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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 $0, \pi \in \mathbb R.$

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    $\begingroup$ A set of real numbers can be bounded above or below by a real number, not by a set of real numbers. $\endgroup$
    – Bernard
    Commented May 7, 2020 at 22:24
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    $\begingroup$ We say $ A $ is bounded above by AN ELEMENT. not a set. $\endgroup$ Commented May 7, 2020 at 22:26
  • $\begingroup$ @Bernard, To clarify, when I say $B$ is bounded above by $A$, I mean $B$ is bounded above by every element in $A$. Abuse of notation :) $\endgroup$
    – croc
    Commented May 7, 2020 at 22:28
  • $\begingroup$ Please specify if you are talking about a partially ordered, linearly ordered , well-ordered etc. set S. $\endgroup$ Commented May 7, 2020 at 22:37

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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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  • $\begingroup$ I see. Suppose $A = (0, \pi) \cap \mathbb Q.$ Then it looks like the examples above hold, but if $A = (0, \pi) \subset \mathbb R$, then $A \not \subset \mathbb Q$, so we cannot even talk about $\mathbb Q$ as being the "ambient superspace" of $A$. Does that make sense? $\endgroup$
    – croc
    Commented May 7, 2020 at 23:03
  • $\begingroup$ Yes, so the point is that you should talk first about what big space everything is in: the subsets you want to bound, and the $\underline{bounds \,themselves}$. I would advise against your abuse of notation. It is ambiguous when you write "$B$ is an upper bound for $A$ in $\mathbb{Q}$" whether you mean that all the sets are ranges or real numbers, and only the rationals in $B$ are upper bounds, or whether you are talking only about the rationals in $A$ being bounded by the rationals in $B$. $\endgroup$ Commented May 7, 2020 at 23:52
  • $\begingroup$ The big picture of notation and convention will become clearer when you read up to suprema and infima. $\endgroup$ Commented May 7, 2020 at 23:55

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