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Given nonempty set $A$ of positive real numbers, and define $$\frac{1}{A}=\left\{z=\frac{1}{x}:x\in A \right\}$$ Show that $$\sup\left(\frac{1}{A}\right)=\frac{1}{\inf A}$$


let $\sup\left(\frac{1}{A}\right)=\alpha$ and $\inf A = \beta$. Apply the definition of supremum, $z<\alpha$, then there exist $z'\in 1/A$ such that for every $\epsilon>0$, $z'>\alpha-\epsilon$

And the definition of infimum, $x>\beta$ and there exists $x'\in A$ such that for every $\epsilon>0$, $x'<\epsilon+\beta$.

At this step, I don't see how to relate $\beta$ with $\alpha$ which is $\alpha=\frac{1}{\beta}$, can anyone give me a hit or suggestion. Thanks.

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    $\begingroup$ $1/x$ has a property that it reverses inequalites: $x \leq y$ if and only if $1/x \geq 1/y$. So you can look at the definition of sup and inf and see that they're the same, but with all the inequalities reversed. hope this helps $\endgroup$
    – user40167
    Jun 21 '15 at 3:58
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The crucial facts here are that if $x \in A$, then $x>0$ and the function $x \mapsto {1 \over x}$ reverses order in $A$, that is, if $x,y \in A$, then $x<y$ iff${1 \over x } > {1 \over y}$.

We have $\sup_{x' \in A} {1 \over x'} \ge {1 \over x}$ for all $x \in A$. Now let $x_n \in A$ such that $x_n \to \inf A$. Then this gives $\sup_{x' \in A} {1 \over x'} \ge {1 \over \inf A}$.

Since $x \mapsto {1 \over x}$ reverses order in $A$, we have ${1 \over \inf A} \ge {1 \over x}$ for all $x \in A$. Taking the $\sup$ yields the desired answer, ${1 \over \inf A} \ge \sup_{x \in A} {1 \over x}$.

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Let Inf$(A)=\beta(\neq0) \implies \forall\ a\in A,\ a\geq\beta$ and $\alpha=\sup\left(\frac{1}{A}\right)$. Since, $\frac{1}{A}$ is defined which means that $\forall\ a\in A,\ a\neq0. $

Claim: $\alpha=\frac{1}{\beta}$

$(i)\frac{1}{\beta}$ is an UB of $\frac{1}{A}.$

Since, $\beta $ is LB of $A\ \implies$ $$ \forall\ a\in A,\ a\geq\beta \implies \forall\ a\in A,\ \frac{1}{a}\leq \frac{1}{\beta}\implies \frac{1}{a} $$ So, $\frac{1}{\beta}$ is an UB of $A.$

Now we have to prove that $\frac{1}{\beta} $ is the LUB of $\frac{1}{A}\ i.e.\ \forall\ \epsilon>0\ \exists\ b\in \frac{1}{A}$ s.t. $\frac{1}{\beta}-\epsilon>b.$ Without loss of generality we can assume that $\epsilon $ is such that $\beta\epsilon<1.$

Let $\delta=\frac{\beta}{1-\beta\epsilon}-\beta$. Clearly, $\delta>0.$ Since, $\beta $ is the GLB of $A\ \implies \beta+\delta$ is not a LB of $A \implies \exists\ a\in A$ s.t $$\beta+\delta>a\geq\beta.$$ $$\implies \beta+\frac{\beta}{1-\beta\epsilon}-\beta>a\geq\beta. $$ $$\implies \frac{\beta}{1-\beta\epsilon}>a\geq\beta. $$ $$ \implies\frac{1-\beta\epsilon}{\beta}<\frac{1}{a}\leq \frac{1}{\beta} .$$ Since, $\frac{1}{a}\in \frac{1}{A} \implies \frac{1}{\beta}$ is the LUB of $\frac{1}{A}.$

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  • $\begingroup$ It is possible that $\inf A = 0$. Take $A= (0,\infty)$ for example. $\endgroup$
    – copper.hat
    Jun 21 '15 at 16:21
  • $\begingroup$ Then the sup($A$) does not exist. Because according to the definition of LUB "a non-empty set which is bounded above has the LUB"$\in \mathbb{R}$". So, inf $A$ must be non-zero, as I think. $\endgroup$ Jun 22 '15 at 5:24
  • $\begingroup$ A fairly standard convention is that if $A$ is unbounded above, then we write $\sup A = \infty$. $\endgroup$
    – copper.hat
    Jun 22 '15 at 5:58
  • $\begingroup$ Okay, thanks for the information. $\endgroup$ Jun 22 '15 at 6:02
  • $\begingroup$ Please can you say if we take $\epsilon \beta \geq 1$ , then what will happen here? please say $\endgroup$ Aug 24 '17 at 20:11

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