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Math Help - Monotone Functions

  1. #1
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    Monotone Functions

    I have attempted this proof and would appreciate any feedback. Thanks.

    Let I be some interval. If f:I\rightarrow \mathbb{R} is increasing and a\in I, prove that f(a) is an upper bound for \{f(x)|x<a\} and lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}.

    Proof: Let f be an increasing function. Then, by definition, for all x,y\in I, x<y\Rightarrow f(x)<f(y). Then, let x<a for a\in I and all x\in I. Thus, since x<a\Rightarrow f(x)<f(a) for all a,x\in I. Hence, a is an upper bound for \{f(x)|x<a\}.

    Then, since x<a for all x,a\in I, then a must be the least upper bound for \{f(x)|x<a\}. Hence, lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}.
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  2. #2
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    You have some confused notation here.
    You have a in two different ways.
    What you want is to note that f(a) is an upper bound for \{f(x):x<a\}.
    Now consider A=\sup\{f(x):x<a\} so that A\le f(a).
    Then show that  \displaystyle \lim _{x \to a^ -  } f(x) = A
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  3. #3
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    Major remarks.

    Hence, a is an upper bound for \{f(x)|x<a\}.
    Should be f(a), not a.

    The second claim is not really proved.
    Then, since x<a for all x,a\in I, then a must be the least upper bound for \{f(x)|x<a\}.
    Even the claim that f(a) (rather than a) is the least upper bound for \{f(x)\mid x<a\} is incorrect. Consider

    f(x)=<br />
\begin{cases}<br />
x & x<a\\<br />
x+1 & x\ge a<br />
\end{cases}<br />

    Then \sup\{f(x)\mid x<a\}=a, but f(a)=a+1.

    Even if f(a)=\sup\{f(x)\mid x<a\}, how does it help prove the following?
    Hence, lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}.
    After all, \lim_{x\nearrow a}f(x)=f(a) only if f is left-continuous at a. In my opinion, there is at least not enough details here that are expected from a proof of such a simple fact.

    Minor remarks.

    Then, let x<a for a\in I and all x\in I.
    You can't expect that x < a for all x in I. Since a is already fixed in the problem statement, this phrase should say, "Let x < a for some x in I".

    Thus, since x<a\Rightarrow f(x)<f(a) for all a,x\in I.
    Again, it is not the case that f(x) < f(a) holds for all a, x in I.
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  4. #4
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    Alright, so from what I gather, most of the first part was ok besides the fact that f(a) is the upper bound and not a.

    As for the second part...
    Let A=\sup\{f(x):x<a\} and A\leq f(a). Then,
    since f(a) is an upper bound for \{f(x)|x<a\} and f is increasing then there is no x>a such that f(x)>f(a). Thus, lim_{x\rightarrow a^-}f(x)\leq f(a). If we let this limit be called A then we have that lim_{x\rightarrow a^-}f(x)=A where A\leq f(a). Hence, we have that lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}.

    I'm not sure this is very strong but its my second attempt.
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  5. #5
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    Sorry, too many flaws. The second part of the problem -- proving \lim_{x\to a^-}f(x)=\sup\{f(x)\mid x<a\} -- has nothing to do with f(a). In the example I gave above, f(a) is different from \sup\{f(x)\mid x<a\}.

    Let A=\sup\{f(x):x<a\} and A\leq f(a).
    Are you assuming A\leq f(a)? In this case, since this assumption is not in the problem statement, you later need to consider the possibility that A>f(a). However, I thought that you have proved this in part 1. Therefore, you should say, "Let A=\sup\{f(x):x<a\}. We have already shown that A\leq f(a)." (To repeat, the second sentence is irrelevant.)

    Then, since f(a) is an upper bound for \{f(x)|x<a\} and f is increasing then there is no x>a such that f(x)>f(a).
    What happens to the right of a is irrelevant to the problem.

    Thus, lim_{x\rightarrow a^-}f(x)\leq f(a).
    I don't see how this follows from the previous discussion.

    If we let this limit be called A
    This is redefining A.

    then we have that \lim_{x\rightarrow a^-}f(x)=A where A\leq f(a). Hence, we have that \lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}.
    It looks like this conclusion is based on confusing two meanings of A.

    Let us define A=\sup\{f(x)\mid x<a\}, as you have done. Here is how I would prove that \lim_{x\to a^-}f(x)=A. We show that for every \varepsilon>0 there exists a \delta>0 such that when a-\delta<x<a and x\in I we have A-\varepsilon<f(x)\le A. Note that the second inequality, i.e., f(x)\le A, holds by the definition of \sup.

    Let \varepsilon be given. Also by the definition of \sup, A-\varepsilon is not an upper bound of \{f(x)\mid x<a\}; therefore, there exists an x_0<a such that f(x_0)>A-\varepsilon. But then, since f is increasing, for every x such that x_0<x<a we have A-\varepsilon<f(x_0)<f(x). Therefore, we can take \delta=a-x_0.
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    A quasi-continuous function is one for which the right and left hand limits exist at each point. It is a standard theorem in analysis that any monotonic function is quasi-continuous. That is the essence of this problem. You are asked to show that the limit on the left exist at each point.
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