# 1.1: Basic Concepts of Set Theory

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Intuitively, a set is a collection of objects with certain properties. The objects in a set are called the **elements **or **members of the set**. We usually use uppercase letters to denote sets and lowercase letters to denote elements of sets. If \(a\) is an element of set \(A\), we write \(a \in A\). If \(a\) is not an element of a set \(A\), we write \(a \notin A\). To specify a set, we can list all of its elements, if possible, or we can use a defining rule. For instance, to specify the fact that a set \(A\) contains four elements \(a, b, c, d\), we write

\[A=\{a, b, c, d\}.\]

To describe the set I containing all even integers, we write

\[E=\{x: x=2 k\text{ for some integer } k \}.\]

We say that a set \(A\) is a *subset* of a set \(B\) if every element of \(A\) is also an element of \(B\), and write \[A \subset B \text { or } B \supset A.\]

Two sets are *equal* if they contain the same elements. If \(A\) and \(B\) are equal, we write \(A=B\). The following result is straightforward and very convenient for proving equality between sets.

Two sets \(A\) and \(B\) are equal if and only if \(A \subset B\) and \(B \subset A\).

If \(A \subset B\) and \(A\) does not equal \(B\), we say that \(A\) is a *proper* subset of \(B\), and write \(A \subsetneq B\).

The set \(\boldsymbol{\theta}=\{x: x \neq x\}\) is called the *empty set*. This set clearly has no elements. Using **Theorem 1.1.1**, it is easy to show that all sets with no elements are equal. Thus, we refer to *the *empty set.

Throughout this book, we will discuss several sets of numbers which should be familiar to the reader:

- \(\mathbb{N}=\{1,2,3, \ldots\}\), the set of
*natural numbers*or*positive integers*. - \(\mathbb{Z}=\{0,1,-1,2,-2, \ldots\}\), the set of
*integers*(that is, the natural numbers together with zero and the negative of each natural number). - \(\mathbb{Q}=\{m / n: m, n \in \mathbb{Z}, n \neq 0\}\), the set of
*rational numbers*. - \(\mathbb{R}\), the set of
*real numbers*. - Intervals, for \(a, b \in \mathbb{R}\), we have

\([ a, b]=\{x \in \mathbb{R}: a \leq x \leq b\}\),

\((a, b]=\{x \in \mathbb{R}: a<x \leq b\}\),

\([ a, \infty)=\{x \in \mathbb{R}: a \leq x\}\),

\((a, \infty)=\{x \in \mathbb{R}: a<x\}\),

and similar definitions for \((a,b)\), \([a,b)\), \((-\infty,b]\), and \((-\infty,b)\). We will say more about the symbols \(\infty\) and \(-\infty\) in Section 1.5.

Since the real numbers are central to the study of analysis, we will discuss them in great detain in Sections 1.4, 1.5, and 1.6.

For two sets \(A\) and \(B\), the *union*, *intersection*, *difference*, and *symmetric difference *of \(A\) and \(B\) are given respectively by

\(A \cup B=\{x: x \in A \text { or } x \in B\}\)

\(A \cap B=\{x: x \in A \text { and } x \in B\}\)

\(A \backslash B=\{x: x \in A \text { and } x \notin B\}\), and

\(A \Delta B=(A \backslash B) \cup(B \backslash A)\).

If \(A \cap B=\emptyset\), we say that \(A\) and \(B\) are *disjoint*.

The difference of \(A\) and \(B\) is also called the *complement* of \(B\) in \(A\). If \(X\) is a *universal set*, that is, a set containing all the objects under consideration, then the complement of \(A\) in \(X\) is denoted simply by \(A^{c}\)

Let \(A\),\(B\), and \(C\) be subsets of a universal set \(X\). Then the following hold:

- \(A \cup A^{c}=X\);
- \(A \cap A^{c}=\emptyset\);
- \(\left(A^{c}\right)^{c}=A\);
- \((\mathit{Distributive law}) A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\);

The proofs of the following properties are similar to those in **Theorem 1.1.2**. We include the proof of part (a) and leave the rest as an exercise.

Let \(\left\{A_{i}: i \in I\right\}\) be an indexed family of subsets of a universal set \(X\) and let \(B\) be a subset of \(X\). Then the following hold:

- \(B \cup\left(\bigcap_{i \in I} A_{i}\right)=\bigcap_{i \in I} B \cup A_{i}\);
- \(B \cup\left(\bigcap_{i \in I} A_{i}\right)=\bigcup_{i \in I} B \cup A_{i}\);
- \(B \backslash\left(\bigcap_{i \in I} A_{i}\right)=\bigcup_{i \in I} B \backslash A_{i}\);
- \(B \backslash\left(\bigcup_{i \in I} A_{i}\right)=\bigcap_{i \in I} B \backslash A_{i}\);
- \(\left(\bigcap_{i \in I} A_{i}\right)^{c}=\bigcup_{i \in I} A^{c}\);
- \(\left(\bigcup_{i \in I} A_{i}\right)^{c}=\bigcap_{i \in I} A^{c}\).

**Proof**-
**Proof of (a):**Let \(x \in B \cup\left(\bigcap_{i \in I} A_{i}\right)\). Then \(x \in B\) or \(x \in \bigcap_{i \in I} A_{i}\). If \(x \in B\), then \(x \in B \cup A_{i}\) for all \(i \in I\) and, thus, \(x \in \bigcap_{i \in I} B \cup A_{i}\). If \(x \in \bigcap_{i \in I} A_{i}\), then \(x \in A_{i}\) for all \(i \in I\). Therefore, \(x \in B \cup A_{i}\) for all \(i \in I\) and, hence, \(x \in \bigcap_{i \in I} B \cup A_{i}\). We have thus showed \(B \cup\left(\bigcap_{i \in I} A_{i}\right) \subset \bigcap_{i \in I} B \cup A_{i}\).Now let \(x \in \bigcap_{i \in I} B \cup A_{i}\). Then \(x \in B \cup A_{i}\) for all \(i \in I\). If \(x \in B\), then \(x \in B \cup\left(\bigcap_{i \in I} A_{i}\right)\). If \(x \notin B\), then we must have that \(x \in A_{i}\) for all \(i \in I\). Therefore, \(x \in \bigcap_{i \in I} A_{i}\) and, hence, \(x \in B \cup\left(\bigcap_{i \in I} A_{i}\right)\). This proves the other inclusion and, so, the equality. \(\square\)

We want to consider pairs of objects in which the order matters. Given objects \(a\) and \(b\), we will denote by \((a, b)\) the *ordered pair *where \(a\) is the first element and \(b\) is the second element. The main characteristic of ordered pairs is that \((a, b)=(c, d)\) if and only if \(a=c\) and \(b=d\). Thus, the ordered pair \((0,1)\) represents a different object than the pair \((1,0)\) (while the set \(\{0,1\}\) is the same as the set \(\{1,0\}\))^{1}.

Given two sets \(A\) and \(B\), the *Cartesian product * of \(A\) and \(B\) is the set defined by

\[A \times B:=\{(a, b): a \in A \text { and } b \in B\}.\]

If \(A=\{1,2\}\) and \(B=\{-2,0,1\}\), then

\[A \times B=\{(1,-2),(1,0),(1,1),(2,-2),(2,0),(2,1)\}.\]

If \(A\) and \(B\) are the intervals \([-1,2]\) and \([0,7]\) respectively, then \(A \times B\) is the rectangle

\[[-1,2] \times[0,7]=\{(x, y):-1 \leq x \leq 2,0 \leq y \leq 7\}.\]

We will make use of cartesian products in the next section when we discuss functions.

Exercise \(\PageIndex{1}\)

Prove the remaining items in **Theorem 1.1.2**.

**Answer**-
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Exercise \(\PageIndex{2}\)

Let \(Y\) and \(Z\) be subsets of \(X\). Prove that

\[(X \backslash Y) \cap Z=Z \backslash(Y \cap Z).\]

**Answer**-
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Exercise \(\PageIndex{3}\)

Prove the remaining items in **Theorem 1.1.3**.

**Answer**-
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Exercise \(\PageIndex{4}\)

Let \(A\), \(B\), \(C\), and \(D\) be sets. Prove the following.

- \((A \cap B) \times C=(A \times C) \cap(B \times C)\).
- \((A \cup B) \times C=(A \times C) \cup(B \times C)\)
- \((A \times B) \cap(C \times D)=(A \cap C) \times(B \cap D)\).

**Answer**-
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Exercise \(\PageIndex{5}\)

Let \(A \subset X\) and \(B \subset Y\). Determine if the following equalities are true and justify your answer:

- \((X \times Y) \backslash(A \times B)=(X \backslash A) \times(Y \backslash B)\).
- \((X \times Y) \backslash(A \times B)=[(X \backslash A) \times Y] \cup[X \times(Y \backslash B)]\).

**Answer**-
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^{1} For a precise definition of ordered pair in terms of sets see [Lay13]