0708-1300/Errata to Bredon's Book: Difference between revisions
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=Errata to Bredn's Book= |
=Errata to Bredn's Book= |
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There is a counterexample to the inverse implication in Problem 1, p. 71. |
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(Damir, do you want to start?) |
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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 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. |
Revision as of 10:20, 26 September 2007
Errata to Bredn's Book
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.