Can a set and it's PowerSet have anything in common?

Question

I wanted to know whether for all sets is there anything common between a set S and it's power set?

Is $S \,\cap P(S) \not= \phi$

Also, is an empty set exception for this?

Answer 1

There a sets where this is true and sets where this is false, so you cannot conclude anything in general.

Let $ S = \{ 1 \} $. Then $ \mathcal{P}(S) = \{ \emptyset, \{ 1 \} \} $. $ 1 \neq \emptyset $ and $ 1 \neq \{ 1 \} $, so $ S \cap \mathcal{P}(S) = \emptyset $.

On the other hand, let $ S = \{ \emptyset \} $. Then $ \mathcal{P}(S) = \{ \emptyset, \{ \emptyset \} \} $. Then $ \emptyset \in S $ and $ \emptyset \in \mathcal{P}(S) $, so $ S \cap \mathcal{P}(S) \neq \emptyset $.

Answer 2

$\emptyset\cap A=\emptyset$ for all $A$ and thus for $A=\mathcal P(\emptyset)$ as well. There are some sets such that $A\cap \mathcal P(A)=A$ or, in other words, such that $x\subseteq A$ for all $x\in A$. Such sets are called transitive. $\emptyset$ is one of those and, more generally, ordinals are. $\{\emptyset\}$ is a non-empty transitive set.

Answer 3

It is not true in general that $S\cap\wp(S)=\varnothing$.

The first exception on this that comes to mind is $\{\varnothing\}$.

Note that $\varnothing\in\{\varnothing\}$ and also $\varnothing\in\wp(S)$ for every set $S$, so also for $S=\{\varnothing\}$.

So we have: $$\varnothing\in\{\varnothing\}\cap\wp(\{\varnothing\})$$

This is not the only exception.

If $S$ is a set that contains some set $a$ and also singleton $\{a\}$ as elements (which is quite well possible, and for any set $a$) then we have:$$\{a\}\in S\cap\wp(S)$$