1. ## Transitive set

Prove that $A$ is a transitive set if and only if $PA$ is a transitive set.
I think I was able to prove one direction.
Suppose $A$ is transitive. We need to show that $PA \subseteq PPA$. Let $x \in PA$. Then $x \subseteq A$. Since $A$ is transitive, we know $A\subseteq PA$. Hence, $x \subseteq PA \Rightarrow x \in PPA$.

For the other direction. I need to show that $A \subseteq PA$. I have a hard time of effectively using the hypothesis that $PA$ is a transitive set. Can someone give me a hand here? Thanks.

2. Originally Posted by namelessguy
Prove that $A$ is a transitive set if and only if $PA$ is a transitive set.
I think I was able to prove one direction.
Suppose $A$ is transitive. We need to show that $PA \subseteq PPA$. Let $x \in PA$. Then $x \subseteq A$. Since $A$ is transitive, we know $A\subseteq PA$. Hence, $x \subseteq PA \Rightarrow x \in PPA$.

For the other direction. I need to show that $A \subseteq PA$. I have a hard time of effectively using the hypothesis that $PA$ is a transitive set. Can someone give me a hand here? Thanks.
If X is a transitive set, then the followings hold (You need to verify the below lemmas).

Lemma 1. $x \in X \rightarrow x \subseteq X$.
Lemma 2. $X \subseteq PX$.

Suppose PA is a transitive set.

If $A \in PA$, then $A \subseteq PA$ by Lemma 1.
If $A \subseteq PA$, then A is a transitive set by Lemma 2.

3. Originally Posted by aliceinwonderland
If X is a transitive set, then the followings hold (You need to verify the below lemmas).

Lemma 1. $x \in X \rightarrow x \subseteq X$.
Lemma 2. $X \subseteq PX$.

Suppose PA is a transitive set.

If $A \in PA$, then $A \subseteq PA$ by Lemma 1.
If $A \subseteq PA$, then A is a transitive set by Lemma 2.
Thank you very much for your help Alice. We did prove the lemmas in class, one more lemma we proved is that if $A$ is transitive then $\bigcup A \subseteq A$. I didn't approach this problem as you did, which is why I was stuck I think. I was trying to show that any element in $A$ is also an element in $PA$ by the usual way, but I couldn't get anywhere.