What is the value of the intersection of X and the set containing X? How to calculate X $\cap$ $\{X\}$ for finite sets to develop an intuition for intersections?
If $X$ = $\{$1,2,3$\}$, then what is $X$ $\cap$ $\{X\}$? 
 A: For your example, it is $\emptyset$, because none of elements of $X$ is in $\{X \}$, and none of element me of $\{X\}$ is in $X$.
For general case, one axiom of set theory is that $A \notin A$ for any set (see this post), which means $\{A\}$ does not have any element in $A$, and thus they intersection is $\emptyset$.
A: $\phi$ ... The set $\{1,2,3\} \notin \{1,2,3\}$.
a collection which contains itself is NOT a set...
A: As far as developing intuition for intersection, the idea of $A \cap B$ are the elements that $A$ and $B$ both have in common. So if we're looking at $X \cap \left\{ X \right\}$ where $X = \left\{ 1,2,3 \right\}$ then it is a matter of 
$$X \cap \left\{ X \right\} = \left\{ 1,2,3 \right\} \cap \left\{ \left\{ 1,2,3 \right\} \right\}.$$
However, there is a rather subtle difference here between the left and right side of the intersection. The left side, $\left\{ 1,2,3\right\} = X$, is the set at hand. Where the right side, $\left\{ X \right\}$ is viewing the set $X$ as an element, which is different than $X$ itself, so they have nothing in common. Hence,
$$X \cap \left\{ X \right\} = \emptyset.$$
A: $\{X\}$ contains one element and one element only.  So as $E \cap F \subset F$ we know $E \cap \{X\} \subset \{X\}$.  So either $E \cap \{X\} = \{X\}$ if $X \in E$ or $E \cap \{X\} = \emptyset$ if $X \not \in E$.  
It violates the axioms of set theory to have a set such that $X \in X$ (a set can't be an element of itself). So $X \in X$ so $X \cap \{X\} = \emptyset$.
As per your example.
$\{1,2,3\} \cap \{\{1,2,3\}\}$....  $1 \not \in  \{\{1,2,3\}\}$, $2 \not \in  \{\{1,2,3\}\}$, $3 \not \in  \{\{1,2,3\}\}$, and $\{1,2,3\} \not \in \{1,2,3,\}$.  And $1,2,3,\{1,2,3\}$ are everything in either set and none of them  are in both sets. so $\{1,2,3\} \cap \{\{1,2,3\}\} = \emptyset$.
Note:  It doesn't matter if $X$ is finite or infinite or empty.  $X \cap \{X\} = \emptyset$.  (Unless you ignore the axiom that $A \not \in A$.)
