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Section 1.4 Intersection

Definition 1.4.1.

Let \(S\) and \(T\) be two sets. The intersection \(S \cap T\) is the set consisting of the elements \(S\) and \(T\) have in common. That is, \(x \in S \cap T\) exactly when both \(x \in S\) and \(x \in T\text{.}\) One can write

\begin{equation*} S \cap T = \{x : (x \in S) \wedge (x \in T)\} \end{equation*}

When \(S\) and \(T\) are subsets of some common set \(U\) and are given by set-builder notation, we can use conjunction to describe the intersection. That is, if

\begin{equation*} S = \{x \in U : \phi(x) \} \end{equation*}

for a statement \(\phi(x)\) with free variable \(x\text{,}\) and if

\begin{equation*} T = \{x \in U : \psi(x) \} \end{equation*}

for a statement \(\psi(x)\text{,}\) then the intersection is given by

\begin{equation*} S \cap T = \{x \in U : \phi(x) \wedge \psi(x)\}\text{.} \end{equation*}

Consider the sets

\begin{equation*} S = \{n \in \NN_0 : n\text{ is divisible by } 2\} \text{ and } T = \{n \in \NN_0 : n\text{ is divisible by } 5\}\text{.} \end{equation*}

Then the intersection is

\begin{equation*} S \cap T = \{n \in \NN_0 : n\text{ is divisible by both } 2\text{ and } 5\}\text{.} \end{equation*}

From this you can see that:

\begin{equation*} S \cap T = \{n \in \NN_0 : n\text{ is divisible by } 10\}=\{0,10,20,\dots\}\text{.} \end{equation*}

Consider the sets

\begin{equation*} S = \{x \in \NN_0 : x \geq 5\} \end{equation*}

and

\begin{equation*} T = \{x \in \NN_0 : x \leq 7\}\text{.} \end{equation*}

Then the intersection is

\begin{equation*} S \cap T = \{x \in \NN_0 : 5 \leq x \leq 7 \} = \{5,6,7\}\text{.} \end{equation*}

Consider the sets

\begin{equation*} S = \{x \in \NN_0 : x \geq 7\} \end{equation*}

and

\begin{equation*} T = \{x \in \NN_0 : x \leq 5\}\text{.} \end{equation*}

Then the intersection is

\begin{equation*} S \cap T = \{x \in \NN_0 : (x \geq 7)\wedge(x \leq 5) \} = \varnothing\text{.} \end{equation*}

We have a word for this situation.

Definition 1.4.5.

Let \(U\) be a set, and let \(S \subseteq U\) and \(T\subseteq U\) be subsets. We say that \(S\) and \(T\) are disjoint if and only if \(S \cap T = \varnothing \text{.}\)