Tube lemma: Difference between revisions

Latest revision as of 02:16, 16 November 2015

This article is about the statement of a simple but indispensable lemma in topology

Statement

Let $X$ be a compact space and $A$ any topological space. Consider $X\times A$ endowed with the product topology. Suppose $a\in A$ and $U$ is an open subset of $X\times A$ containing the entire slice $X\times \{a\}$ . Then, we can find an open subset $V$ of $A$ such that:

$a\in V$ , and $X\times V\subseteq U$

In other words, any open subset containing a slice contains an open cylinder that contains the slice.

Proof

Given: A compact space $X$ , a topological space $A$ . $a\in A$ , and $U$ is an open subset of $X\times A$ containing the slice $X\times \{a\}$ .

To prove: There exists an open subset $V$ of $A$ such that $a\in V$ $X\times V$ is contained in $U$ .

Proof:

A collection of open subsets inside $U$ whose union contains $X\times \{a\}$ : For each $x\in X$ , we have $(x,a)\in U$ , so by the definition of openness in the product topology, there exists a basis open subset $M_{x}\times N_{x}\subseteq U$ containing $(x,a)$ . In particular, we get a collection $M_{x}\times N_{x},x\in X$ of open subsets contained in $U$ , whose union contains $X\times \{a\}$ .
This collection yields a point-indexed open cover for $X$ : Note that since $M_{x}\times N_{x}$ is a basis open set containing $(x,a)$ , $M_{x}$ is an open subset of $X$ containing $X$ , so the $M_{x},x\in X$ , form an open cover of $X$ .
(Given data used: $X$ is compact): This cover has a finite subcover: Indeed, since $X$ is compact, we can choose a finite collection of points $\{x_{1},x_{2},\dots ,x_{n}\}\subseteq X$ such that $X$ is the union of the $M_{x_{i}}$ s.
If $V$ is the intersection of the corresponding $N_{x_{i}}$ s, then $V$ is open in $A$ , $a\in V$ , and $X\times V\subseteq U$ : First, $V$ is open since it is an intersection of finitely many open subsets of $Y$ . Second, each $N_{x_{i}}$ contains $a$ , so $a\in V$ . Third, if $(x,v)\in X\times V$ , then there exists $x_{i}$ such that $x\in M_{x_{i}}$ . By definition, $v\in N_{x_{i}}$ , so $(x,v)\in M_{x_{i}}\times N_{x_{i}}\subseteq V$ . Thus, $(x,v)\in U$ .

References

Textbook references

Topology (2nd edition) by James R. Munkres, ^{More info}, Page 168, Lemma 26.8, Chapter 3, Section 26 (the proof is given before the theorem, as Step 1 of the proof of Theorem 26.7 on page 167)

@@ Line 1: / Line 1: @@
-{{factrelatedto|compactness}}
 {{indispensable lemma}}
 ==Statement==
-Let <math>X</math> be a [[compact space]] and <math>A</math> any [[topological space]]. Consider <math>X \times A</math> endowed with the [[product topology]]. Suppose <math>a \in A</math> and <math>U</math> is an [[open subset]] of <math>X \times A</math> containing the entire [[slice]] <math>X \times \{ a \}</math>. Then, we can find an open subset <math>V</math> of <math>A</math> such that:
+Let <math>X</math> be a [[fact about::compact space;1| ]][[compact space]] and <math>A</math> any [[topological space]]. Consider <math>X \times A</math> endowed with the [[fact about::product topology;2| ]][[product topology]]. Suppose <math>a \in A</math> and <math>U</math> is an [[open subset]] of <math>X \times A</math> containing the entire [[slice]] <math>X \times \{ a \}</math>. Then, we can find an open subset <math>V</math> of <math>A</math> such that:
+<math>a \in V</math>, and <math>X \times V \subseteq U</math>
+In other words, any open subset containing a slice contains an [[open cylinder]] that contains the slice.
+==Proof==
+'''Given''': A compact space <math>X</math>, a topological space <math>A</math>. <math>a \in A</math>, and <math>U</math> is an open subset of <math>X \times A</math> containing the slice <math>X \times \{ a \}</math>.
+'''To prove''': There exists an open subset <math>V</math> of <math>A</math> such that <math>a \in V</math> <math>X \times V</math> is contained in <math>U</math>.
+'''Proof''':
+# A collection of open subsets inside <math>U</math> whose union contains <math>X \times \{ a \}</math>: For each <math>x \in X</math>, we have <math>(x,a) \in U</math>, so by the definition of openness in the product topology, there exists a basis open subset <math>M_x \times N_x \subseteq U</math> containing <math>(x,a)</math>. In particular, we get a collection <math>M_x \times N_x, x \in X</math> of open subsets contained in <math>U</math>, whose union contains <math>X \times \{ a \}</math>.
+# This collection yields a point-indexed open cover for <math>X</math>: Note that since <math>M_x \times N_x</math> is a basis open set containing <math>(x,a)</math>, <math>M_x</math> is an open subset of <math>X</math> containing <math>X</math>, so the <math>M_x, x \in X</math>, form an open cover of <math>X</math>.
+# ('''Given data used''': <math>X</math> is compact): This cover has a finite subcover: Indeed, since <math>X</math> is compact, we can choose a finite collection of points <math>\{ x_1, x_2, \dots, x_n \} \subseteq X</math> such that <math>X</math> is the union of the <math>M_{x_i}</math>s.
+# If <math>V</math> is the intersection of the corresponding <math>N_{x_i}</math>s, then <math>V</math> is open in <math>A</math>, <math>a \in V</math>, and <math>X \times V \subseteq U</math>: First, <math>V</math> is open since it is an intersection of finitely many open subsets of <math>Y</math>. Second, each <math>N_{x_i}</math> contains <math>a</math>, so <math>a \in V</math>. Third, if <math>(x,v) \in X \times V</math>, then there exists <math>x_i</math> such that <math>x \in M_{x_i}</math>. By definition, <math>v \in N_{x_i}</math>, so <math>(x,v) \in M_{x_i} \times N_{x_i} \subseteq V</math>. Thus, <math>(x,v) \in U</math>.
-<math>a \in V</math>, and <math>X \times V \subseteq A</math>
+==References==
-In other words, any open subset containing a slice, contains an [[open cylinder]] that contains the slice.
+===Textbook references===
+* {{booklink-proved|Munkres}}, Page 168, Lemma 26.8, Chapter 3, Section 26 (the proof is given before the theorem, as Step 1 of the proof of Theorem 26.7 on page 167)