1
$\begingroup$

How do I show this sequence is unbounded.

${b_j=j}$ from j=1 to infinity

By using the following definition.

${b_j}$ is called bounded if there exist $M>0$

such that $b_j<M$ for all $j\in$ Natural numbers.

So the sequence is

$1,2,3,4,5,......$

It seem obvious that it is unbounded because not matter what you pick for M

there will be an integer bigger than it.

But how do you show this?

$\endgroup$
1
  • 1
    $\begingroup$ This follows directly from the Archimedean property of $\Bbb{R}$. Have you ever seen it? $\endgroup$
    – Crostul
    Commented Aug 30, 2015 at 19:22

3 Answers 3

5
$\begingroup$

To prove that the sequence defined by $b_j = j$ is unbounded, let $M$ be an arbitrary natural number and observe that $M \not> M + 1 = b_{M+1}$ so that $M$ is not an upper bound of the $b_j$.

$\endgroup$
2
  • $\begingroup$ I see and "Any" can be all natural numbers M. $\endgroup$ Commented Aug 30, 2015 at 19:38
  • $\begingroup$ I wouldn't say it like that. I've reworded my answer. Does that help? $\endgroup$
    – Rob Arthan
    Commented Aug 30, 2015 at 19:44
1
$\begingroup$

For any $b_j=j$ exists N such that $N>b_j$. You can take $N=j+1$

P.S. You wrote a definition of bounded sequence.

$\endgroup$
3
  • $\begingroup$ yes I made mistake. $\endgroup$ Commented Aug 30, 2015 at 19:23
  • $\begingroup$ @Pacman: you need to show that there is no $N$ that is greater than every $b_j$. Your proof shows that for every $b_j$, some $N$ is greater than that $b_j$. $\endgroup$
    – Rob Arthan
    Commented Aug 30, 2015 at 19:32
  • $\begingroup$ Yeah! You are right. $\endgroup$
    – RFZ
    Commented Aug 30, 2015 at 19:34
1
$\begingroup$

Assume that the sequence $\mathbb{N}=\{1, 2, 3, \ldots\}$ is bounded by some real number $M$, that is, for all $n \in \mathbb{N},$ $n \leq M$. But notice that $M<\lfloor{M}\rfloor+1$, where $\lfloor{M}\rfloor \in \mathbb{N}$ is the greatest integer less than or equal to $M$. But this is a contradiction, since $\lfloor{M}\rfloor \in \mathbb{N}$ implies $\lfloor{M}\rfloor+1 \in \mathbb{N}$.

$\endgroup$
1
  • 1
    $\begingroup$ Your proof uses the fact that the floor function is well defined. But this is a consequence of what we want to prove, so it cannot be used. $\endgroup$
    – Crostul
    Commented Aug 30, 2015 at 19:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .