# Partial order on power sets

Let's say $A = {1, 2, 3, 4}$

How can I show that $⊆$ is a partial order on $P(A)$?

Would I still have to show that it's reflexive/anti-symmetric/transitive?

I'm just confused on how to apply partial orders on power sets.

• You have to show the definition is satisfied. Whatever your definition may be. – David Peterson May 18 '17 at 4:57
• You are not applying partial order on power sets. You are applying partial order on the elements of the power set. Does that make things clearer? – Juanito May 18 '17 at 5:11
• Partial orders on power sets work exactly the same way as partial orders on any other sets. It's just that those elements you compare happen to themselves be sets. – celtschk May 18 '17 at 5:12

$$\forall B \in P(A), B \subseteq B$$
$$\forall B, C \in P(A), B \subseteq C \wedge C \subseteq B \implies B=C.$$
$$\forall B, C, D \in P(A), B \subseteq C \wedge C \subseteq D \implies B \subseteq D.$$