This is one of exercise questions in our textbook. Is the sequence $3n$ is bounded? Prove or disprove. It first seems that I should use Archimedean property. Let $B\in \mathbb R$, and suppose $3n$ is bounded. Then, $3n \le B$. But, I am stuck here because to use the property, I think that the inequality should be $\ge$.
Could you give some hint to prove this?
Thank you in advance!
On
We will first introduce some common notation. To express the sequence $(3,6,9,\ldots)$, we typically write $\{3n\}_{n\in\mathbb{N}}$ where $\mathbb{N}=\{1,2,3,\ldots\}$ is the set of all natural numbers.
Now we will prove that the sequence $\{3n\}_{n\in\mathbb{N}}$ is not bounded. Suppose, for a contradiction, that $\{3n\}_{n\in\mathbb{N}}$ is bounded. By definition, there exists a positive real number $B$ such that $$|3n|\le B \mbox{ for all } n\in\mathbb{N}.$$ However, taking $n=\lceil B/3\rceil+1$, where $\lceil x\rceil$ is the smallest integer $\ge x$, violates the above inequality. This gives us our desired contradiction. Hence the sequence $\{3n\}_{n\in\mathbb{N}}$ is not bounded.
Archimedean says that $1/n_{0}<3/B$ for some $n_{0}$, so $B<3n_{0}$, so $3n\leq B$ for all $n$ is not valid then.