Supremum and infinum of a set

Question

Find supremum and infimum of the set: $B={ \frac{x}{1+ \mid x \mid }} \ for \ x\in \mathbb{R}$ For me it is visible that it will be 1 and -1 respectively but how to prove it properly?

Answer 1

Let :

$$ B = \left\{ f(x), \; x \in \mathbb{R} \right\}. $$

First, note that $B$ is bounded and not empty, which ensures that $\inf(B)$ and $\sup(B)$ exist. In order to prove that $\sup(B) = 1$, you need to prove that : $\forall y \in B, \, y \leq 1$ and either :

• $1 \in B$ and in this case, $\sup(B) = \max(B) = 1$.
• OR : find a sequence $(u_{n})_{n \in \mathbb{N}}$ of elements of $B$ which converges to $1$.

Here, $1 \notin B$. The sequence $\displaystyle \Big( \frac{n}{1+n} \Big)_{n \in \mathbb{N}} = \big( f(n) \big)_{n \in \mathbb{N}}$ converges to $1$, which proves that $\sup(B) = 1$.

You can do the same to prove that $\inf(B) = -1$.

Answer 2

For all $x\in \mathbb R$ we have $\frac{x}{1+|x|} < 1$, hence the supremum is less than or equal to 1. Suppose the supremum is smaller than 1, then we can write it as $1-\epsilon$ for some $\epsilon > 0$. Can you find an $x\in \mathbb R$ such that $\frac{x}{1+|x|} \in (1-\epsilon, 1)$?

Answer 3

You can do both in one move. Show for any $x \in \Bbb{R}$ that $$\left |\frac{x}{1+|x|} \right| \leq 1 $$ Then suppose there is some $c>0$ such that $$\left |\frac{x}{1+|x|} \right|\leq c < 1 $$ for all $x$. Show this case is a contradiction, as you can always find some $x$ where $$\left|\frac{x}{1+|x|} \right | >c$$

Answer 4

It is completely apparent that the magnitude of this function is less than one. Now we want to prove that its supremum is $1$, for this issue, suppose that the supremum is $a$, which according to the previous statement it must be less than or equal to 1. Hence for all values of $x$, $\frac{x}{1+|x|}\leq a\rightarrow \lim_{x\rightarrow \infty}\frac{x}{1+|x|}\leq a\rightarrow 1\leq a\rightarrow a=1$. For the infimum the procedure is similar.