0708-1300/Errata to Bredon's Book - Revision history

Franklin at 15:58, 21 October 2007

2007-10-21T15:58:12Z

← Older revision		Revision as of 11:58, 21 October 2007
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	'''Problem 4, p. 88.'''		'''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 ~~example~~ by your self.		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.

Franklin at 15:50, 21 October 2007

2007-10-21T15:50:40Z

← Older revision		Revision as of 11:50, 21 October 2007
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	{{0708-1300/Navigation}}		{{0708-1300/Navigation}}

			'''Problem 1, p. 71.'''

	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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	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.		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.


			'''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 example by your self.

Drorbn at 15:02, 27 September 2007

2007-09-27T15:02:00Z

← Older revision		Revision as of 11:02, 27 September 2007
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			{{0708-1300/Navigation}}
	=Errata to Bredn's Book=

	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.

	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\ 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.		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.
	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.		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.

Franklin at 13:59, 27 September 2007

2007-09-27T13:59:21Z

← Older revision		Revision as of 09:59, 27 September 2007
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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.

	Let <math>X=\mathbb{B}</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\ 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.		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\ 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.
			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.

			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.

Franklin at 13:43, 27 September 2007

2007-09-27T13:43:39Z

← Older revision		Revision as of 09:43, 27 September 2007
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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.

	Let <math>X=\mathbb{B}</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\~~text{~~ 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.		Let <math>X=\mathbb{B}</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\ 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.

Franklin at 13:41, 27 September 2007

2007-09-27T13:41:49Z

← Older revision		Revision as of 09:41, 27 September 2007
Line 3:		Line 3:
	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.

	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.		Let <math>X=\mathbb{B}</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\text{ 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.

Franklin at 13:39, 27 September 2007

2007-09-27T13:39:09Z

← Older revision		Revision as of 09:39, 27 September 2007
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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.

	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 ~~neighbourhood~~ (we take an open interval containing x) such that there exists a function f \in F_X(U) (we define it to be identically equal to 1) such that a function g:U \to \mathbb R is in F_X(U) (it is identically equal to a constant by our definition) if and only if there exists a smooth function h such that g=h \circ f (if g is given, then we define h(x)=g for all x, if f is given, then we take arbitrary smooth h:\mathbb R \to \mathbb R, since h \circ f is identically equal to constant and, thus, is in F_X(U)). Clearly, (X,F_X) is not a smooth manifold.		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.

Dkinz: /* Errata to Bredn's Book */

2007-09-26T15:20:48Z

Errata to Bredn's Book

← Older revision		Revision as of 11:20, 26 September 2007
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	=Errata to Bredn's Book=		=Errata to Bredn's Book=

			There is a counterexample to the inverse implication in Problem 1, p. 71.
	(Damir, do you want to start?)

			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 neighbourhood (we take an open interval containing x) such that there exists a function f \in F_X(U) (we define it to be identically equal to 1) such that a function g:U \to \mathbb R is in F_X(U) (it is identically equal to a constant by our definition) if and only if there exists a smooth function h such that g=h \circ f (if g is given, then we define h(x)=g for all x, if f is given, then we take arbitrary smooth h:\mathbb R \to \mathbb R, since h \circ f is identically equal to constant and, thus, is in F_X(U)). Clearly, (X,F_X) is not a smooth manifold.

Drorbn: 0708-1300/Errata to Bredn's Book moved to 0708-1300/Errata to Bredon's Book

2007-09-25T13:59:45Z

0708-1300/Errata to Bredn's Book moved to 0708-1300/Errata to Bredon's Book

← Older revision

Revision as of 09:59, 25 September 2007

Franklin at 13:06, 25 September 2007

2007-09-25T13:06:02Z

New page

=Errata to Bredn's Book=

(Damir, do you want to start?)