I am confused about the following.
Could you explain me why if $A=\varnothing$,then $\cap A$ is the set of all sets?
Definition of $\cap A$:
For $A \neq \varnothing$:
$$x \in \cap A \leftrightarrow (\forall b \in A )x \in b$$
EDIT:
I want to prove that $\cap \varnothing$ is not a set.
To do that, do I have to begin, supposing that it is a set?
See Herbert Enderton, Elements of Set Theory (1977), page 24 :
Suppose we want to take the intersection of infinitely many sets $b_0, b_1, \ldots$. Then where $A = \{ b_0, b_1,\ldots \}$ the desired intersection can be informally characterized as $\cap A = \cap_i b_i = \{ x | x$ belongs to every $b_i \in A \}$.
In general, we define for every non empty set $A$, the intersection $\cap A$ of $A$ by the condition
$x \in \cap A \Leftrightarrow x$ belongs to every member of $A$.
What happens if $A = \emptyset$ ? For any $x$ at all, it is vacuously true that $x$ belongs to every member of $\emptyset$. (There can be no member of $\emptyset$ to which $x$ fails to belong.) Thus it looks as if $\cap \emptyset$ should be the class $V$ of all sets. By Theorem 2A [page 22 : There is no set to which every set belongs], there is no set $C$ such that for all $x$,
$x \in C \Leftrightarrow x$ belongs to every member of $\emptyset$
since the right side is true of every $x$. This presents a mild notational problem: How do we define $\cap \emptyset$ ? The situation is analogous to division by zero in arithmetic. How does one define $a/0$ ? One option is to leave $\cap \emptyset$ undefined, since there is no very satisfactory way of defining it. This option works perfectly well, but some logicians dislike it. It leaves $\cap \emptyset$ as an untidy loose end, which they may later trip over. The other option is to select some arbitrary scapegoat (the set $\emptyset$ is always used for this) and define $\cap \emptyset$ to equal that object.
Either way, whenever one forms $\cap A$ one must beware the possibility that perhaps $A = 0$. Since it makes no difference which of the two options one follows, we will not bother to make a choice between them at all.
Usually, $\cap A$ is defined as the class of all things that are in every element of $A$.
No matter what $x$ is, $\forall y \in \varnothing: x \in y$ is vacuously true, therefore, all sets are members of the class $\cap \varnothing$.
Intuitively, if $A\subseteq B$ then $\bigcap B\subseteq \bigcap A$. Now, for any set $X$, let $B=\{\{X\}\}$. Then $\emptyset = A\subseteq B$ and $\{X\}=\bigcap B \subseteq \bigcap A$, so $X\in\bigcap A$.
But that definition cannot actually be done - there is no set of all sets.
I believe that the statement you want us to prove is wrong. For one thing, there is (in the versions of set theory that I know) no set $R$ of all sets, for if there were, you could form $$ Q = \{X \in R : X \notin X\} $$ the set of all sets that are not elements of themselves. The statement $Q \in Q$ then becomes neither true nor false, which is problematic.
To do that, do I have to begin, supposing that it is a set?
– evinda Oct 05 '14 at 12:47