Date: Tue, 29 Nov 1994 13:52:29 -0500 (EST) From: MRI@ids.net Subject: help with problems In analysis: If f:[0,1] is continuous. Show that there is an x in [0,1] such that f(x) = x. Problem #2: If A and B are open and closed sets respectively, of R^n, show B\A is closed and A\B is open. Definition: A\B are all objects that belong to A but do not to B. If B is a subset of A, A\B is called the complement of B in A.
Date: Tue, 29 Nov 1994 15:09:33 -0500 (EST) From: Sydney Foster <email@example.com> Subject: Re: help with problems Hello there! Thanks for writing Dr. Math. I just LOVE analysis (I'm taking a course in it now, too!), so I am happy to help you with these problems. First question: Do you mean f:[0,1] --> [0,1]? If the function maps from the interval [0,1] to [0,1] then the function will indeed have a fixed point. If not we can't conclude that there must be a fixed point. Can you think of a function that maps from [0,1] to the real numbers that doesn't have a fixed point? I'll assume you did mean for the function to map to [0,1]. I don't want to completely give this one away, because it would be fun for you to think of it yourself, but I'll head you in the right direction, and we'll see if you can take it from there. First, let's start with the obvious. If f(0) = 0 or f(1) = 1, we are done. So let's assume f(0)=a, where a is not 0 and f(1) = b where b is not 1. a is bigger than 0 and b is less than 1. Now consider the function g(x) = f(x) - x. Is this function continuous? What happens when g(x) = 0? See if you can play around with g(x) and also use the Intermediate Value Theorem to prove that f does have a fixed point. Second question: Since A is open then the complement of A is closed, right? Similarly, since B is closed, the complement of B is open. From your definition of A\B, A\B is the intersection of A and the complement of B, thus it is the intersection of two open sets. How does that help? I bet you can figure out the end of the argument. The argument for B\A is similar. Can you see how you would do it? I hope this helps. Please do write back if you have any questions about what I said or anything else (especially if it is analysis! :) ) --Sydney, "Dr. Analysis"
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