Let S be an infiinite set of positive real numbers. Is there exist a minimum in S ? I am not just interested in the YES or NO, I am most interested in the REASON behind
Thanks a lot
Here is one of my favorite examples. Suppose that $\mathcal{S}=\left\{\dfrac{1}{n}: n\in\mathbb{N}^+\right\}$
It is easy to show that $\mathcal{S}$ has no first term. All we need is to grant that $\mathbb{N}^+$ is not bounded above.
Suppose that $t\in\mathcal{S}$ is the smallest term. Then $\dfrac{1}{t}$ is not an upper bound for $\mathbb{N}^+$. So $\exists T\in\mathbb{N}^+$ and $T>\dfrac{1}{t}$.
Well that means that $\dfrac{1}{T}<t$ which is a contradiction to $t$ being the first term in $\mathcal{S}$
Here is another fact. Suppose that
$ \begin{align*}a &<b \\\dfrac{a}{2}&<\dfrac{b}{2}\\\dfrac{a}{2}+\dfrac{ a}{2}&<\dfrac{a}{2}+\dfrac{b}{2}\\a&<\dfrac{a+b}{2 } \end{align*}$ LIKEWISE $\dfrac{a+b}{2}<b$ together $a<\dfrac{a+b}{2}<b$
That tells us that the average of two numbers is always between them.
Moreover, in any non-degenerate finite open interval of real numbers there cannot be a smallest.
If $a~\&~b$ are two real numbers and $t\in(a,b)$ then $a<\dfrac{a+t}{2}<t$ in$(a,b))$.