This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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?...
View Full
Document
 Spring '08
 PLOTKIN
 Integers

Click to edit the document details