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Discrete metric and continuity equivalence
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[QUOTE="pasmith, post: 4505861, member: 415692"] You don't need to derive a contradiction; you can show directly that f as defined is not continuous at x: If [itex]\epsilon < d'(0,1)[/itex] then for every [itex]\delta > 0[/itex] there exists [itex]y \in B(x,\delta)[/itex] with [itex]d'(f(x),d(y)) > \epsilon[/itex] and thus [itex]f[/itex] is not continuous at [itex]x[/itex]. That's sufficient to complete the proof: the object is to show that if X is not discrete then there exists a metric space Y such that there exists a function from X to Y which is not continuous, which is equivalent to showing that if every function from X to an arbitrary metric space is continuous then X is discrete. A direct proof would probably start with the observations that if every function from X to an arbitrary metric space is continuous then every function from X to {0,1} is continuous, and that every such function can be written as [tex] f : x \mapsto \left\{\begin{array}{rc} 0 & \mbox{ if } x \in U \\ 1 & \mbox{ otherwise} \end{array} \right.[/tex] for some [itex]U \subset X[/itex]. [/QUOTE]
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Discrete metric and continuity equivalence
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