\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Set Specification
Needed by:
Operations
Set Dualities
Set Intersections
Set Unions and Intersections
Uncertain Outcomes
Venn Diagrams
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Pair Intersections

Why

Does a set exist containing the elements shared between two sets? How might we construct such a set?

Definition

Let $A$ and $B$ denote sets. Consider the set $\Set{x \in A}{x \in B}$. This set exists by the principle of specification (see Set Specification). Moreover $(y \in \Set{x \in A}{x \in B}) \iff (y \in A \land y \in B)$. In other words, $\Set{x \in A}{x \in B}$ contains all the elements of $A$ that are also elements of $B$.

We can also consider $\Set{x \in B}{x \in A}$, in which we have swapped the positions of $A$ and $B$. Similarly, the set exists by the principle of specification (see Set Specification) and again $y \in \Set{x \in B}{x \in A} \iff (y \in B \land y \in B)$. Of course, $y \in A \land y \in B$ means the same as1 $y \in B \land y \in A$ and so by the principle of extension (see Set Equality)

\[ \Set{x \in A}{x \in B} = \Set{x \in B}{x \in A}. \]

We call this set the pair intersection of the set denoted by $A$ with the set denoted by $B$.

Notation

We denote the intersection fo the set denoted by $A$ with the set denoted by $B$ by $A \cap B$. We read this notation aloud as “A intersect B”.

Basic properties

All the following results are immediate.2

$A \cap \varnothing = \varnothing$
$A \cap B = B \cap A$
$(A \cap B) \cap C = A \cap (B \cap C)$
$A \cap A = A$
$(A \subset B) \iff (A \cap B = A)$.

  1. Future editions will name and cite this rule. ↩︎
  2. Proofs of these results will appear in the next edition. ↩︎
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