Chapter 5: Problem 6
Let \(a
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Chapter 5: Problem 6
Let \(a
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Let \(\delta\) be the gauge on \([0,1]\) defined by \(\delta(0):=\frac{1}{4}\) and \(\delta(t):=\frac{1}{2} t\) for \(t \in(0,1]\). (a) Show that \(\dot{\mathcal{P}}_{1}:=\left\\{\left(\left[0, \frac{1}{4}\right], 0\right) \cdot\left(\left[\frac{1}{4} \cdot \frac{1}{2}\right], \frac{1}{2}\right),\left(\left[\frac{1}{2}, 1\right], \frac{3}{4}\right)\right\\}\) is \(\delta\) -fine. (b) Show that \(\dot{P}_{2}:=\left\\{\left(\left[0, \frac{1}{4}\right], 0\right),\left(\left[\frac{1}{4}, \frac{1}{2}\right), \frac{1}{2}\right),\left(\left[\frac{1}{2}, 1\right], \frac{3}{5}\right) \mid\right.\) is not \(\delta\) -fine.
If \(I:=[a, b]\) is an interval and \(f: I \rightarrow \mathbb{R}\) is an increasing function, then the point \(a\) [respectively, \(b\) ] is an absolute minimum [respectively, maximum] point for \(f\) on \(I\). If \(f\) is strictly increasing, then \(a\) is the only absolute minimum point for \(f\) on \(I\).
Prove that if \(f\) is uniformly continuous on a bounded subset \(A\) of \(\mathbb{R}\), then \(f\) is bounded on \(A\).
If \(f:[0,1] \rightarrow \mathbb{R}\) is continuous and has only rational [respectively, irrational] values, must \(f\) be constant? Prove your assertion.
If \(g(x):=\sqrt{x}\) for \(x \in[0,1]\), show that there does not exist a constant \(K\) such that \(|g(x)| \leq K|x|\) for all \(x \in[0,1]\). Conclude that the uniformiy continuous \(g\) is not a Lipschitz function on \([0,1]\).
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