Let A be a set of positive numbers. Define $\frac{1}{A}=\{\frac{1}{a}, a\in A\}$

Now the first part is to prove that if $\inf A>0$ then $\sup(\frac{1}{A})=\frac{1}{\inf A}$

Here's my attempt for it:

At first we set $\alpha=\inf(A).$ At first we'll want to show that $\frac{1}{A}$ is bounded above. for some $b\in \frac{1}{A} \Rightarrow \frac{1}{b}\in A \Rightarrow \frac{1}{b} \geq\alpha \Rightarrow b\leq \frac{1}{\alpha} $

$\frac{1}{A}$ Is a non empty, bounded above set then by the completeness axiom it has a supremum.

we'll choose $\beta=\sup\frac{1}{A}. \ $ since $\frac{1}{\alpha}$ is an upper bound of $\frac{1}{A}$ we get $\beta \leq \frac{1}{\alpha}.$ In a similar form for A: $ \ \alpha$ is the infimum of A, meaning $\alpha \leq \frac{1}{\beta}$

By the last two inequalities we have: $\alpha=\frac{1}{\beta}\ , \ \beta=\frac{1}{\alpha} \ $ as required.

I'd love to get constructive comments on how to improve my style or corrections if the proofs need them. And for the second part of the problem which im having trouble with:

Suppose now that $ \inf A=0. \ $ Prove that $\sup \frac{1}{A}=+\infty$

I'm having trouble about how to put this in mathematical terms. Any help is appreciated

  • $\begingroup$ You can typeset $\sup$ and $\inf$ by \sup and \inf. $\endgroup$ – Theoretical Economist Nov 16 '19 at 19:59
  • $\begingroup$ Thanks for the tip :) still getting used to mathjax $\endgroup$ – Zappa Nov 16 '19 at 20:00
  • $\begingroup$ If $\inf A=0$ then there is a sequence $x_n$ of elements of $A$ with $x_n<\frac1n$. Then $\frac1{x_n}$ is a sequence of elements of $\frac1A$, showing what about $\sup\frac1A$? $\endgroup$ – Math1000 Nov 16 '19 at 20:08
  • $\begingroup$ that he's infinitely growing? $\endgroup$ – Zappa Nov 16 '19 at 20:11

First Part:

You are right that $\frac{1}{\alpha}$ is an upper bound for $\frac{1}{A}$, so $\beta\leq\frac{1}{\alpha}$. However, you also say that $\alpha\leq\frac{1}{\beta}$, but this just implies that $\frac{1}{\alpha}\geq \beta$, which is the same as the previous inequality.

The other inequality that you should use is based on the following. For each $a\in A$ we have $\frac{1}{a}\leq\beta\iff \frac{1}{\beta}\leq a$. Now $\alpha$ is the greatest lower bound for $A$, so we must have $\frac{1}{\beta}\leq\alpha$. This gives $\beta\geq\frac{1}{\alpha}$ which is what we want (combine this with $\beta\leq\frac{1}{\alpha}$ to get $\beta=\frac{1}{\alpha}$).

Second Part: Now if $\alpha=0$, then there is a sequence $(a_n)_{n\geq 1}$ in $A$ with $a_n\searrow 0$. Now $\frac{1}{a_n}\nearrow\infty$ so $\sup(\frac{1}{A})=\infty$.

  • $\begingroup$ Thanks for this quick qnswer! $\endgroup$ – Zappa Nov 16 '19 at 20:18
  • $\begingroup$ My pleasure. If you have any further questions I'd be happy to help. $\endgroup$ – Dave Nov 16 '19 at 20:20

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