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I was recently introduced to partial order relations. Although I understand the concept of a relation, I do not understand how subsets can have a sequence, or what that has to do with the phrase "ordering of sets by inclusion." Can anyone help me clarify the concept with an intuitive example?

Matt Raymond
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3 Answers3

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Let $\mathcal{S}$ be a collection of sets. Then for sets $S_1, S_2 \in \mathcal{S}$, we can define $S_1 \prec S_2$ to mean $S_1 \subset S_2$.

So for example, considering sets of integers, we can say $\{1, 2, 3\} \prec \{1, 2, 3, 4\}$. This is a partial order because not all sets are comparable. For instance $\{1, 2, 3\}$ and $\{1, 3, 5\}$ are not comparable because neither is a subset of the other.

augurar
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  • Thanks, sir. +1 for this. At the first sight, it nailed the doubt. –  May 30 '15 at 07:56
  • Sir, one question; by ordering what do we get? The definition of set in naive set theory is that it is an unordered collection of mathematical objects. So, what does this "ordering' all mean? –  May 30 '15 at 08:17
  • @user36790 It's a way of giving the set additional structure. A set along with a partial order is called a partially ordered set. We can then study the properties of this new object. – augurar May 30 '15 at 08:26
  • A set along with a partial order means? Suppose ${a,b,c,d}$ where $a < b< c$. Then the set along with partial order relation is meaning ${(a,b),(a,c),(b,c)}$ this? I am new, so I face problems with these phrases like set along with a relation:| –  May 30 '15 at 08:40
  • @user36790 It just means we think about the set and the relation together as a single mathematical object. In your comment you provide a set $S$ and a relation $R$ on the set. If we consider these together, we can think of the pair as representing a partially ordered set. – augurar May 30 '15 at 09:26
  • Then you are saying ${{a,b,c,d}, R}$ is the partial order set? Elsewhere, I saw $({a,b.c,d} R)$ as termed as loop-digraph. Is it different from the above? –  May 30 '15 at 09:41
  • @user36790 A set and a relation can also be used to represent a digraph, it all depends on the interpretation you attach to it. If the relation is a partial order, then the corresponding digraph has no directed cycles. – augurar May 30 '15 at 09:55
  • So, a poset is the power set basically & not every element of the power set can be related by partial order, right? Then how can I use this to define the supremum & infimum of a set, can you tell? BTW, can you tell me what directed cycles mean? –  Jun 01 '15 at 11:52
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Suppose you have a set $\mathtt{S}$. Now denote power set of $\mathtt{S}$ by $\mathcal{P}(\mathtt{S})$. Then $\mathcal{P}(\mathtt{S})$ is a poset under the relation $\subseteq$ (inclusion).

  1. Reflexive- As $A \subseteq A$ for all $A \in \mathcal{P}(\mathtt{S})$
  2. AntiSymmetric- As $A\subseteq B$ and $B \subseteq A$ $\implies A=B$
  3. Transitive- You can see it now, I am sure.

So it is a partial order on $\mathcal{P}(\mathtt{S})$.

On any set of sets, inclusion is a partial order.

Bhaskar Vashishth
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Given any family of sets $\mathcal{F}$ there is a poset $P=(\mathcal{F},\{(A,B)\in\mathcal{F}^2:A\subseteq B\})$ corresponding to that family ordered by inclusion. Now by an "inclusion maximal/maximum/minimal/minimum" set in $\mathcal{F}$ what is meant is simply a maximal/maximim/minimal/minimum element of $\mathcal{F}$. Also its worth noting that every partial order is isomorphic to a family of sets ordered by inclusion, in particular the principal lower sets of any partial order when ordered by set inclusion are always isomorphic to said partial order.

Ethan Splaver
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