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Let $r$ be a strictly positive real number. How to prove that ${\{x \in \mathbb R_{\ge0}:x^2

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For instance $\max(1,r)$ is an upper bound.

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    And how would I prove that? – 2017-02-03
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    Because, if $ r\ge 1$, $r\le r^2$, so $x^2$x$. And if $r\le1$, $x^2 – 2017-02-03
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Note that $f(x)=x^2$ is a strictly increasing function on $(0,\infty)$, and $x^2\to\infty$ as $x\to\infty$. So, for any $r$, you can make $x$ "big enough" that $x^2$ exceeds $r$; and, once that happens, any $y>x$ also satisfies $y^2>r$. Hence the set is bounded above by such an $x$.