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I got stuck in proving that $A=\{yn:n\in\mathbb N,y\in(1,\infty)\}$ is not bounded above? (without of course using lim)?

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  • $\begingroup$ Suppose $M$ was its upper bound. What can you do with that? $\endgroup$ – John Douma Apr 10 '20 at 20:34
  • $\begingroup$ Well.... do you know how to show $(1,\infty)$ is not bounded above? If $y\in (1,\infty)$ and $n \in \mathbb N$ then $ny \ge y$. $\endgroup$ – fleablood Apr 10 '20 at 21:11
  • $\begingroup$ Assume bounded . Archimedean principle? Then? $\endgroup$ – Peter Szilas Apr 10 '20 at 21:12
  • $\begingroup$ If $a \in A$ then $a > 0$ and if $M \ge a$ then $M > 0$ so $2M \in A$. But $M < 2M$. So $M$ can not be an upper bound. $\endgroup$ – fleablood Apr 10 '20 at 21:26
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If $a \in A$ then $a = ny$ for some $n \ge 1$ and $y > 0$ so $a =ny>0$.

Suppose $M$ is an upper bound of $A$. Then for any $a \in A$ we have $M \ge a > 0$ so $M$ is a positive real number. So $M \in (0, \infty)$ and so $2M \in A$. But $M < 2M$ so $M$ can not be an upper bound.

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