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Math Help - Is this proof correct?

  1. #1
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    Is this proof correct?

    prove that if f(x) is monotone increasing on an open interval (a,b) and x_0 is any point in (a,b), then lim_{x \to x_0+} f(x) exists and satisfies lim_{x to x_0+} f(x) >= f(x_0). (from the right side)

    Is this proof correct? if not could someone please correct it/ explain what I'm doing wrong?

    Fix x_0 in (a,b). First consider x in ( x_0,b] then f(x)>= f( x_0) so inf f(x) ( x_0<x<b) >=f( x_0)

    Claim: lim f(x)=inf f(x) ( x_0<=x<b) . Let S=inf f(x). Let \epsilon>0. Then there exists x \epsilon in ( x_0,b) such that f(x \epsilon)<S- \epsilon.

    But if x is in (x \epsilon, x_0) then f(x \epsilon)>=f(x) so S<=f(x)<S- \epsilon

    Hence |f(x)-s|< \epsilon for all x in (x \epsilon, x_0).

    therefore, lim f(x) (x-> x_0+) exists and equals S since S>=f( x_0) we get lim (x-> x_0+) >= f( x_0)
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  2. #2
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    Frankly, I find your post very hard to read. Let me give you some alternative ways to see this.

    If a<c<b then let f(c+)=\inf\{f(x):x\in (c,b]\} and f(c-)=\{\sup\{f(x):x\in [a,c)\}.
    It is not hard the see that for an increasing function each of these exists.

    Clearly  \lim _{x \to c^ +  } f(x) = f(c + ) \ge f(c) and likewise Clearly  \lim _{x \to c^ -  } f(x) = f(c - ) \le f(c).

    I hope this helps you with notation.
    I will be glad to help with questions you may have.
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  3. #3
    Member mabruka's Avatar
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    I think the property yo uare using in your wrong has a little mistake:

    Claim: lim f(x)=inf f(x) (<=x<b) . Let S=inf f(x). Let >0. Then there exists x in (,b) such that f(x)<S-.
    I think it should say:


    For ever \epsilon there exists x_\epsilon \in  (x_0,b) such that  f(x_\epsilon) <\inf\{f(x):x_0\leq x \leq b \}+\epsilon


    what do you think?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    Whatever you said..
    I think you mean to prove that if a,b)\mapsto\mathbb{R}" alt="fa,b)\mapsto\mathbb{R}" /> is increasing then \lim_{x\to x_0^-}f(x)=\sup_{x\in (a,x_0)}f(x). To see this let \varepsilon>0 be given. By virtue of the definition of supremum we know that there exists some f(\xi)\in f((a,x_0)) such that \sup_{x\in(a,x_0)}f(x)-\varepsilon<f(\xi)\implies \sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon. Notice though by definition we have that \sup_{x\in(a,x_0)}f(x)-f(\xi)=\left|\sup_{x\in(a,x_0)}f(x)-f(\xi)\right|. But, by f's monotonicity we have that for any \xi<\xi'<\xi that f(\xi)<f(\xi')<\sup_{x\in(a,x_0)}f(x)\implies \sup_{x\in(a,x_0)}f(x)-f(\xi')<\sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon. In particular if x\in(\xi,x_0)\implies \left|\sup_{x\in(a,x_0)}f(x)-f(x)\right|=\sup_{x\in(a,x_0)}f(x)-f(x)<\sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon. It follows that \lim_{x\to x_0^-}f(x)=\sup_{x\in(a,x_0)}f(x).
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