# Thread: Equivalence of the statements

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. Originally Posted by Mathick
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 $C$.

Originally Posted by Evgeny.Makarov
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?

4. Yes, I do.