# $A$ is the power set of set $C$, and $S$ is the relation on $A$ defined by $x S y$ iff $x \subseteq y$.

The following is an exercise from my textbook:

Let $C$ be a set and let $A=\mathcal{P}(C)$, the set of all subsets of $C$. Let $S$ be the relation on $A$ defined by $x S y$ iff $x \subseteq y$. Then $S$ is a partial order relation on $A$.
(a) Find the least and greatest element of $A$.
(b) Let $B=\{x \in A$: $x \ne ∅\}$. Suppose $C$ has two or more elements. Show that $B$ has no least element.

I'm confused about the definition of least and greatest element as applied to this exercise.

From the my textbook, least element is defined as such:

Let $(A,\leq)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.

Now, referring back to the exercise, $A$ consists of sets, not merely numbers; thus, to say that some set $x_1$ is less than or equal to some other set $x_2$ doesn't make sense. We don't relate sets in that way do we? (Unless, we're talking about order? But even so, I don't think the exercise is asking us to consider that.)

Now, part (b) leads me to believe that a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$. Is that correct?

• Note in your definition $(A,\leq)$ is just a partially ordered set. There is no reference to numbers. You need to find the $S$-least element of $A$ and $B$. – James Aug 19 '15 at 23:04
• The exercise is indeed abour inclusion as an order relation . As for part (b) what you believe is right. – Bernard Aug 19 '15 at 23:05

The notation of $\leq$ is often used in mathematics as a general partial order. Not just the usual ordering on the natural/integers/rationals/reals. But in this context, $\leq$ is really just $\subseteq$.

So you are being asked to find the maximum and minimum elements of $(\mathcal P(C),\subseteq)$.

• Okay, so I'm right to say a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$? – Jordan Miller Aug 19 '15 at 23:13
• Yes. That is the case here. And note that in the definition of least element, $\leq$ is said to be a partial order, not anything else. – Asaf Karagila Aug 19 '15 at 23:15
• I must get clarification. The following in italics is also an exercise in my textbook: $(A,\leq)$ is a partially ordered set. That each finite non-empty subset of $A$ has a $maxB$ implies $(A,\leq)$ is totally ordered. Then does $\le$ refer again to some general partial order relation? – Jordan Miller Aug 19 '15 at 23:27
• Yes, a general partial order. Mathematics is much much much much more than just real numbers. – Asaf Karagila Aug 19 '15 at 23:34

In your case, the least element is an element of $A$, hence a subset of $C$ that is contained in all subsets of $C$. (Apply your definition in the second highlighted paragraph with $A=\mathcal{P}(C)$ and $B=A$), and so this set, I guess, is the empty set. Now, for the greatest element of $A$, what is the subset of $A$ that contains any subset of $A$?

• So, I am right to think that a least element, with respect to the set $A$, is a set such that it is a subset of each set in $A$? Does that make sense why I was a bit confused. To me, the relation of $\le$ is not the same as $\subseteq$ – Jordan Miller Aug 19 '15 at 23:11
• $\leq$ is just a symbol! – mich95 Aug 19 '15 at 23:12

Okay, I know that what follows basically repeats what responders have been trying to tell me, but hitherto, I was still misunderstanding the notation of $\le$. Thus, I merely post this answer as an expression of my understanding (finally!).

I refer to the definiton:

Let $(A,\le)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.

The $\le$ in the ordered pair $(A,\le)$ does not refer to the relation of less than or equal to, but rather serves as a "variable" for some partial order relation on A. With this in mind, the condition for each $x \in B$, $c \le x$ doesn't mean that $c$ must be less than or equal to $x$, but rather that $c$ must be in relation $\le$ to $x$.

Now, the partial order relation specified in the above exercise is the relation of set inclusion. Then, by definition, the least element of $A$ must be some set in $A$ such that it is a subset of each set in $A$.