# Subset of a Power set

• Oct 18th 2010, 07:45 AM
novice
Subset of a Power set
I am a little confused by this question: Suppose $A \subseteq \mathcal{P}(A)$, Prove that $\mathcal{P}(A) \subseteq \mathcal{P}(\mathcal{P}A)$.

By definition, I know that $A\in \mathcal{P}(A)$ but not $A \subseteq \mathcal{P}(A)$. I believe only when $A=\varnothing$ that it can be a subset of $\mathcal{P}(A)$. I am right?
• Oct 18th 2010, 08:35 AM
emakarov
Quote:

Originally Posted by novice
By definition, I know that $A\in \mathcal{P}(A)$ but not $A \subseteq \mathcal{P}(A)$.

In general, it is not the case that $A \subseteq \mathcal{P}(A)$; however, it is possible. Such sets are called transitive. Suppose $A\subseteq\mathcal{P}(A)$. Then for every $x\in A$, $x\in\mathcal{P}(A)$, i.e., $x\subseteq A$. This in turn means that for every $y\in x$, $y\in A$, which explains the name.

Quote:

Originally Posted by novice
I believe only when $A=\varnothing$ that it can be a subset of $\mathcal{P}(A)$. I am right?

Other examples include $\{\emptyset,\{\emptyset\}\}$, $\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\ }\}\}$ and other von Neumann ordinals.