why is the intersection of two sets with no common element equal to the subset containing the empty set and not simply the empty set? working through some notes and, in essence, the following is said: 
"$A=\{1,2\}$ and $B=\{3,4\}$
$A\cap B = \{\emptyset\}$"
Is this a mistake? Should it not be $\emptyset$?
 A: Definitely a written/typed error.
If we consider the "$\{$" "$\}$" as notational placeholders to indicate "the list/description between them is the list/description of elements in a set then:
$\emptyset = \{\}$ is the set that when you list its elements.... doesn't have any.
$\{\emptyset\}=\{\{\}\}$ is a set that when you list its elements contains as its single element: the emptyset.
$A\cap B = \{x| x\in A; x \in B\}$.  And there is nothing that is in both $A$ and $B$ so $A\cap B=$ a set who if you listed its elements would contain.... nothing $= \{\}=\emptyset$.
If we had a set $C= \{\emptyset, Barbara, \text{a mushy banana}\}$ then $\emptyset$ is an an elements of $C$ so $\emptyset \in C$ but as $A=\{1,2\}$ and neither $1$ nor $2$ is the same thing as the empty set, so $\emptyset$ is NOT an elements of $A$ and $\emptyset \not \in A$.  Likewise  if $B = \{3,4\}$ then $\emptyset \not \in B$.
And if $\emptyset$ is in neither $A$ nor $B$ it certainly can't be in both $A$ and $B$.
But, you knew all that reasoned it out yourself.  I just want to pound home that this is utterly consistent and logical and you should trust that if you look down you will see that both your feet actually are firmly on the ground.
A: Indeed it is a mistake. $\emptyset$ doesn't belong to $A$ (nor to $B$) and therefore can't be an element of $A \cap B$ which is equal to $\emptyset$.
A: Hint,  The set $\{\emptyset \}$ has one element, but the set $\emptyset$ has no element. 
A: Yes, this is a mistake, it should be $A\cap B=\emptyset$.
