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=Errata to Bredn's Book= |
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'''Problem 1, p. 71.''' |
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(Damir, do you want to start?) |
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There is a counterexample to the inverse implication in Problem 1, p. 71. |
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Let <math>X=\mathbb{R}</math> 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\mbox{ 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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Even taking <math>X</math> as any <math>T_2</math> second countable topological space with the functional structure of constant functions will do the work. |
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Adding to the statement of the problem that the <math>F=(f_1,\ldots,f_n)</math> function is invertible we get a correct theorem. Maybe other weakening of this condition works. |
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'''Problem 4, p. 88.''' |
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Last line of problem 4 says "Also show that '''XY''' itself is not a vector field." and should say "Also show that '''XY''' itself is not ''always'' a vector field." There are trivial examples in which '''XY''' is a vector field. For example if '''X''' is identically zero. There are non-trivial examples too but lets give them after the due day of Homework III because I'm sure you will enjoy finding those examples by your self. |
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Latest revision as of 10:58, 21 October 2007
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Problem 1, p. 71.
There is a counterexample to the inverse implication in Problem 1, p. 71.
Let [math]\displaystyle{ X=\mathbb{R} }[/math] 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\mbox{ 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. Even taking [math]\displaystyle{ X }[/math] as any [math]\displaystyle{ T_2 }[/math] second countable topological space with the functional structure of constant functions will do the work.
Adding to the statement of the problem that the [math]\displaystyle{ F=(f_1,\ldots,f_n) }[/math] function is invertible we get a correct theorem. Maybe other weakening of this condition works.
Problem 4, p. 88.
Last line of problem 4 says "Also show that XY itself is not a vector field." and should say "Also show that XY itself is not always a vector field." There are trivial examples in which XY is a vector field. For example if X is identically zero. There are non-trivial examples too but lets give them after the due day of Homework III because I'm sure you will enjoy finding those examples by your self.
