Prove that the binary relation "is a subset of" is a...

Prove that the binary relation "is a subset of" is a partial order (POSET)?

Should I try to prove this in reference to the power set of a general set?

When is this relation a total order?

• A poset is a set with a relation, so the relation is not enough. What is the relation defined on? It has to be a set, so it cannot be "all sets." It has to be some set of sets. We'd say "is a subset of" is a partial order of some set, not just that it is a partial order. Sep 4, 2014 at 14:07
• The relation $\subseteq$ is a partial order on the class of all sets, you can immediately verify the defining conditions. As the class of all sets fails to be a set, we do not get a poset this way. However, if we work only with subsets of a "unversal" set, then the universal set together with $\subseteq$ is a poset. Sep 4, 2014 at 14:15

To prove that $R \subseteq \mathcal{P}(X) \times \mathcal{P}(X)$ defined by $(A,B) \in R \Leftrightarrow A\subseteq B$ is a partial order you need to show:

• Reflexivity: for all $A\in \mathcal{P}(X)$ we have $A\subseteq A$;
• Anti-symmetry: if $A,B \in \mathcal{P}(X)$ and $A\subseteq B$ and $B\subseteq A$ then $A=B$;
• Transitivity: if $A,B,C \in \mathcal{P}(X)$ and $A\subseteq B$ and $B\subseteq C$ then $A\subseteq C$.

It's very easy to prove these three items just using the definition of $\subseteq$.

Moreover, $\mathcal{P}(X)$ is totally ordered if and only if $X$ has at most 1 element.

See Partially ordered set or poset :

A (non-strict) partial order is a binary relation "≤" over a set $$P$$ which is reflexive, antisymmetric, and transitive, i.e., which satisfies for all $$a, b, c \in P$$ :

• $$a ≤ a$$ (reflexivity);

• if $$a ≤ b$$ and $$b ≤ a$$ then $$a = b$$ (antisymmetry);

• if $$a ≤ b$$ and $$b ≤ c$$ then $$a ≤ c$$ (transitivity).

We have to show that the $$\subseteq$$ relation over the power set $$\mathcal P(X)$$ has the three above properties, using the definition :

$$A \subseteq B$$ iff for all $$x$$, if $$x \in A$$, then $$x \in B$$.