Hei.
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Show that the requirement in the definition of uniform continuity can be rephrased as follows, in terms of diameters of sets: To every [tex]\epsilon > 0[/tex] there exists a [tex]\delta > 0[/tex] such that diam[tex]f(E) < \epsilon[/tex] for all [tex]E \subset X[/tex] with diam [tex]E < \delta[/tex].
LØSNINGSFORSLAG (på engelsk):
Assume [tex]f[/tex] is a uniformly continuous function. Then there exists a [tex]\epsilon > 0[/tex] and a [tex]\delta > 0[/tex]such that [tex]|f(x) - f(y)| < \epsilon[/tex] for all [tex]x[/tex] and [tex]y[/tex] where [tex]|x - y| < \delta[/tex]. Now, given the metric space [tex]X[/tex], suppose there exists a subset [tex]E \subset X[/tex] with the property that diam[tex]E < \delta[/tex]. It then follows that for all [tex]x,y \in E[/tex], [tex]|x - y| <[/tex]diam [tex]E = \delta[/tex]. So if [tex]f[/tex] is uniformly continuous on [tex]E[/tex], this implies that since [tex]|f(x) - f(y)| < \epsilon[/tex], we have that diam[tex]f(E) < \epsilon[/tex].
Conversely, given a subset [tex]E \subset X[/tex] where [tex]X[/tex] is a metric space. Let [tex]\epsilon > 0[/tex], and choose [tex]\delta > 0[/tex] such that diam[tex]E < \delta[/tex] and diam[tex]f(E) < \epsilon[/tex]. Fix two arbitrary points [tex]x, y \in E[/tex]. Then, since diam[tex]E < \delta[/tex], it follows that [tex]|x-y| < \delta[/tex], and since diam[tex]f(E) < \epsilon[/tex], it follows that [tex]|f(x) - f(y)| < \epsilon[/tex]. Since [tex]x[/tex] and [tex]y[/tex] were chosen arbitrarily, it follows that [tex]f[/tex] is uniformly continuous.
QED.
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