P.5. Problem Set (Challenging)

# P.5. Problem Set (Challenging) - #1 We need some background...

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Unformatted text preview: #1 We need some background before we introduce the problem. We define a rational number to be any number of the form where p and q are integers such that p and q are relatively prime (the fraction is in lowest terms) and q is nonzero. Now, we define a function by Show that f is continuous at every irrational point yet discontinuous at every rational point. Remark: The above problem challenges our normal notion of what it means for a function to be “continuous.” Most people consider a continuous function as a “smooth” curve. For those interested, you can see Principles of Mathematical Analysis by Walter Rudin or (for a more friendly treatment) Understanding Analysis by Stephen Abbott for an example of a continuous, yet nowhere differentiable function. Below, we outline the proof of the task above. Proof Throughout the proof, we will be using the sequential criterion for continuity at a point . The theorem states: Theorem: Let be a function and let p be some point in D . Then f is continuous at p if and only if for every sequence of points in D satisfying , it follows that This is one of the few cases we can use the sequential criterion to produce positive results. Exercise: First, we show that f is discontinuous at every rational point. This is the easier part. Fix some rational number p in Exhibit two sequences in , and , both of which converge to p , but . This implies that f is not continuous at p . Now, let r be an irrational number. Then . Consider any sequence in that converges to r . We must show that . Of course, if this sequence contains irrational numbers, they can be disregarded as they are points where f is zero. Therefore, it is fine to simply consider sequence of rational numbers which converge to r . That is, where and are relatively prime nonnegative integers and is nonzero. Exercise: Prove that if , then . Note that this would imply the desired result. To prove this, suppose otherwise. It follows that is bounded. Show why this leads to a contradiction. Another interesting result is that this function is Riemann integrable on , but that is a bit too advanced to prove at the moment. #2 In this problem, we assume knowledge of the Nested Interval Theorem, which we state below: Theorem: Define a sequence of closed intervals, which satisfies for each positive integer n . Then is nonempty. If additionally, , then where c is some real number. That is, the intersection is a single point. Our overall goal is to prove the Bolzano Weierstrass Theorem, which states that every bounded set with infinitely many elements must have at least one accumulation point. Exercise: Let S be a subset of real numbers. Define what it means for a real number p to be an accumulation point of S . (I would accept either of two equivalent definitions; the sequential definition is left as an exercise below, and so please do note state it here.) Exercise: Is the set empty? How about the set ? Prove your assertions. If either of these sets is empty, have we found an exception to the Nested Interval Theorem? Explain your answer.found an exception to the Nested Interval Theorem?...
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P.5. Problem Set (Challenging) - #1 We need some background...

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