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Math Help - increasing function

  1. #1
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    increasing function

    Let f be strictly increasing on a subset S of R. If f(S) is open. Prove that f is continuous on S.
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    Quote Originally Posted by Kat-M View Post
    Let f be strictly increasing on a subset S of R. If f(S) is open. Prove that f is continuous on S.
    This took a bit of thought, but I figured an epsilon chase works best.

    Let s_0\in S and \epsilon>0 be given. Then f(s_0) is in f(S) which is open and so there exists an \epsilon>\gamma>0 such that (f(s_0)-\gamma, f(s_0)+\gamma)\subseteq f(S).

    Now for the trickier bits, I'll give a few helpful nudges. What we want to show is that there exists a delta such that |f(s)-f(s_0)|<\epsilon whenever |s-s_0|<\delta. We already found an epsilon neighborhood of f(s_0), so we want to construct some points in S whose images lie in this neighborhood. This is how we want to go about picking our delta.

    Big idea 1:
    Spoiler:
    Given points in the neighborhood of f(s_0), then they are in f(S). What is the definition of being in f(S)?


    Big idea 2:
    Spoiler:
    If we construct points x,y such that f(x)<f(s_0)<f(y), what can we say about the relationship of x and y?


    Big idea 3:
    Spoiler:
    What about images of points between x and y?
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  3. #3
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    Quote Originally Posted by Kat-M View Post
    Let f be strictly increasing on a subset S of R. If f(S) is open. Prove that f is continuous on S.
    Let x\in S, \underline{x}_n\in S be a (weakly) increasing sequence with \lim_{n\to \infty}\underline{x}_n=x, and \overline{x}_n\in S be a (weakly) decreasing sequence with \lim_{n\to \infty}\overline{x}_n=x.
    The images f(\underline{x}_n) and f(\overline{x}_n) of these two sequences are (weakly) increasing and decreasing in f(S), respectively; and both are bounded by f(x), therefore the limits \underline{y}:=\lim_{n\to \infty}f(\underline{x}_n) and \overline{y}:=\lim_{n\to \infty}f(\overline{x}_n) both exist and we have \underline{y}\leq f(x)\leq \overline{y}.

    But the assumption that \underline{y}<f(x) or f(x)<\overline{y} contradicts openness of f(S) combined with monotonicity of f. (Why?) Therefore \underline{y}=\lim_{n\to\infty}f(\underline{x_n})=  f(x)=\lim_{n\to\infty}f(\overline{x}_n)=\overline{  y}, which means that f is continuous at x.

    Since x\in S was arbitrary, continuity of f on the whole of S follows.
    Last edited by Failure; August 11th 2009 at 01:01 AM.
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    openness

    But the assumption that \underline{y}<f(x) or f(x)<\overline{y} contradicts openness of f(S) combined with monotonicity of f. (Why?) Therefore \underline{y}=\lim_{n\to\infty}f(\underline{x_n})=  f(x)=\lim_{n\to\infty}f(\overline{x}_n)=\overline{  y}, which means that f is continuous at x.

    Since x\in S was arbitrary, continuity of f on the whole of S follows.
    why assuming \underline{y}<f(x) or f(x)<\overline{y} contradicts the openness and monotonisity?
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  5. #5
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    Quote Originally Posted by Kat-M View Post
    why assuming \underline{y}<f(x) or f(x)<\overline{y} contradicts the openness and monotonisity?
    Well, suppose for example that \underline{y}<f(x). If this were the case, then because of f(x)\in f(S) and the openness of f(S) there would exist an x_0\in S with \underline{y}<f(x_0)<f(x). By monotonicity of f it follows that \underline{x}_n< x_0 < x for all n, and therefore we get the contradiction x=\lim_{n\to\infty}\underline{x}_n\leq x_0<x (which is to say x<x).

    So, \underline{y}< f(x) cannot be true, and since we know that \underline{y}\leq f(x), the only remaining possiblity is that \underline{y}=f(x).
    An almost identical argument shows that f(x)=\overline{y}.
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    open set

    Let s_0\in S and \epsilon>0 be given. Then f(s_0) is in f(S) which is open and so there exists an \epsilon>\gamma>0 such that (f(s_0)-\gamma, f(s_0)+\gamma)\subseteq f(S)
    f(S) is open so i understand that for every point f(s_0) of f(S), there is an open set that lies in f(S). but how do we know that the radius \gamma is less than the given \epsilon? i thought what we know is there is at least one open set centered at f(s_0) with some radius that lies completely in f(S). i didnt know we could pick any radius we want
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  7. #7
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    Quote Originally Posted by Kat-M View Post
    f(S) is open so i understand that for every point f(s_0) of f(S), there is an open set that lies in f(S). but how do we know that the radius \gamma is less than the given \epsilon? i thought what we know is there is at least one open set centered at f(s_0) with some radius that lies completely in f(S). i didnt know we could pick any radius we want
    Suppose that \varepsilon >0 is given. You seem to agree that because f(s_0)\in f(S) and f(S) is open, that there exists at least one \gamma >0 such that (f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S).

    Your only worry seems to be that \varepsilon>\gamma>0 might not be true. Indeed, that might be a problem. But any \gamma' that satisfies 0<\gamma'\leq \gamma also has the property that (f(s_0)-\gamma',f(s_0)+\gamma')\subseteq f(S), simply because (f(s_0)-\gamma',f(s_0)+\gamma')\subseteq (f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S).
    Thus, the possibility that our initial value for \gamma might in fact not be smaller than the given \varepsilon is not a particularly big worry, because we can always replace it with \min(\gamma,\varepsilon/2) if need be. Indeed, we would do this in a "preventive" fashion, just to make sure \varepsilon>\gamma>0 holds (afterwards).
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    open set

    So if there is an open set centered at s_0 with radius \gamma, then any \gamma ' \leq \gamma can be a radius of an open set around s_0? which means that there are infinitely many open sets with radius less than \gamma.
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  9. #9
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    Quote Originally Posted by Kat-M View Post
    So if there is an open set centered at s_0 with radius \gamma, then any \gamma ' \leq \gamma can be a radius of an open set around s_0? which means that there are infinitely many open sets with radius less than \gamma.
    In \mathbb{R} there is always an infinite number of open intervalls around any given number. For example, all the intervalls (f(s_0)-\tfrac{1}{n},f(s_0)+\tfrac{1}{n}) are open and centered around the same number f(s_0). They can get arbitrarily small - and yet they are all open.
    The particular problem in your case was that those intervalls had to be contained in a given open set f(S). But once one such open intervall, centered around a given number, is completely contained in that given open set f(S), say, then an infinite number of smaller open intervalls, all centered around the given number, are also contained in that given set, f(S).
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