Show that there is no positive real number which is less than every positive real number. I think what would happen if there is less than every positive real numbers
How can I prove that?
 A: If $x$ is a real number then $x/2$ is also a real number $<  x$ with a contradiction.
If the real numbers are made up of the classes of limits of converging Cauchy series then there is a series
$a_1, a_2, a_3, …$ with the limit  $x$
$x/2$  is a real number since there is a Cauchy sequence 
 $a_1/2, a_2/2, a_3/2 , …$ which has the limit  $x/2$ 
A: Let $r$ be a positive real. Consider the set $S=\{x\in\mathbb{R}~:~x<r\}$. Now this set $S$ is non-empty, as $0\in S$. So it has a lowest upper bound, say $r^*$. $r^*\leq r$ as $r\not<r$. If $r^*<r$, then $r^*<\frac{r+r^*}{2}<r$. Hence $\frac{r+r^*}{2}\in S$, which contradicts our assumption that $r^*$ is the lowest upper bound. So $r^*=r$, but $r\not\in S$.
A: If $0 < \epsilon \in \Bbb R$, and $\epsilon < r$ for every positive $r \in \Bbb R$, then
$\epsilon < \epsilon; \tag 1$
that will obviously not work!
Even if we relax the condition from $\epsilon < r$ to $\epsilon \le r$ for all positive $r \in \Bbb R$, it still won't work, since
$0 < \dfrac{\epsilon}{2} < \epsilon. \tag 2$
A: Suppose that there does exist some real $y > 0$ that is less than every other positive real number. Then $y < \frac{1}{n}$ for every natural number $n$. This would then imply that $n < \frac{1}{y}$ for every natural number (the natural numbers would then have an upper bound!) and so we have a contradiction. 
