Theorem: Suppose f: D => R (all real numbers) is continuous with D closed and bo
ID: 1891236 • Letter: T
Question
Theorem: Suppose f: D => R (all real numbers) is continuous with D closed and bounded. Then f is uniformly continuous on D.a. Assume f is not uniformly continuous on D. Then there exist E > 0 such that for each n, there are an, bb E D. such that l an - bn l < 1/n and L f(an)- f(bn) L > or equal to E. Explain why this must be true if f is not uniformly continuous on D. Show complete proof.
b. There is a sequence of positive integer {k} such that the sub-sequence {ak} and {bk} both converge, call the limits a and b respectively. Explain why this must be true. Show complete proof.
c. Both a and b belong to D and a = b. Explain why this is true and why it leads to a contraction. Finish the proof.
Explanation / Answer
$00$. Let $E_0=Eigcap[-k_0,k_0]^n$, $E_1=[-k_0,k_0]^n-E_0$. By definition of Lebesgue outer measure, for all $ arepsilon>0$ we have a sequence of open intervals, ${I_k}$, covering $E_1$, satisfying $sum_{k=1}^{infty}mI_kRelated Questions
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