set V := \bigcap_{a \in A_0} V_a. It then follows by the above reasoning that A \times B \subseteq U \times V \subseteq N, and U \subseteq X and V \subseteq Y are respectively open, which completes the proof. == See also ==