Supremum of a very simple sequence of functions

Question

Given $f_n(x)=nx$ with $x \in \mathbb{R}$ and $n \in \mathbb{N}$, what is $\sup_n\{f_n(x)\}$.

I argue that the $\sup$ is $+\infty$. My friend argue that it is

$$ \sup= \begin{cases} -\infty, & \mbox{if } x<0 \\ 0, & \mbox{if } x=0 \\ +\infty, &\mbox{if } x>0 \end{cases} $$

We can't seem to understand this concept.

Answer 1

We have that, assuming $0\not \in \mathbb N$

• for $x>0 \implies f_n(x)=nx \to \infty$
• for $x=0 \implies f_n(x)=nx =0$
• for $x<0 \implies \max(f_n(x))=-x \implies \sup(f_n(x))=-x$

assuming $0 \in \mathbb N$ we would obtain

• for $x<0 \implies \max(f_n(x))=0 \implies \sup(f_n(x))=0$

Refer also to the related

• Is $0$ a natural number?

Answer 2

Let us define for $x \in \mathbb R$:

$$A_x:=\{nx: n \in \mathbb N\}.$$

1. If $x=0$, then $A_0=\{0 \}$, hence $ \sup A_0=0.$

2. If $x>0$, then $A_x$ is not bounded from above, hence $\sup A_x= \infty.$

3. If $x<0$, then $t <0$ for all $t \in A_x.$ Hence $0$ is an upper bound of $A_x.$ It is your turn to show that $\sup A_x=0$ , if $x<0.$