If $A\subset\mathbb{R}$ is nonempty, show that $A$ is not bounded above if $r\in A$ implies $\sqrt{r^2+1}\in A$.

Question

Let $r\in A$. Then form the sequence $$ r_1=r,r_2=\sqrt{r^2+1},\ldots,r_n=\sqrt{r_{n-1}^2+1},\ldots $$ Observe $\sqrt{r^2+1}>r$ for all $r\in\mathbb{R}$. Therefore this sequence is bounded below by the sequence $x_n=n$. Since $x_n\to\infty$ as $n\to\infty$, we must also have $r_n\to\infty$ as $n\to\infty$. Thus $A$ cannot be bounded above. Is this right?

Answer 1

Suppose $\;A\;$ is bounded above, and let $\;M:=\sup A\;$, and let $\;n\in\Bbb N\;$. For simplicity assume $\;M\ge0\;$ Then there exists $\;a_n\in A\;\;$ s.t. $\;M-\frac1n<a_n<M\;$ (this follows at once from the definition of supremum), and passing to the limit and using the squeeze theorem we get that $\;\lim\limits_{n\to\infty}a_n=M\;$.

But then, by arithmetic of limits, $\;\lim\limits_{n\to\infty}\sqrt{a_n^2+1}=\sqrt{M^2+1}>M\;$ , but this is impossible since $\;\sqrt{a_n^2+1}\in A\;$ and thus $\;\forall\,n\in\Bbb N\,,\,\,\sqrt{a_n^2+1}\le M\;$ ...