0708-1300/Errata to Bredon's Book: Difference between revisions
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There is a counterexample to the inverse implication in Problem 1, p. 71. |
There is a counterexample to the inverse implication in Problem 1, p. 71. |
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Let X=\mathbb R be endowed with the ordinary topology (thus, it is Hausdorff and second countable). Let U be an arbitrary connected open set in X (that is, an interval). Let F_X(U) consists of all functions identically equal to constant. If U is an arbitrary open set, then by theorem on structure of open sets in \mathbb R it is a union of countably many open intervals. We define F_X(U) to be the set of all real-valued functions which are constant on open intervals forming U. The family F=\{F_X(U):U is open in X\} forms a functional structure, as one can check. Furthermore, it satisfies the hypothesis of the theorem: every point x \in X has a |
Let <math>X=\mathbb</math> R be endowed with the ordinary topology (thus, it is Hausdorff and second countable). Let <math>U</math> be an arbitrary connected open set in <math>X</math> (that is, an interval). Let <math>F_X(U)</math> consists of all functions identically equal to constant. If <math>U</math> is an arbitrary open set, then by theorem on structure of open sets in <math>\mathbb R</math> it is a union of countably many open intervals. We define <math>F_X(U)</math> to be the set of all real-valued functions which are constant on open intervals forming <math>U</math>. The family <math>F=\{F_X(U):U is open in X\}</math> forms a functional structure, as one can check. Furthermore, it satisfies the hypothesis of the theorem: every point <math>x \in X</math> has a neighborhood (we take an open interval containing <math>x</math>) such that there exists a function <math>f \in F_X(U)</math> (we define it to be identically equal to <math>1</math>) such that a function <math>g:U \to \mathbb R</math> is in <math>F_X(U)</math> (it is identically equal to a constant by our definition) if and only if there exists a smooth function <math>h</math> such that <math>g=h \circ f</math> (if <math>g</math> is given, then we define <math>h(x)=g</math> for all <math>x</math>, if <math>f</math> is given, then we take arbitrary smooth <math>h:\mathbb R \to \mathbb R</math>, since <math>h \circ f</math> is identically equal to constant and, thus, is in <math>F_X(U)</math>). Clearly, <math>(X,F_X)</math> is not a smooth manifold. |
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Revision as of 08:39, 27 September 2007
Errata to Bredn's Book
There is a counterexample to the inverse implication in Problem 1, p. 71.
Let [math]\displaystyle{ X=\mathbb }[/math] R be endowed with the ordinary topology (thus, it is Hausdorff and second countable). Let [math]\displaystyle{ U }[/math] be an arbitrary connected open set in [math]\displaystyle{ X }[/math] (that is, an interval). Let [math]\displaystyle{ F_X(U) }[/math] consists of all functions identically equal to constant. If [math]\displaystyle{ U }[/math] is an arbitrary open set, then by theorem on structure of open sets in [math]\displaystyle{ \mathbb R }[/math] it is a union of countably many open intervals. We define [math]\displaystyle{ F_X(U) }[/math] to be the set of all real-valued functions which are constant on open intervals forming [math]\displaystyle{ U }[/math]. The family [math]\displaystyle{ F=\{F_X(U):U is open in X\} }[/math] forms a functional structure, as one can check. Furthermore, it satisfies the hypothesis of the theorem: every point [math]\displaystyle{ x \in X }[/math] has a neighborhood (we take an open interval containing [math]\displaystyle{ x }[/math]) such that there exists a function [math]\displaystyle{ f \in F_X(U) }[/math] (we define it to be identically equal to [math]\displaystyle{ 1 }[/math]) such that a function [math]\displaystyle{ g:U \to \mathbb R }[/math] is in [math]\displaystyle{ F_X(U) }[/math] (it is identically equal to a constant by our definition) if and only if there exists a smooth function [math]\displaystyle{ h }[/math] such that [math]\displaystyle{ g=h \circ f }[/math] (if [math]\displaystyle{ g }[/math] is given, then we define [math]\displaystyle{ h(x)=g }[/math] for all [math]\displaystyle{ x }[/math], if [math]\displaystyle{ f }[/math] is given, then we take arbitrary smooth [math]\displaystyle{ h:\mathbb R \to \mathbb R }[/math], since [math]\displaystyle{ h \circ f }[/math] is identically equal to constant and, thus, is in [math]\displaystyle{ F_X(U) }[/math]). Clearly, [math]\displaystyle{ (X,F_X) }[/math] is not a smooth manifold.
