Find the Supremum and Infimum of a set.

Question

I have been tasked with finding both the supremum and infimum of the following set: $E := \{ \frac{x}{x + 1} : x \in (3,5)\}$.

I notice that the function $f:\mathbb{R}_{>0} \to \mathbb R$ is differentiable on $\mathbb R_{>0}$ with $f^\prime (x) = \frac{1}{(x+1)^2}$. Moreover the deriative $f^\prime (x)$ is positive on the closed interval $[3,5]$. So $f(3) \leq f(x)\leq f(5)$ for all $x \in [3,5]$

So clearly $\inf \{x \in \mathbb R : 3 \leq x \leq 5 \} = f(3)$ and $\sup \{ x \in \mathbb R : 3 \leq x \leq 5 \} = f(5)$. So for any $\varepsilon > 0$ there exists $x_1, x_2 \in [3,5]$ such that $f(x_1) > f(5) - \varepsilon $ and $f(x_2) < f(3) + \varepsilon $.

I am having a bit of trouble, rigorously showing that there are points $x^\prime_a, x^\prime_b \in (3,5)$ such that $f(x^\prime_a) > f(5) - \varepsilon$ and $f(x^\prime_b) < f(3) + \varepsilon $. Plus I am not convinced that this is most concise argument. So any improvements and/or suggestions are welcome

Answer 1

An easy argument goes like this: Let $f:[3, 5]\to\mathbb [3/4, 5/6], f(x) = x/(1+x)$. Then $f$ is bijective, since it is monotone (as you showed), and $f(3) = 3/4, f(5) = 5/6$*.

So, $$E=f((3,5)) = f([3,5]\setminus\{3,5\}) = f([3,5])\setminus\{f(3),f(5)\}\\= \text{Im}(f)\setminus\{3/4,5/6\} = [3/4, 5/6] \setminus\{3/4,5/6\} = (3/4, 5/6)$$

So $E$ is an open interval, and the $\sup$ and $\inf$ are the respective interval borders.

* _{Or, just use the inverse mapping $x = y/(1-y)$ to show bijectivity}

Answer 2

$$E := \{ \frac{x}{x + 1} : x \in (3,5)\}$$ Then using the fact that the function $f$ as you defined is increasing, $$E=[f(3),f(5)]=[\frac{3}{4},\frac{5}{6}] $$ Now it is obvious, since $E$ is an interval, that, $$\sup E=\frac{5}{6} $$ and,$$\inf E=\frac{3}{4} $$

Answer 3

The function $f(x)=\displaystyle\frac{x}{x+1}$ is a continuous, monotone and increasing on $]3,5[$. So it takes the infimum at $3$ and $f(3)=\frac{3}{4}$ and it takes the supremum at $5$ and $f(5)=\frac{5}{6}$.