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Thread: Sequence

  1. #1
    Oct 2008


    Suppose $\displaystyle f $ is a real valued function whose domain contains the interval $\displaystyle [a,b] $. Assume that $\displaystyle f $ is unbounded on $\displaystyle [a,b] $.

    (a) Prove that there exists a sequence $\displaystyle (y_n) $ in $\displaystyle [a,b] $ that converges to $\displaystyle y \in [a,b] $ such that for every $\displaystyle n \in \mathbb{N} $, $\displaystyle |f(y_n)| > n $.

    Isn't this just the definition of an unbounded sequence? Do we have to actually construct one? What would be the easiest way to do this?

    (b) Using this unbounded sequence how would we show that the set of Riemann sums of $\displaystyle f $ corresponding ot $\displaystyle P $ is an unbounded set of real numbers?
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  2. #2
    Mar 2009
    GENERALLY speaking ,usually in forming sequences in analysis there are two basic ways:

    1)By using the axiom of Choice.

    2)By using the inductive method

    The following steps are a summary of the process in forming the sequence in concern,by using the axiom of Choice.

    step 1) PROVE: For all nεΝ there exists an ,xε [a,b] and such that |f(x)|>n,by using the fact that f is unbounded over [a,b]

    step 2) Form the sets $\displaystyle A_{n}$={x:xε[a,b],|f(x)|>n,nεN}

    Step 3) By using step 1 prove : For all neΝ, $\displaystyle A_{n}\neq\emptyset$.

    step 4) Now use the axiom of Choice to form a sequence {$\displaystyle x_{n}$} in [a,b] and such that |f($\displaystyle x_{n}$)|>n ,nεN.

    Step 5) By the Bolzano Weiestrass theorem ,since the sequence {$\displaystyle x_{n}$} is bounded in [a,b] ,there exists an accumulation point ,v in [a,b].

    step 6)Prove:For all nεN ,THERE exists a ,$\displaystyle y\neq v$,yε{$\displaystyle x_{n}:n\in N$} and such that |y-v|<1/n,by using the fact that v is an accumulation point.

    step 7)Form the sets, $\displaystyle S_{n}$ ={y:|y-v|<1/n,nεΝ,$\displaystyle y\neq v$,yε{$\displaystyle x_{n}$:nεN}}

    Step 8) Prove: for all nεN $\displaystyle S_{n}\neq\emptyset$ by using step 6.

    step 9) Use again the axiom of Choice to form the sequence {$\displaystyle y_{n}$} in [a,b],such that |f($\displaystyle y_{n}$)|>n ,$\displaystyle y_{n}\neq v$ and |$\displaystyle y_{n}$-v|<1/n for all nεΝ

    step 10) prove ,$\displaystyle \lim_{n\rightarrow\infty}{y_{n}}=v$.

    ALTERNATIVELY,by using the inductive process we can say:

    After we have proved that ,for all n belonging to natural Nos ,there exists an ,x belonging to [a,b] and such that |f(x)|>n.

    For n= 1 ,we pick $\displaystyle x_{1}$,such that |f($\displaystyle x_{1}$)|>1.

    For n=2 ,we pick $\displaystyle x_{2}>x_{1}$,such that |f($\displaystyle x_{2}$)|>2.

    Suppose we have found:

    $\displaystyle x_{1}<x_{2}<x_{3}.......................<x_{n}$ all in [a,b] and such that |f($\displaystyle x_{n}$)|>n .

    We select $\displaystyle x_{n+1}>x_{n}$.We can thus proceed inductively to construct the sequence {$\displaystyle x_{n}$},which is increasing and bounded from above by ,a.Hence there exists a,y in [a,b] and such,$\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=y$.

    So we have proved the existence of a sequence {$\displaystyle x_{n}$} in [a,b] such that |f($\displaystyle x_{n}$|>n and $\displaystyle \lim_{n\rightarrow\infty}{x_{n}}=y$
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