Problem from Real Analysis
Date: 10/05/2002 at 13:54:28 From: Anonymous Subject: Problem from Real Analysis This is the problem: Let X = A U B where A and B are subspaces of X. Let f:X->Y. Suppose that the restricted functions f|A:A->Y and f|B:B->Y are continuous. Show that if A and B are closed in X, then f is continuous. I am completely stuck on this problem. Any help would be appreciated. Thank you.
Date: 11/13/2002 at 20:10:03 From: Doctor Nitrogen Subject: Re: Problem from Real Analysis Hello, Stacy: This might be of some help. In many opening chapters in books on real Analysis or Point Set Topology, the following identity appears as an exercise:  "Let A and B be subsets of a set X. Prove that f(AUB) = f(A)Uf(B)." If f|A:--> Y is a continuous map of elements in A to image elements in Y, then the elements from A mapped into Y are actually mapped to a subset of image elements in Y, or mapped to f(A). So f maps those elements in A continously to f(A), a subset of Y. Now, if f|B:--> Y is a continuous map of elements of B to image elements in Y, then the elements from B mapped into Y are actually mapped to a subset of image elements in Y, or to f(B). So f maps those elements in B continuously to f(B), a subset of Y also. But the set f(A)Uf(B) is also a subset of Y, because the image elements in both f(A) and in f(B) are both in Y. But then the statement: "Either f continuously maps elements from A to f(A), or f continuously maps elements from B to f(B), or both," must be a true statement, meaning f(A)Uf(B) must be image elements in Y continuously mapped from X to Y. This being true, consider the identity f(AUB) = f(A)Uf(B), with X = AUB, and ask yourself the question: What does the set of all the inverse elements in X, for the elements in f(A)Uf(B) look like in X? Note that the inverse of f(A)Uf(B) (note I did NOT say inverse function here, just inverse) is the set of all elements in X whose images under f all lie in f(A)Uf(B), a subset of Y. Now ask yourself, what does that "preimage" inside X look like? If you can show it is X itself, then you will have proved that all elements in X get continuously mapped into Y, or, specifically, get mapped to f(A)Uf(B), meaning f: X --> Y = f:AUB --> Y is a continuous map. Of course, you will have to show that f(AUB) = f(A)Uf(B) One hint to do this is to note that f preserves set inclusion and set union, meaning  If a set M is a subset of a set N, then f(M) is a subset of f(N).  If a is an element of MUN, then f(a) is an element of f(M)Uf(N). I hope this helped answer the questions you had concerning your mathematics problem. You are welcome to return to The Math Forum/Doctor Math whenever you have any math-related questions. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/
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