Understanding the definition of upper/lower bound
Definition: Let $E$ be a subset of an ordered set $S$. If there exists an $\alpha \in S$ such that every element of $E$ is less than or equal to $\alpha$, then $\alpha$ is an upper bound of $E$, and $E$ is bounded above. Lower bounds are defined similarly (just replace $\le$ with $\ge$).
One thing I realized is that sets cannot simply be bounded above or below. They require an "ambient" superset.
Here below I made up a few examples to see if I am reading the definition above correctly. Please, see if they make sense.
Let $A = (0, \pi).$
$A$ is bounded below by $(-3, 0]$ and unbounded above in $(-3, \pi)$. The set of lower bounds of $A$ does not include $-3$ because $-3 \not \in (-3, \pi)$ and the set of upper bounds is empty because $(-3, \pi)$ has nothing greater than $\pi$ (not even $\pi$ itself).
$A$ is bounded below by $(-\infty, 0]$ and bounded above by $(\pi, \infty)$ in $\mathbb Q$. The set of lower bounds include $0$ because $0 \in \mathbb Q$ and the set of upper bounds exclude $\pi$ because $\pi \not \in \mathbb Q.$
$A$ is bounded below by $(-\infty, 0]$ and bounded above by $[\pi, \infty)$ in $\mathbb R$. The numbers $0, \pi$ are included because both $0, \pi \in \mathbb R.$
1 Answer
$\begingroup$If we accept the abuse of notation, as discussed in the comments, then your understanding is fine. Note however that it's not only the existence of upper and lower bounds that depends on the "ambient" superspace, as you put it. Even writing $A=(0,\pi)$ is ambiguous unless you say of which set you consider it to be a subset!
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