0
$\begingroup$

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.

$\endgroup$
  • 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
0
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.