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?
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$\begingroup$ It means containment. Llike, $A <B$ if $A \subset B$ $\endgroup$– Bhaskar VashishthMay 30, 2015 at 7:46
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$\begingroup$ @Bhaskar Vashishth: Can you elaborate, please? $\endgroup$– user142971May 30, 2015 at 7:48
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$\begingroup$ Can you provide a quote of where this phrase appears? I'm not sure what you mean by "ordering of sequence of subset". $\endgroup$– augurarMay 30, 2015 at 7:49
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$\begingroup$ The story is long @augurar; I'm also confused whether to include the word 'sequence'. So, I gave ?? $\endgroup$– user142971May 30, 2015 at 7:51
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3 Answers
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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.
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$\begingroup$ Thanks, sir. +1 for this. At the first sight, it nailed the doubt. $\endgroup$– user142971May 30, 2015 at 7:56
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$\begingroup$ 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? $\endgroup$– user142971May 30, 2015 at 8:17
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$\begingroup$ @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. $\endgroup$– augurarMay 30, 2015 at 8:26
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$\begingroup$ 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:| $\endgroup$– user142971May 30, 2015 at 8:40
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$\begingroup$ @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. $\endgroup$– augurarMay 30, 2015 at 9:26
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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).
- Reflexive- As $A \subseteq A$ for all $A \in \mathcal{P}(\mathtt{S})$
- AntiSymmetric- As $A\subseteq B$ and $B \subseteq A$ $\implies A=B$
- 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.
answered May 30, 2015 at 7:56
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$\begingroup$ If set $S$ has less than 3 elements (or even less than 2 elements), can we infer that inclusion is a partial order? $\endgroup$ Oct 16, 2020 at 23:04
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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.
answered Feb 13, 2019 at 11:33