2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $B$. $\endgroup$ – James Aug 19 '15 at 23:04
  • 1
    $\begingroup$ The exercise is indeed abour inclusion as an order relation . As for part (b) what you believe is right. $\endgroup$ – Bernard Aug 19 '15 at 23:05
1
$\begingroup$

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 $(\mathcal P(C),\subseteq)$.

$\endgroup$
  • $\begingroup$ 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 $D$? $\endgroup$ – Jordan Miller Aug 19 '15 at 23:13
  • 1
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Aug 19 '15 at 23:15
  • $\begingroup$ 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? $\endgroup$ – Jordan Miller Aug 19 '15 at 23:27
  • 1
    $\begingroup$ Yes, a general partial order. Mathematics is much much much much more than just real numbers. $\endgroup$ – Asaf Karagila Aug 19 '15 at 23:34
0
$\begingroup$

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 $A$?

$\endgroup$
  • $\begingroup$ 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$ $\endgroup$ – Jordan Miller Aug 19 '15 at 23:11
  • $\begingroup$ $\leq$ is just a symbol! $\endgroup$ – mich95 Aug 19 '15 at 23:12
0
$\begingroup$

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 $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.