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Thread: Getting a minimum in an infinite set

  1. #1
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    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
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    Re: Getting a minimum in an infinite set

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

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

    the answer to your question is Yes and No
    Last edited by Idea; Apr 5th 2019 at 12:01 AM.
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  4. #4
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    Re: Getting a minimum in an infinite set

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

    $1$ is a minimum in $S$

    combine this with the previous example (romsek )

    the answer to your question is Yes and No
    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.
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    Re: Getting a minimum in an infinite set

    Quote Originally Posted by KLHON View Post
    [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$(a,b))$.
    Thanks from topsquark and MacstersUndead
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