# What does "ordering of sets by inclusion" mean?

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?

• It means containment. Llike, $A <B$ if $A \subset B$ May 30, 2015 at 7:46
– user142971
May 30, 2015 at 7:48
• Can you provide a quote of where this phrase appears? I'm not sure what you mean by "ordering of sequence of subset". May 30, 2015 at 7:49
• The story is long @augurar; I'm also confused whether to include the word 'sequence'. So, I gave ??
– user142971
May 30, 2015 at 7:51

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.

• Thanks, sir. +1 for this. At the first sight, it nailed the doubt.
– user142971
May 30, 2015 at 7:56
• 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?
– user142971
May 30, 2015 at 8:17
• @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. May 30, 2015 at 8:26
• 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:|
– user142971
May 30, 2015 at 8:40
• @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. May 30, 2015 at 9:26

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.

• If set $S$ has less than 3 elements (or even less than 2 elements), can we infer that inclusion is a partial order? Oct 16, 2020 at 23:04

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.