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How to calculate X $\cap$ $\{X\}$ for finite sets to develop an intuition for intersections?

If $X$ = $\{$1,2,3$\}$, then what is $X$ $\cap$ $\{X\}$?

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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 $\emptyset$.

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$\phi$ ... The set $\{1,2,3\} \notin \{1,2,3\}$.

a collection which contains itself is NOT a set...

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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, $$X \cap \left\{ X \right\} = \emptyset.$$

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$\{X\}$ 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 $A \not \in A$.)

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