Q: Upper bound of the set I am struggling with understanding of the concept of the upper bound of the set:
in Rudin, it says: Suppose, $S$ is an ordered set and $E$ is a subset of $S$. If there exists $b \in S$ such that $x\le b$ for any $x \in E$, than $E$ is bounded above, and $b$ is an upper bound of $E$. But what about $S$ itself? What if $S=(0,1)$, so there is nothing else. My prof says that if the set is given, assume nothing else exists, but this set. Can we consider $S$ bounded in $S$ or not? There is no such $b \in S$ that for any $x \in S x\le b$. Or the concept of bound is defined for the proper subset only?
Thank you very much in advance.

A: Boundedness (above or below) need not be defined for proper subsets only. Consider instead the real interval $(0,1],$ which is bounded above in itself, but not below.

A: If there is no such $b \in S$ such that $x \leq b$ for all $x \in E$ then $E$ is unbounded from above.  In this definition, you always need to define a parent set, $S$.  That is the place of potential upper bounds.  In your example of $S = (0,1)$.  Then $S \subseteq S$ is an unbounded (from above) subset of $S$.  However, it is true that $(0,1) \subset (0,2)$ is bounded from above, by namely $b = 1$, or any other element larger than $1$ in $(0,2)$.
Does this clarify anything?
