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  1. MHB Apprentice

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    #1
    Let $A$ be the statement 'The sequence $(x_n)$ is bounded'. Use your mathematical intuition to decide which of the following statements are equivalent to $notA$. You don't need to give a justification.

    (a) There exists $C > 0$ such that $\left| x_n \right| > C$ for some $n \in \Bbb{N}$.
    (b) For all $C > 0$ there exists an index $n \in \Bbb{N}$ such that $\left| x_n \right| \ge C$.
    (c) For infinitely many $n \in \Bbb{N}$ we have $\left| x_n \right| \ge n$.
    (d) For all $C > 0$ there exist infinitely many terms of $(x_n)$ with $\left| x_n \right| > C$.

    By definition of bounded sequence: $\exists \; C > 0 \quad \forall \; n \in \Bbb{N} \quad \left| x_n \right| \le C$

    I know that negation is: $\forall \; C > 0 \quad \exists \; n \in \Bbb{N} \quad \left| x_n \right| > C$

    So I think that all these statements are NOT equivalent to $notA$. But I am not sure. Can you help me?

  2. MHB Master
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    #2
    Quote Originally Posted by Mathick View Post
    By definition of bounded sequence: $\exists \; C > 0 \quad \forall \; n \in \Bbb{N} \quad \left| x_n \right| \le C$

    I know that negation is: $\forall \; C > 0 \quad \exists \; n \in \Bbb{N} \quad \left| x_n \right| > C$
    How is this different from (b)? If you mean the difference is in $\ge$ vs $>$, then note that your statement of negation obviously implies (b). But suppose (b) holds and you want to prove
    \[
    \forall C > 0\;\exists n \in \Bbb{N}\; |x_n| > C.\qquad(*)
    \]
    Fix an arbitrary $C$ and feed $C+1$ to (b) (i.e., replace $C$ in (b) with $C+1$ where $C$ is the value you fixed). It will spit out some $n$ such that $|x_n|\ge C+1$. So $|x_n|>C$ for the same $n$.

    It is also trivial that (d) implies (b) and (*). But if (b) holds and you want to prove (d), you fix an arbitrary $C$ and then instantiate the universally quantified $C$ in (b) with $C+1$, $C+2$, $C+3$ and so on. This way you get infinitely many sequence members that exceed $C$.

  3. MHB Apprentice

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    Quote Originally Posted by Evgeny.Makarov View Post
    How is this different from (b)? If you mean the difference is in $\ge$ vs $>$, then note that your statement of negation obviously implies (b). But suppose (b) holds and you want to prove
    \[
    \forall C > 0\;\exists n \in \Bbb{N}\; |x_n| > C.\qquad(*)
    \]
    Fix an arbitrary $C$ and feed $C+1$ to (b) (i.e., replace $C$ in (b) with $C+1$ where $C$ is the value you fixed). It will spit out some $n$ such that $|x_n|\ge C+1$. So $|x_n|>C$ for the same $n$.

    It is also trivial that (d) implies (b) and (*). But if (b) holds and you want to prove (d), you fix an arbitrary $C$ and then instantiate the universally quantified $C$ in (b) with $C+1$, $C+2$, $C+3$ and so on. This way you get infinitely many sequence members that exceed $C$.

    So you claim that (b) and (d) are equivalent to $ notA $ and (a) and (c) are not, don't you?

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    #4
    Yes, I do.

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