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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


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

[2] If a set M is a subset of a set N, then f(M) is a subset of f(N).

[3] 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 
Associated Topics:
College Analysis
College Logic

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