Consider the set $\;A=\left\{3-\dfrac{4}{x}:x\ge2\right\}$.
I want to prove that Set A has a minimum or doesn't.
$x \ge 2$
$\dfrac1{x} \le \dfrac1{2}$
$\dfrac4{x} \le \dfrac4{2}$
$-\dfrac4{x}\ge-2$
$3-\dfrac4{x}\ge1$
Thus, I have found that any element in A is always bigger or equal to 1. Thus, I know that 1 is a lower bound. But is that enough to prove that it is the minimum of A since it 1 is in the set A. if it is enough, why? I then tried to prove by contradiction that 1 is the minimum, but I went nowhere. I said :
Assume that there is $b>0$ such that $1<b<3-\dfrac4{x}$ for all $x
\ge 2$.
Can someone please show me how to show 1 is the minimum by contradiction? it should be possible since it's true that 1 is the minimum.
We first have to prove that $1$ belongs to $A$, and then that it's the smallest element in $A$.
To prove $1\in A$ it's sufficient to notice that for $x = 2$ we indeed get $1$.
To prove $x = 1$ is the minimum by contradiction, we assume that $\exists s \in A$ such that $s < 1$. Then we substitute, getting
$$3 - \dfrac{4}{x} = s$$
$$\dfrac{4}{x} = 3-s$$
Now since $s < 1$ it holds that $3-s > 2$, therefore $\dfrac{4}{x} > 2$
But the set $A$ is defined for those $x$ that are greater or equal than $2$, and hence $\frac{4}{x} > 2$ implies $x < 2$ which is a contradiction.