If $\dfrac{x}{y}\in A$ for $x,~y\in A,~x<y$ and $\sup A<1$, prove $\sup A\in A.$

Question

Let $A$ be a non empty subset of positive reals, such that $\dfrac{x}{y}\in A$ whenever $x,~y\in A,~x<y.$ If $\sup A<1$, prove that $\sup A\in A.$

Attempt. If $\sup A\notin A$, then we could find a sequence $0<a_1<a_2<\ldots<1$ of $A$ such that $a_n \to \sup A$. Of course, for all $k$ we have $\dfrac{a_{k}}{a_{k+1}}\in A$, so I thought by telescoping to reach $\sup A$, but I got nowhere.

Thanks in advance for the help.

Answer 1

Suppose that $\sup A\notin A$. Then $A$ has some sequence $(a_n)_{n\in\mathbb N}$ such that $\lim_{n\to\infty}a_n=\sup A$. We can assume, without loss of generality (since $\sup A\notin A$) that the sequence is strictly increasing. Then$$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=\frac{\sup A}{\sup A}=1.\tag1$$But the sequence $\left(\frac{a_n}{a_{n+1}}\right)_{n\in\mathbb N}$ is a sequence of elements of $A$ and therefore its limit should be smaller than or equal to $\sup A$. And we are assuming that $\sup A<1$. So, the equality $(1)$ allows us to reach a contradiction.