\newlec

\note $\{ \}$ and $\emptyset$ are the empty set, a set with no elements.
\vskip1em
\note $\{\emptyset\}$ is NOT the empty set. It is a set with one element, the element that is the empty set.
\vskip1em
\ex In set notation, write the set of positive integers less than $1000$ and which are multiples of $7$.
\ic{Might be good to give students a minute to try this}
\soln $\{n \in \mathbb{N} : n < 1000 \wedge 7 \mid n\}$. Another answer is given by
\[
\{7k: k \in \mathbb{N} \wedge k \leq 142\}
\]
\note The $:$ symbol means ``such that''. Sometimes $|$ is used as well (though because we use it for divisibility, we won't use it in this context very often if at all).
\ic{This is the 7 minute mark}
\newpage
\hdc
Describe the following sets using set-builder notation:

\begin{enumerate}
\item Set of even numbers between 5 and 14 (inclusive).
\vskip2.75em
\item All odd perfect squares.
\vskip2.75em
\item Sets of three integers which are the side lengths of a (non-trivial) triangle.
\vskip2.75em
\item All points on a circle of radius 8 centred at the origin.
\end{enumerate}
\vskip3em
\ic{5 minutes to try on their own and 5 to take up}

\soln
\begin{enumerate}
\item $\{6,8,10,12,14\}$ or $\{n \in \mathbb{N}: 5 \leq n \leq 14 \wedge 2 \mid n\}$
\item $\{(2k+1)^{2} : k \in \mathbb{Z}\}$ (or $\mathbb{N}$ overlap doesn't matter!)
\item $\{(a,b,c) : a,b,c\in \mathbb{N} \wedge a < b+c \wedge b < a+c \wedge c < a+b\}$
\item $\{(x,y) : x,y \in \mathbb{R} \wedge x^{2} + y^{2} = 8^{2}\}$
\end{enumerate}

\ic{This is the 17 minute mark}

\newpage

\noindent \textbf{Set Operations}. Let $S$ and $T$ be sets. Define
\begin{enumerate}
\item $\#S$ or $|S|$. Size of the set $S$.
\item $S \cup T = \{x : x \in S \vee x \in T\}$ (Union)
\item $S \cap T = \{x : x \in S \wedge x \in T\}$ (Intersection)
\item $S-T = \{x \in S : x \notin T\}$ (Set difference)
\item $\bar{S}$ or $S^{c}$ (with respect to universe $U$) the complement of $S$, that is
\[
S^{c} = \{x \in U : x \notin S\} = U-S
\]
\item $S \times T = \{(x,y) : x \in S \wedge y \in T\}$ (Cartesian Product)
\end{enumerate}

\ex $(1,2) \in \mathbb{Z}\times \mathbb{Z}$, $(2,1) \in \mathbb{Z}\times \mathbb{Z}$, BUT $(1,2) \neq (2,1)$.
\vskip1em
\note $\mathbb{Z} \times \mathbb{Z}$ and $\{(n,n) : n \in \mathbb{Z}\}$ are different sets!!!
\vskip1em
\ex
\begin{align*}
\mathbb{Z} &= \{m \in \mathbb{Z} : 2 \mid m\} \cup  \{2k+1 : k \in \mathbb{Z}\}\\
\emptyset &= \{m \in \mathbb{Z} : 2 \mid m\} \cap  \{2k+1 : k \in \mathbb{Z}\}
\end{align*}

\ic{This is the 30-33 minute mark}

\defn Let $S$ and $T$ be sets. Then
\begin{enumerate}
\item $S \subseteq T$: $S$ is a subset of $T$. Every element of $S$ is an element of $T$.
\item $S \subsetneq T$: $S$ is a proper/strict subset of $T$. Every element of $S$ is an element of $T$ and some element of $T$ is not in $S$.
\item $S \supseteq T$: $S$ contains $T$. Every element of $T$ is an element of $S$.
\item $S \supsetneq T$: $S$ properly/strictly contains $T$. Every element of $T$ is an element of $S$ and some element of $S$ is not in $T$.
\end{enumerate}

\defn $S=T$ means $S \subseteq T$ and $T \subseteq S$. 
\vskip1em
\ex $\{1,2\} = \{2,1\}$
\vskip1em
\ex Prove $\{n \in \mathbb{N} : 4 \mid (n+1)\} \subseteq \{2k +1 : k \in \mathbb{Z}\}$
\vskip1em
\pf Let $m \in \{n \in \mathbb{N} : 4 \mid (n+1)\} $. Then $4 \mid (m+1)$. Thus, $\exists \ell \in \mathbb{Z}$ such that $4\ell = m+1$. Now 
$$m = 2(2\ell)-1 = 2(2\ell)-2+2-1 = 2(2\ell-1)+1.$$
Hence $m \in \{2k+1 : k \in \mathbb{Z}\}$. \qed

\ic{This is the 40-43 minute mark. You might run out of time in the next example. Carry forward to Lecture 7 as need be.}

\ex Show $S=T$ if and only if $S \cap T = S \cup T$.
\vskip1em
\pf Suppose $S = T$. To show $S \cap T = S \cup T$ we need to show that $S \cap T \subseteq S \cup T$ and that $S \cap T \supseteq S \cup T$
\vskip1em
First suppose that $x \in S \cap T$. Then $x \in S$ and $x \in T$. Hence $x \in S \cup T$.
\vskip1em
Next, suppose that $x \in S \cup T$. Then $x \in S$ or $x \in T$. Since $S=T$ we have in either case that $x \in S$ and $x \in T$. Thus $x \in S\cap T$. This shows that $S \cap T = S \cup T$ and completes the forward direction.
\vskip1em
Now assume that $S \cap T = S \cup T$. We want to show that $S=T$ which we do by showing that $S \subseteq T$ and $T \subseteq S$. 
\vskip1em
First, suppose that $x \in S$. Then $x \in S \cup T = S \cap T$. Hence $x \in T$.
\vskip1em
Next, suppose that $x \in T$. Then $x \in S \cup T = S \cap T$. Hence $x \in S$. Therefore, $S=T$. \qed

\ic{The last two points give a good learning moment to explain when the word `similarly' can be used. This is the 50 minute mark.}
