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.

  • 1
    $\begingroup$ You have to show the definition is satisfied. Whatever your definition may be. $\endgroup$ – David Peterson May 18 '17 at 4:57
  • $\begingroup$ 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? $\endgroup$ – Juanito May 18 '17 at 5:11
  • $\begingroup$ 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. $\endgroup$ – celtschk May 18 '17 at 5:12

$$\forall B \in P(A), B \subseteq B$$

Hence it is reflexive.

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

Hence it is antisymmetry.

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

Hence it is transitive.


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