# 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 $\displaystyle A \subseteq \mathcal{P}(A)$, Prove that $\displaystyle \mathcal{P}(A) \subseteq \mathcal{P}(\mathcal{P}A)$.

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

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

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

Quote:

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

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