Query on supremums of sets of real numbers. I would like to know if the following reasoning is correct:
If we have $a, b \in \mathbb{R}$ and $a < b$ then $\exists c \in \mathbb{R}$ such that $a < c < b$. From this, it follows that if $A$ is a set of real numbers then $\sup(A) = x$ for some $x \in A$. For if this were not the case, we would have $\sup(A) > x \quad \forall x \in A$. Hence we would have some $y$ such that $x \le y < \sup(A)$, which is clearly a contradiction.
 A: Let $A=(0,1)$. Then $\sup A=1$, but $1\not\in A$.
It is true that for any $x\in A$, there is a $y\in A$ such that $x\lt y\lt 1$. But that does contradict the fact that $\sup A=1$.
A: No, the reasoning is not correct. Consider the set $A=[0,1)$: $\sup A=1$, but $1\notin A$. To see that $\sup A$ really is $1$, note first that $1$ is clearly an upper bound for $A$. Suppose that $u<1$. Then $\max\left\{0,\frac12(u+1)\right\}$ is an element of $A$ that is greater than $u$, so $u$ is not an upper bound for $A$. Thus, $1$ must be the least upper bound for $A$, which is $\sup A$.
It’s perfectly true that for each $a\in A$ there is an $x\in\Bbb R$ such that $a<x<1$; the problem with your argument is that it can’t be the same $x$ for every $a$. For instance, if $a=\frac34$, you can take $x=\frac78$, but that won’t work if you now pick a new $a\in\left(\frac78,1\right)$: you’ll have to pick a new, larger $x$. Your argument implicitly assumes that you can find one $x$ that is simultaneously between every element of $A$ and $1$.
