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Advanced Calc 1 Question about Closed Sets

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose f: [a,b] -> R and g: [a,b] -> R are both continuous. Let T = {x: f(x)=g(x)}. Prove that T is closed.


    2. Relevant equations

    If a function f: D -> R is continuous, then for each E > 0 there exists a delta >0 such that if |x-x0| < delta (for x, x0 in D) then |f(x)-f(x0)| < E.

    Definition of closed: A set is closed if it contains all of its accumulation points.

    Definition of open: A set is closed if for each element of the set, there exists a neighborhood Q of the element such that Q is a subset of the set.

    Theorem: T is closed iff R\T is open.

    3. The attempt at a solution

    There are three possibilities for the set T = {x:f(x)=g(x)}.

    1) If f and g are equal then f(x) = g(x) for all x in [a,b], so T = [a,b]. In this case, R/T = (-infinity, a) U (b, infinity).

    For any x in (-infinity, a), take E such that 0 < E < (a-x). Then x+E < a, and clearly -infinity< x-E; so (x-E,x+E) is a subset of (-infinity,a). So for every x in (-infinity,a) there is a neighborhood Q of x such that Q is a subset of (-infinity,a); so (-infinity,a) is open.

    For any x in (b, infinity) take E such that 0 < E < (x-b). Then E-x < -b and dividing through by -1 gives b < x-E, and clearly x+E < infinity; so (x-E,x+E) is a subset of (b, infinity). So (b, infinity) is open. The union of open sets is open, so (-infinity,a) U (b,infinity) is open; so R\T is open which means that T is closed.

    2) If f and g never intersect then T = null set. In this case T has no accumulation points and therefore contains all of its accumulation points; so T is closed.

    3) If T is nonempty but not equal to [a,b], then T = {x1,x2,...,xn} (arranged from smallest to biggest element). If T has one element then T has no accumulation points and thus is closed. So assume T has two or more elements. Then R\T = (-infinity, x1) U (x1,x2) U ... U (xn, infinity).

    For any x in (-infinity, x1), take E such that 0 < E < (x1-x). Then x+E < x1, and clearly -infinity< x-E; so (x-E,x+E) is a subset of (-infinity,x1). So for every x in (-infinity, x1) there is a neighborhood Q of x such that Q is a subset of (-infinity, x1); so (-infinity, x1) is open.

    For any x in (x1, x2), take E such that 0 < E < min{(x-x1, x2-x)}; then E < x-x1 and E < x2-x . Then x1 < x-E and x+E < x2, so (x-E,x+E) is a subset of (x1, x2) and thus (x1, x2) is open.

    Use the same method that was used to show that (x1, x2) is open for each following interval up to (xn, infinity).

    For any x in (xn, infinity), take E such that 0 < E < (x-xn). Then xn < x-E and clearly x+E < infinity, so (x-E,x+E) is a subset of (xn, infinity), so (xn, infinity) is open. So R\T is equal to a union of open sets, which is open; so R\T is open, and therefore T is closed.


    **I was just wondering if this was the way I was supposed to do the proof, because I was given the fact that f and g are continuous functions and I don't use that fact in my proof. Also, are the three possibilities I have for T correct, and do the other parts of my proof seem valid?

    Thank you!
     
  2. jcsd
  3. Oct 15, 2008 #2

    Dick

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    Show that the set where f(x) is NOT equal to g(x) is open, because f and g are continuous. The inverse image of that open set is open. The complement of that open set is where f(x)=g(x). The complement of an open set is closed. So the set of x where f(x)=g(x) is closed. Case closed.
     
  4. Oct 16, 2008 #3
    That makes sense, but how do I show that the set where f(x) is not equal to g(x) is open?
     
  5. Oct 16, 2008 #4

    Dick

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    epsilon-delta sort of arguments? Pick epsilon=|f(x)-g(x)|/2.
     
  6. Oct 16, 2008 #5
    If I pick epsilon = |f(x)-g(x)|/2 then that would help me show that the set |f(x):f(x) is not equal to g(x)| is open; and then can I just say that because f and g are continuous the inverse image set |x:f(x) is not equal to g(x)| is open? Which theorem says that the inverse of an open set is open?
     
  7. Oct 16, 2008 #6

    Dick

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    If you don't have that theorem you can prove is easily (use epsilons and deltas again). It's so basic it's sometimes used as the definition of continuity.
     
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