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Topology and Analysis
Proving a function f is continuous given A U B = X
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[QUOTE="pasmith, post: 6552595, member: 415692"] [itex]X[/itex] is open by definition of being a topological space. Again by definition, [itex]Y[/itex] has at least two open subsets: [itex]Y[/itex] and [itex]\emptyset[/itex]. That is the idea, yes. What is the complement of [itex]X[/itex]? It must be with respect to some set, and that set must be [itex]X[/itex]. So the complement of [itex]X[/itex] is empty. As this is open, [itex]X[/itex] itself is closed. (Remember that "closed" is not the logical negation of "open" and that a set can be both.) If [itex]V \subset Y[/itex] is closed, then [itex]Y \setminus V[/itex] is open. Now [itex]f^{-1}(Y \setminus V)[/itex] is everything in the domain of [itex]f[/itex] whose image is not in [itex]V[/itex], which is exactly the complement of [itex]f^{-1}(V)[/itex]. I assume you mean in the case that [itex]A[/itex] and [itex]B[/itex] are not both open or both closed, since we know the result holds for those cases. Consider a non-constant function [itex]f: [0,1] \to \{0,1\}[/itex]. Then [itex]f[/itex] is necessarily continuous when restricted to [itex]A = f^{-1}(\{0\})[/itex] and [itex]B = f^{-1}(\{1\})[/itex] respectively, and clearly [itex][0,1] = A \cup B[/itex]. Can [itex]f[/itex] be continuous on [itex][0,1][/itex]? [/QUOTE]
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Proving a function f is continuous given A U B = X
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