# Thread: Getting a minimum in an infinite set

1. ## Getting a minimum in an infinite set

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

2. ## Re: Getting a minimum in an infinite set

No there is no minimum. The set of positive reals is an open set.

$\forall x >0,~ \exists~ y \ni 0 < y < x$

3. ## Re: Getting a minimum in an infinite set

$\displaystyle S=[1,\infty )$

$1$ is a minimum in $S$

combine this with the previous example (romsek )

4. ## Re: Getting a minimum in an infinite set

Originally Posted by Idea
$\displaystyle S=[1,\infty )$

$1$ is a minimum in $S$

combine this with the previous example (romsek )

Yes, my mistake. OP asks refers to "an" infinite set of positive numbers, not "the" infinite set of positive numbers.

If the set you specify is closed, such as the example Idea stated, then a minimum does exist.

5. ## Re: Getting a minimum in an infinite set

Originally Posted by KLHON
[LEFT]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
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$(a,b))$.