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Real Analysis: Continuity & Intervals

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    If the domain of a continuous function is an interval, show that the image is an interval.



    2. Relevant equations
    Theorem from book: f is a cont. function with compact domain D, then f is bounded and there exists points y and z such that f(y) = sup{ f(x) | x ∈ D} and f(z) = inf{ f(x) | x ∈ D}
    The intermediate value theorem.


    3. The attempt at a solution

    Case 1: D=[a,b]
    Since D is compact and by hypothesis, f is continuous we know that there exist points y and z in D such that f(y) = sup{ f(x) | x ∈ D} and f(z) = inf{ f(x) | x ∈ D}. By definition of sup and inf these are the greatest and least points of the image of f. And so, by the intermediate value theorem, we know that any t in between f(z) < t < f(y), there exists an x in D such that f(x)=t. So, we can write the interval of the image as [f(z),f(y)].

    Case 2: D=(a,b)
    We can write (a,b) as an infinite union of closed sets. For example ∪[a+1/m, b-1/m] from m=1 to infinity.
    By the previous case we know that each [a+1/m, b-1/m] attains a sup and inf value for all 1/m. Now we can simply take the max( sup{ f(x) | x ∈ [a+1, b-1]}, sup{ f(x) | x ∈ [a+(1/2), b-(1/2)]}, sup{ f(x) | x ∈ [a+(1/3), b-(1/3)}, ... ) and call it p. Similarly we can take the min( inf{ f(x) | x ∈ [a+1, b-1]}, inf{ f(x) | x ∈ [a+(1/2), b-(1/2)]}, inf{ f(x) | x ∈ [a+(1/3), b-(1/3)}, ... ) and call it q. This will give us an open interval in which all values of the image lie. And so, the interval of the image of f is (q,p).

    Is this the right way to prove this? It seems right to me but I also have my doubts haha. Thanks!
     
  2. jcsd
  3. Oct 29, 2011 #2
    What about half open intervals?
     
  4. Oct 30, 2011 #3
    Suppose for a second that I did not need to worry about half open intervals is my above proof right then?

    Now suppose I do need it in a more general case. What about this.

    Given any interval besides the trivial case [x,x], there exits two points, y and z, within the interval such that y < z

    So, using these two points we know that there are infinitely many p's within y < p < z. By the definition of continuity we can find points, all in the image of the function such that f(y)=a and f(z)=b, using the intermediate value theorem we know that for any k in between a and b we can find points within y and z such that f(p)=k for all k in a<k<b and for some p in the domain. Thus the image of f is an interval.
     
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