Given an ordered set, find the infimum and supremum of one of its subsets

Question

I am given a an ordered set $(\mathcal{P}(\mathbb{N}), \subseteq)$ I need to find the infimum and supremum of its subset $A= \{ \{ 3,5,8 \}, \{1,2,3,5 \}, \{5,10 \} \}$.

Now, I am having quite a dilemma. From what I know, the element $S \in \mathcal{P}(\mathbb{N})$ is the minorant of $A$ if for each $y \in A$ we have that $S \subseteq y$.

1. Regarding infimum: From what I understand, the lower bounds would be for example $ \{ 5\}$ and $\emptyset$. But now, the definition of infimum is the "greatest" out of those. How can I "measure" which one out of $ \{ 5\}$ and $\emptyset$ is greater?

2. Regarding supremum, I need upper bounds. An upper bound would be an element from $\mathcal{P}(\mathbb{N})$ so that every element of $A$ is a subset of it. But still, I can think of many subsets of N for which this holds, the first one would be $ \{ 1,2,3,5,8,10 \} $. But still, how can I be certain that this is the supremum?

Answer 1

In any power set ordering $(P(S), \subseteq) $:

• $Z$ being an "upper bound" of a collection $\mathcal A\subseteq P(S)$ means for all $X\in \mathcal A,$ $X\subseteq Z.$
• This means the same thing as $\bigcup_{X\in A} X\subseteq Z$
• $Y$ is the least upper bound for $\mathcal A$ iff $Y\subseteq Z$ for every upper bound $Z$ for $\mathcal A$
• So, $Y=\bigcup_{X\in A} X$ is the least upper bound of $\mathcal A.$

Similarly, the infimum of $\mathcal A$ is $\bigcap_{X\in \mathcal A} X.$

---

More generally:

If $\mathcal B$ is any collection of sets whatsoever, and $\mathcal A\subseteq \mathcal B,$ then in $(\mathcal B,\subseteq),$ we have $\bigcup_{X\in \mathcal A}X = \sup\mathcal A$ provided $\bigcup_{X\in \mathcal A}X\in\mathcal B$.

If $\bigcup_{X\in \mathcal A}X\notin\mathcal B,$ then there could be a smallest set in $\mathcal B$ containing it, which would be the sup, or there could not, in which case the sup doesn't exist.

(And similarly for intersection/inf.)