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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?

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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.

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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 $\cap \varnothing$.

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  • $\begingroup$ So, do we use the fact that the implication $p \to q$ is always true, if $p$ is wrong? $$$$ And how could I use this to prove that $\cap \varnothing$ is not a set? $\endgroup$ – evinda Oct 5 '14 at 12:56
  • $\begingroup$ The usual proof that the class of all sets is not a set is the argument from Russell's paradox. $\endgroup$ – user14972 Oct 5 '14 at 13:11
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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.

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  • $\begingroup$ Why do we take $B=\{ \{ X \} \}$ and not $B=\{ X \}$? Is $X$ not a set? $\endgroup$ – evinda Oct 5 '14 at 12:54
  • $\begingroup$ Also, how could we prove it formally? $\endgroup$ – evinda Oct 5 '14 at 12:54
  • $\begingroup$ Because choosing $B=\{X\}$ would yield $X\subseteq \bigcap A$, and we want $X\in \bigcap A$. As for proving it formally, it depends on how you define $\bigcap A$. $\endgroup$ – Thomas Andrews Oct 5 '14 at 13:28
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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.

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  • $\begingroup$ 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? $\endgroup$ – evinda Oct 5 '14 at 12:47
  • $\begingroup$ @Mauro's answer is perfect: the thing you want to prove is not a set is actually undefined. Suppose that I wanted to prove that $ANSBD$ is not a set. How could I do that? Well, first, I'd have to know what that collection of symbols denotes. When we discover it doesn't denote anything -- its semantics are undefined -- we pretty much have to give up proving statements about it. $\endgroup$ – John Hughes Oct 5 '14 at 17:54

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