In Section 4.1, we explored some elementary properties of sets, such as element membership, subset relationships, and cardinality. In this section, we explore operations with sets, and common ways of combining sets with other sets. We will establish our results via the element-chasing method of proving set equality.
Let \(A\) and \(B\) be sets. The union of \(A\) and \(B\text{,}\) denoted \(A\cup B\text{,}\) is the set of all elements \(x\) which are in \(A\) or \(B\) (or both):
\begin{equation*}
A\cup B := \setof{x}{x\in A \lor x\in B}.
\end{equation*}
The intersection of \(A\) and \(B\text{,}\) denoted \(A\cap B\text{,}\) is the set of all elements \(x\) which are in both \(A\) and \(B\text{:}\)
\begin{equation*}
A\cap B := \setof{x}{x\in A\land x\in B}.
\end{equation*}
If \(A\cap B = \emptyset\text{,}\) we say \(A\) and \(B\) are disjoint.
Let \(A = \set{5,10,15}\text{,}\)\(B = \set{\emptyset, 5, \pi}\text{,}\)\(C = \set{ 2, \pi, -\sqrt{7}, \set{2, - \sqrt{7}}}\text{,}\) and \(D = \set{2, -\sqrt{7}}\text{.}\) Give each of the following sets in roster form.
Given the way set theory has reshaped mathematics over the past 150 years, many mathematical proofs rely on the equality of certain sets. While Axiom 4.1.7 gives a precise meaning to set equality, we typically verify set equality using a technique sometimes called element chasing.
The typical way to prove that sets \(A\) and \(B\) are equal via Axiom 4.1.7 is to show that \(A\subseteq B\) and \(B\subseteq A\text{.}\) We generally do this in two steps: first by letting \(x\in A\) and demonstrating via mathematics and logical deduction that \(x\in B\text{,}\) and then letting \(x\in B\) and demonstrating that \(x\in A\text{.}\)
Let \(x\in A\cup B\text{.}\) Then \(x\in A\) or \(x\in B\text{,}\) so \(x\in B\) or \(x\in A\text{.}\) But this means that \(x\in B\cup A\text{,}\) so \(A\cup B\subseteq B\cup A\text{.}\)
Now suppose \(x\in B\cup A\text{.}\) Then \(x\in B\) or \(x\in A\text{,}\) which means \(x\in A\) or \(x\in B\text{.}\) This means that \(x\in A\cup B\text{,}\) so \(B\cup A\subseteq A\cup B\text{.}\)
We next wish to introduce the notion of the complement of a set \(A\text{,}\) which we will denote \(A^c\text{.}\) Informally, \(A^c\) consists of all elements which are not in \(A\text{...}\)but, wait —really? All elements? So if \(A\) simultaneously does not contain a \(5\times 5\) Rubik’s cube and the Lincoln Memorial, then \(A^c\) does?
This is clearly untenable. Until now, we’ve been playing a little fast and loose with one thing: the idea of a larger universe \(U\) in which our sets live. In mathematics we often consider sets of numbers; if we are interested in a set of integers, the universe might be \(\Z\) (or maybe \(\N\) if we know they’re positive). This was implicit in our definition of the set \(B\) in Example 4.1.4, as the first half of the set-builder form specified that we were considering only \(x\in \Z\text{.}\)
This universe is typically implied (as it was in Example 4.1.4), but in defining and using both the complement and relative complement (the set difference), our definitions and theorems will name the universal set \(U\text{.}\)
Let \(A\) be a set contained in some universe \(U\text{.}\) Then the complement of \(A\), denoted \(A^c\text{,}\) is the set of elements \(x\in U\) with \(x\notin A\text{.}\) That is:
Let \(A\) and \(B\) be sets. The (set) difference of \(A\) and \(B\) (sometimes called the relative complement of \(A\) in \(B\)), denoted \(A\setminus B\text{,}\) is the set of elements \(x\) such that \(x\in A\) and \(x\notin B\text{:}\)
The Cartesian product of \(A\) and \(B\text{,}\) denoted \(A\times B\text{,}\) is the set of all ordered pairs whose first coordinate is from \(A\) and whose second coordinate is from \(B\text{:}\)
We include the definition of ordered pairs in Definition 4.2.23 for completeness, but in practice we rarely use it, instead relying on our intuition for how ordered pairs work in the \(xy\)-plane, \(\R\times \R\text{.}\)
