Infimum and supremum of subset of inclusion Power set

Question

I'm having trouble understanding the following exercise:

Given $U=\{1,2,3,4\}$ with $A=P(U)$ the power set of the elements of $U$ and $R$ the inclusion relation over $A$. Determine the infimum and supremum of $B$, being $B$ a subset of $A$.

a) $B=\{\{1\},\{2\}\}$

b) $B=\{\{1\},\{2\},\{3\},\{1,2\}\}$

I can't find a definition for infimum and supremum of a set where its elements aren't connected by a inclusion relation. For example on the part (a) 1 isn't included on 2 and neither 2 is included on 1, so are they both supremum and infimum at the same time? Isn't the infimum and supremum unique?

If the case was $\{\{2\},\{1,2\}\}$ I believe one could say $\{2\}$ is the infimum and $\{1,2\}$ the supremum since the first is included on the second.

Thanks in advance

Answer 1

The relationships amongst the elements of $B$ don’t actually matter. For the infimum you’re looking for the largest subset of $U$ that is a subset of every member of $B$, and for the supremum you’re looking for the smallest subset of $U$ that contains each member of $B$ as a subset. (Here largest and smallest refer to the subset relation: $X$ is smaller than $Y$ in this sense if and only if $X\subsetneqq Y$.)

As an example, if $B=\big\{\{3\},\{4\}\big\}$, the only subset of $U$ that is a subset of both $\{3\}$ and $\{4\}$ is $\varnothing$, the empty set, so the infimum of $B$ must be $\varnothing$. There are four subsets of $U$ that have both $\{3\}$ and $\{4\}$ as subsets: $\{3,4\},\{1,3,4\},\{2,3,4\}$, and $U$ itself. $\{3,4\}$ is contained in each of the others, so it’s the smallest of them in the sense of $\subseteq$ (as well as being the smallest in cardinality, though that’s not relevant). Thus, the supremum of $B$ is $\{3,4\}$.

Now apply these ideas to your sets $B$. You might notice that the infimum and supremum can be described rather simply in terms of familiar basic set operations. Do you see how?

Answer 2

Your answer for the example in your last paragraph is correct.

The "inclusion relation" on the set of subsets of a set defines just a partial order, not a total order. Many (most) pairs of sets aren't related.

Hint for a): what is the largest set that's a subset of both $\{1\}$ and $\{2\}$? What is the smallest subset that has both as subsets?

Similarly for b).

Answer 3

Let $A$ be a partially ordered set. For $B\subseteq A$ we say that $a=\sup B$ if for every $b\in B$, $b\leq a$, and for every $c\in A$, if every $b\in B$ satisfies that $b\leq c$, then $a\leq c$. Namely, amongst the set of upper bounds for $B$, $a$ is the minimum.

The infimum is defined similarly, reversing the order and taking the maximum of lower bounds.

This is the same definition as you know from the real numbers. Only now it might be that a set might not have a supremum or infinmum, and it might have several upper bounds which are minimal upper bounds, but no upper bound is the minimum upper bound.