What is ${\cal P}(A ) \cap A$? I need some clarification
I am not sure but what is :
${\cal P}(A ) \cap  A$ equal to the empty set? For any $A$ ? If not how can we prove it ?
Dir example $A=\{1,2\}$,
${\cal P}(A) = \{ \{1\},\{2\},\{1,2\}, \emptyset \}$.
Then ${\cal P}(A) \cap A = A$ ?
 A: Specific Sets are specific sets and the element elements in the sets are elements in the sets.  They are not the same.   $1\in \{1,2\}$ because $1$ is an element in $\{1,2\}$ but $\{1\} \not \in \{1\}$ because $1$ is the number $1$ and $\{1\}$ is the set containing the number $1$.  The set containing the numbers $1$ is not the same thing as the number one and of the two elements in $\{1,2\}$ neither one of them are are the set containing $1$.
If $A= \{1,2\}$ then $A$ is a set with two elements.  They are the number $1$ and the number $2$.  All of its elements are numbers.  None of its elements are sets.
And $P(A) = \{\emptyset, \{1\},\{2\},\{1,2\}\}$.  That is a set with four elements.  They are: a set with no elements, a set with the element $1$, a set with the elements 2, and the set $A$.  All if it's elements are sets.  None of its elements are numbers.
$P(A)\cap A = \{w| w\in A; w\in P(A)\} = ???$.  $A$ has two elements.  They are $1$ and $2$.  $1$ is not an element of $P(A)$.  (Note: $\{1\}$ and $\{1,2\}$ are elements of $P(A)$ but neither of those are the number $1$).  ANd $2$ is not an element of $P(A)$.  So none of the elements of $A$ are elements of $P(A)$.
ANd likewise all the elements of $P(A)$ are not elements of $A$.  So $P(A)$ and $A$ have no elements in common.  SO $P(A)\cap A = \{w|w\in A;w\in P(A)\} = \{w| w$ is an element that is common to both $A$ and $P(A)\} = \{w|w $ does not exist$\} = \{\} =\emptyset$.
....
But that doesn't mean $P(A)\cap A = \emptyset$ always.
Consider $A = \{1,2, \{1\}\}$ that is a set with three elements: the number $1$, the number $2$, and the set $\{1\}$.  And $P(A)=\{\emptyset, \{1\},\{2\}, \{\{1\}\}, \{1,2\}, \{1,\{1\}\},\{2,\{1\}\}, \{1,2,\{1\}\}$ is a set with nine elements all of them sets.  Of the three elements in $A$ and the nine elements of $P(A)$ note that $\{1\}$ is an element the both contain.  And it is the only element they both contain.
So $\{1\} \in A$ and $\{1\} \in P(A)$ and $P(A)\cap A = \{\{1\}\}$.
That's it.
Another example is if $B= \{\emptyset\}$, the set that has a single element, that element being the empty set.  An if $A = \{\emptyset, B\} = \{\emptyset, \{\emptyset\}\}$ then $P(A) = \{\emptyset, \{\emptyset\}, \{B\},\{\emptyset,B\}\} = \{\emptyset, B,\{B\},A\}$.  And
$P(A) \cap A =  \{\emptyset, B,\{B\},A\} \cap \{\emptyset, B\}=$
$\{\emptyset, B\} = A$.
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I think the confusion come that if $A=\{babar, tantor, pink honk-honk\}$ is a set containing three elephants, that a student naively thinks that because the set is made of three elephants, that the set must be three elephants.
This is just not the case.  A set is a collection of things and what the set is the idea of the collection.  It is distinct from the things in the collection.
And then I guess the student often finds this complicated and figures it must have hard to understand rules.  But the rules are simple. SETS are collections of things, but the things inside the collections are not the same thing at all and have a completely separate "existence".  $\{A\}\ne A$ and $\{1,2\} \ne $"then numbers $1$ and $2$" and $\{1,2\}$ is not an element in the set $\{1,2,3\}$.  $1$ is an element.  And $2$ is an element. But the set containing $1$ and $2$ is not.
It's not a complicated rule.  It is a simple rule of looking through a microscope and never looking outside the range of the slide.  Ever.
A: An example where $P(A)\cap A=A$ and $A\neq\emptyset$:
Let $A=\{\emptyset\}$. Then $P(A)=\{\emptyset,\{\emptyset\}\}$, so $P(A)\cap A=\{\emptyset\}=A$.
An example where $P(A)\cap A=\emptyset$:
Let $A=\{a\}$, $a\neq\emptyset$. Then $P(A)=\{\emptyset,\{a\}\}$, so $P(A)\cap A=\emptyset$.
You can also construct situations where $\emptyset\neq P(A)\cap A\subsetneq A$, for instance by letting $A=\{\emptyset,a\}$, $a\neq\emptyset$. Essentially, make $A$ contain subsets of itself as elements.
A: ${\cal P}(A)$ is the set of subsets of $A$. One of its elements is $A$. Thus it is correct to write $A \in {\cal P}(A)$, $\{A\} \subset {\cal P}(A)$ and ${\cal P}(A) \cap \{A\} = \{A\}$. This can be disturbing as, in some context, a singleton $\{m\}$ is sometimes simply denoted by $m$.
In your example, the intersection is empty.
A: $P(A)\cap A$ consists of all the elements of $A$ that are also subsets of $A.$ What that is depends heavily on what $A$ is. It could be $\emptyset,$ it could be $A,$ it could be something in between. If $P(A)\cap A=A,$ we say $A$ is transitive.
In order to answer your question, we need to know what $1$ and $2$ are. If the natural numbers are to be viewed as non-sets (aka urelements), then certainly none of them are subsets and $P(A)\cap A=\emptyset,$ and it would be so for any subset of the naturals.
However, a lot of times, we define the natural numbers as sets $0=\emptyset, 1=\{0\}, 2=\{0,1\},\ldots$ in which case $P(A)\cap A$ for $A=\{1,2\}$ is still $\emptyset,$ but the argument is a bit more complicated (for instance, you can notice that both $1$ and $2$ have the empty set as an element, but $A$ does not, so $1$ and $2$ can't be subsets of $A$). And for instance, if we had $A=\{0,1\}=2,$ then $P(A)\cap A=A.$
