Let $a_{n} = 1/n$. Thus $\sup(a_{n})_{n=1}^{\infty} = 1$ and $\inf(a_{n})_{n=1}^{\infty}$ = 0.

Question

Let $a_{n} = 1/n$. Thus $\sup(a_{n})_{n=1}^{\infty} = 1$ and $\inf(a_{n})_{n=1}^{\infty}$ = 0.

MY ATTEMPT

Indeed, one has that $a_{n} \leq 1$. Otherwise, we should have \begin{align*} a_{n} > 1 \Rightarrow \frac{1}{n} > 1 \Rightarrow 0 < n < 1 \end{align*} which is impossible. Besides this fact, for every $a < 1$ there exists an $N\geq 1$ such that $a < a_{N} \leq 1$: it suffices to take $N = 1$.

On the other hand, $a_{n} > 0$ given that it is the quotient of two positive natural numbers.

Moreover, for every $a > 0$ there exists an $N\geq 1$ such that $0 \leq a_{N} < a$ due to the archimedean property, and we are done.

I am little bit new to this. Can someone please check if I am reasoning correctly?

Answer 1

For all $n>0, $ $ a_n\le 1$ and $\; a_1=1$ thus

$$\sup\{a_n,n>0\}=\max\{a_n ,n>0\}=1$$

$(a_n) $ is a positive decreasing sequence thus it converges to its infimum.

$$\inf \{a_n,\; n>0\}=\lim_{n\to +\infty}a_n=0$$

But your reasoning is correct.