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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 <syd@sccs.swarthmore.edu>
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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