0708-1300/Errata to Bredon's Book: Difference between revisions
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=Errata to Bredn's Book= |
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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 <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 |
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. |
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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Revision as of 10:02, 27 September 2007
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There is a counterexample to the inverse implication in Problem 1, p. 71.
Let be endowed with the ordinary topology (thus, it is Hausdorff and second countable). Let be an arbitrary connected open set in (that is, an interval). Let consists of all functions identically equal to constant. If is an arbitrary open set, then by theorem on structure of open sets in it is a union of countably many open intervals. We define to be the set of all real-valued functions which are constant on open intervals forming . The family forms a functional structure, as one can check. Furthermore, it satisfies the hypothesis of the theorem: every point has a neighborhood (we take an open interval containing ) such that there exists a function (we define it to be identically equal to ) such that a function is in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_X(U)} (it is identically equal to a constant by our definition) if and only if there exists a smooth function such that (if is given, then we define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h(x)=g} for all , if is given, then we take arbitrary smooth Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h:\mathbb{R} \to \mathbb{R}} , since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h \circ f} is identically equal to constant and, thus, is in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F_X(U)} ). Clearly, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (X,F_X)} is not a smooth manifold. Even taking as any Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_2} second countable topological space with the functional structure of constant functions will do the work.
Adding to the statement of the problem that the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle F=(f_1,\ldots,f_n)} function is invertible we get a correct theorem. Maybe other weakening of this condition works.
