{{Short description|Lemma in topology}} In [[mathematics]], particularly [[topology]], the '''tube lemma''', also called Wallace's theorem, is a useful tool in order to prove that the [[product topology|product]] of finitely many [[compact spaces]] is compact.
== Statement ==
The lemma uses the following terminology:
* If <math>X</math> and <math>Y</math> are [[topological space]]s and <math>X \times Y</math> is the product space, endowed with the [[product topology]], a '''slice''' in <math>X \times Y</math> is a set of the form <math>\{ x \} \times Y</math> for <math>x \in X</math>. * A '''tube''' in <math>X \times Y</math> is a subset of the form <math>U\times Y</math> where <math>U</math> is an open subset of <math>X</math>. It contains all the slices <math>\{x\} \times Y</math> for <math>x \in U</math>.
{{Math theorem|name=Tube Lemma|math_statement= Let <math>X</math> and <math>Y</math> be topological spaces with <math>Y</math> compact, and consider the [[Product topology|product space]] <math>X \times Y.</math> If <math>N</math> is an open set containing a slice in <math>X \times Y,</math> then there exists a tube in <math>X \times Y</math> containing this slice and contained in <math>N.</math> }}
Using the concept of [[closed map]]s, this can be rephrased concisely as follows: if <math>X</math> is any topological space and <math>Y</math> a compact space, then the projection map <math>X \times Y \to X</math> is closed.
{{Math theorem|name=Generalized Tube Lemma 1|math_statement= Let <math>X</math> and <math>Y</math> be topological spaces and consider the product space <math>X \times Y.</math> Let <math>A</math> be a compact subset of <math>X</math> and <math>B</math> be a compact subset of <math>Y.</math> If <math>N</math> is an open set containing <math>A \times B,</math> then there exists <math>U</math> open in <math>X</math> and <math>V</math> open in <math>Y</math> such that <math>A \times B \subseteq U \times V \subseteq N.</math> }}
{{Math theorem|name=Generalized Tube Lemma 2|math_statement= Let <math>X_i, i\in I</math> be topological spaces and consider the product space <math>\prod_{i\in I} X_i.</math> For each <math>i\in I</math>, let <math>A_i</math> be a compact subset of <math>X_i.</math> If <math>N</math> is an open set containing <math>\prod_{i\in I} A_i,</math> then there exists <math>U_i</math> open in <math>X_i</math> with <math>U_i = X_i</math> for all but finite amount of <math>i\in I</math>, such that <math>\prod_{i\in I} A_i \subseteq \prod_{i\in I} U_i \subseteq N.</math> }}
== Examples and properties ==
1. Consider <math>\mathbb{R} \times \mathbb{R}</math> in the product topology, that is the [[Euclidean plane]], and the open set <math>N = \{ (x, y) \in \mathbb{R} \times \mathbb{R} ~:~ |xy| < 1 \}.</math> The open set <math>N</math> contains <math>\{ 0 \} \times \mathbb{R},</math> but contains no tube, so in this case the tube lemma fails. Indeed, if <math>W \times \mathbb{R}</math> is a tube containing <math>\{ 0 \} \times \mathbb{R}</math> and contained in <math>N,</math> <math>W</math> must be a subset of <math>\left( - 1/x, 1/x \right)</math> for all <math>x>0</math> which means <math>W = \{ 0 \}</math> contradicting the fact that <math>W</math> is open in <math>\mathbb{R}</math> (because <math>W \times \mathbb{R}</math> is a tube). This shows that the compactness assumption is essential.
2. The tube lemma can be used to prove that if <math>X</math> and <math>Y</math> are compact spaces, then <math>X \times Y</math> is compact as follows:
Let <math>\{ G_a \}</math> be an [[Cover (topology)|open cover]] of <math>X \times Y</math>. For each <math>x \in X</math>, cover the slice <math>\{ x \} \times Y</math> by finitely many elements of <math>\{ G_a \}</math> (this is possible since <math>\{ x \} \times Y</math> is compact, being [[homeomorphic]] to <math>Y</math>). Call the union of these finitely many elements <math>N_x.</math> By the tube lemma, there is an [[open set]] of the form <math>W_x \times Y</math> containing <math>\{ x \} \times Y</math> and contained in <math>N_x.</math> The collection of all <math>W_x</math> for <math>x \in X</math> is an open cover of <math>X</math> and hence has a finite subcover <math>\{W_{x_1},\dots,W_{x_n}\}</math>. Thus the finite collection <math>\{W_{x_1}\times Y,\dots,W_{x_n}\times Y\}</math> covers <math>X\times Y</math>. Using the fact that each <math>W_{x_i} \times Y</math> is contained in <math>N_{x_i}</math> and each <math>N_{x_i}</math> is the finite union of elements of <math>\{G_a\}</math>, one gets a finite subcollection of <math>\{G_a\}</math> that covers <math>X \times Y</math>.
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 <math>A = \{ x \}</math> and <math>B = Y.</math> It therefore suffices to prove the generalized tube lemma. By the definition of the product topology, for each <math>(a, b) \in A \times B</math> there are open sets <math>U_{a, b} \subseteq X</math> and <math>V_{a, b} \subseteq Y</math> such that <math>(a, b) \in U_{a, b} \times V_{a, b} \subseteq N.</math> For any <math>a \in A,</math> <math>\left\{ V_{a,b} ~:~ b \in B \right\}</math> is an open cover of the compact set <math>B</math> so this cover has a finite subcover; namely, there is a finite set <math>B_0(a) \subseteq B</math> such that <math>V_{a} := \bigcup_{b \in B_0(a)} V_{a,b}</math> contains <math>B,</math> where <math>V_a</math> is open in <math>Y.</math> For every <math>a \in A,</math> let <math>U_a := \bigcap_{b \in B_0(a)} U_{a,b},</math> which is an open set in <math>X</math> since <math>B_0(a)</math> is finite. Moreover, the construction of <math>U_a</math> and <math>V_a</math> implies that <math>\{ a \} \times B \subseteq U_a \times V_a \subseteq N.</math> We now essentially repeat the argument to drop the dependence on <math>a.</math> Let <math>A_0 \subseteq A</math> be a finite subset such that <math>U := \bigcup_{a \in A_0} U_a</math> contains <math>A</math> and set <math>V := \bigcap_{a \in A_0} V_a.</math> It then follows by the above reasoning that <math>A \times B \subseteq U \times V \subseteq N</math>, and <math>U \subseteq X</math> and <math>V \subseteq Y</math> are respectively open, which completes the proof.
== See also ==
* {{annotated link|Alexander's sub-base theorem}} * {{annotated link|Tubular neighborhood}} * {{annotated link|Tychonoff theorem}}
== References == {{Reflist}}
* {{cite book | author = James Munkres | author-link = James Munkres | year = 1999 | title = Topology | edition = 2nd | publisher = [[Prentice Hall]] | isbn = 0-13-181629-2 }} * {{cite book | author = Joseph J. Rotman | author-link = Joseph J. Rotman | year = 1988 | title = An Introduction to Algebraic Topology | url = https://archive.org/details/introductiontoal0000rotm | url-access = registration | publisher = Springer | isbn = 0-387-96678-1 }} ''(See Chapter 8, Lemma 8.9)''
[[Category:Topology]] [[Category:Lemmas]] [[Category:Articles containing proofs]]