MATH 308 — Lec 01 and Lec 02 — Sets I (Describing, Subsets, Operations)

Lecture 1 --- Sets I: Describing Sets, Subsets, and Set Operations MATH 308  Bridge to Advanced Mathematics  Chartrand §1.1--1.4

Goals for today. By the end of this lecture you should be able to:

• describe a set by roster and by set-builder notation, and use $\in,\notin$∈,∈/ correctly;

• name the standard number systems $\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R}$N,Z,Q,R and state the cardinality of a finite set;

• decide when one set is a subset of another and write the power set of a small set;

• compute unions, intersections, differences, and complements, and read a Venn diagram.

• read an indexed family $\{A_\alpha\}_{\alpha \in I}${Aα​}α∈I​ and compute $\bigcup_{\alpha \in I} A_\alpha$⋃α∈I​Aα​ and $\bigcap_{\alpha \in I} A_\alpha$⋂α∈I​Aα​

Describing a set (§1.1)

A set is a collection of objects, called its elements (or members). A set is completely determined by which objects belong to it --- not by order or repetition. We write $x \in A$x∈A for "$x$x is an element of $A$A" and $x \notin A$x∈/A for its negation.

Example 1. Two ways to describe the same set.

• Roster (list) notation: $A = \{1, 2, 3, 4, 5\}$A={1,2,3,4,5}.

• Set-builder notation: $A = \left\{\, n \in \mathbb{N} \;:\; 1 \le n \le 5 \,\right\}$A={n∈N:1≤n≤5}, read "the set of $n$n in $\mathbb{N}$N such that $1 \le n \le 5$1≤n≤5."

Order and repetition do not matter: $\{1,2,3\} = \{3,1,2\} = \{1,1,2,3\}${1,2,3}={3,1,2}={1,1,2,3}.

Standard number systems.

$\mathbb{N}= \{1,2,3,\dots\}, \quad \mathbb{Z}= \{\dots,-2,-1,0,1,2,\dots\}, \quad \mathbb{Q}= \left\{\, \tfrac{a}{b} \;:\; a,b \in \mathbb{Z},\ b \neq 0 \,\right\}, \quad \mathbb{R}= \text{the real numbers.}$N={1,2,3,…},Z={…,−2,−1,0,1,2,…},Q={ba​:a,b∈Z, b=0},R=the real numbers.

(Whether $0 \in \mathbb{N}$0∈N is a convention; in this course $\mathbb{N}$N starts at $1$1.)

Definition 1 (Empty set and cardinality). The empty set $\varnothing = \{\,\}$∅={} is the set with no elements. For a finite set $A$A, its cardinality $|A|$∣A∣ is the number of elements in $A$A. Thus $|\varnothing| = 0$∣∅∣=0 and $|\{1,2,3\}| = 3$∣{1,2,3}∣=3.

Example 2. $\left\{\, x \in \mathbb{R} \;:\; x^2 = -1 \,\right\} = \varnothing${x∈R:x2=−1}=∅, since no real number squares to $-1$−1. Note the difference between $\varnothing$∅ and $\{\varnothing\}${∅}: the latter has one element (namely $\varnothing$∅), so $|\{\varnothing\}| = 1$∣{∅}∣=1.

Subsets (§1.2)

Definition 2. $A$A is a subset of $B$B, written $A \subseteq B$A⊆B, if every element of $A$A is also an element of $B$B. If $A \subseteq B$A⊆B but $A \neq B$A=B, we call $A$A a proper subset and write $A \subset B$A⊂B.

This is our first "for all" statement, and it is the template for nearly every proof this term:

$A \subseteq B \quad\Longleftrightarrow\quad \text{for every } x,\ \ x \in A \Rightarrow x \in B.$A⊆B⟺for every x,  x∈A⇒x∈B.

Proposition 1. For every set $A$A:  (i) $\varnothing \subseteq A$∅⊆A,  and  (ii) $A \subseteq A$A⊆A.

Proof. (ii) is immediate. For (i), we must show $x \in \varnothing \Rightarrow x \in A$x∈∅⇒x∈A for every $x$x. There is no $x$x with $x \in \varnothing$x∈∅, so the implication is vacuously true. (We will make "vacuously true" precise when we study logic.) ◻

Definition 3 (Set equality). $A = B$A=B means $A \subseteq B$A⊆B and $B \subseteq A$B⊆A. This "double-containment" is the standard strategy for proving two sets are equal.

Definition 4 (Power set). The power set of $A$A, written $\mathcal{P}(A)$P(A), is the set of all subsets of $A$A:  $\mathcal{P}(A) = \left\{\, X \;:\; X \subseteq A \,\right\}$P(A)={X:X⊆A}.

Example 3. For $A = \{1, 2\}$A={1,2},

$\mathcal{P}(A) = \bigl\{\, \varnothing,\ \{1\},\ \{2\},\ \{1,2\} \,\bigr\}, \qquad |\mathcal{P}(A)| = 4.$P(A)={∅, {1}, {2}, {1,2}},∣P(A)∣=4.

In general, if $|A| = n$∣A∣=n then $|\mathcal{P}(A)| = 2^n$∣P(A)∣=2n --- a fact we will prove by induction later in the course.

Set operations (§1.3)

Fix sets $A, B$A,B (thought of as living inside a universal set $U$U when we take complements).

Definition 5.

$\begin{aligned} A \cup B &= \left\{\, x \;:\; x \in A \ \text{or}\ x \in B \,\right\} && \text{(union)}\\ A \cap B &= \left\{\, x \;:\; x \in A \ \text{and}\ x \in B \,\right\} && \text{(intersection)}\\ A - B &= \left\{\, x \;:\; x \in A \ \text{and}\ x \notin B \,\right\} && \text{(difference)}\\ A^{c} &= U - A = \left\{\, x \in U \;:\; x \notin A \,\right\} && \text{(complement)}\end{aligned}$A∪BA∩BA−BAc​={x:x∈A or x∈B}={x:x∈A and x∈B}={x:x∈A and x∈/B}=U−A={x∈U:x∈/A}​​(union)(intersection)(difference)(complement)​

Sets $A, B$A,B are disjoint if $A \cap B = \varnothing$A∩B=∅.

Notation. The textbook writes $\overline{A}$A for the complement of $A$A. We will write $A^{c}$Ac instead, and reserve the overline $\overline{\phantom{A}}$A for the closure of a set, which you will meet when we study the real numbers.

Example 4. Let $A = \{1,2,3,4\}$A={1,2,3,4} and $B = \{3,4,5\}$B={3,4,5} with universe $U = \{1,\dots,6\}$U={1,…,6}. Then

$A \cup B = \{1,2,3,4,5\},\quad A \cap B = \{3,4\},\quad A - B = \{1,2\},\quad A^{c} = \{5,6\}.$A∪B={1,2,3,4,5},A∩B={3,4},A−B={1,2},Ac={5,6}.

The basic set operations, shown by shading (inside the universe $U$U).

A first taste of a set identity we will prove carefully once we have logic:

$(A \cup B)^{c} = A^{c} \cap B^{c} \qquad\text{(one of De Morgan's laws).}$(A∪B)c=Ac∩Bc(one of De Morgan’s laws).

Indexed collections of sets (§1.4)

Often we have not two or three sets but a whole family of them, one for each element of an index set $I$I. We write $\{A_\alpha\}_{\alpha \in I}${Aα​}α∈I​ (read "the sets $A_\alpha$Aα​, as $\alpha$α ranges over $I$I"). Union and intersection extend naturally:

$\bigcup_{\alpha \in I} A_\alpha = \left\{\, x \;:\; x \in A_\alpha \text{ for \emph{some} } \alpha \in I \,\right\}, \qquad \bigcap_{\alpha \in I} A_\alpha = \left\{\, x \;:\; x \in A_\alpha \text{ for \emph{every} } \alpha \in I \,\right\}.$α∈I⋃​Aα​={x:x∈Aα​ for some α∈I},α∈I⋂​Aα​={x:x∈Aα​ for every α∈I}.

Notice the quiet appearance of some and every --- exactly the quantifiers we take up next week.

Example 5. Let $A_n = [0, \tfrac{1}{n}]$An​=[0,n1​] for each $n \in \mathbb{N}$n∈N. Every $A_n$An​ contains $0$0, and as $n$n grows the intervals shrink toward $\{0\}${0}:

$\bigcup_{n \in \mathbb{N}} A_n = [0,1], \qquad \bigcap_{n \in \mathbb{N}} A_n = \{0\}.$n∈N⋃​An​=[0,1],n∈N⋂​An​={0}.

The intersection is not empty: $0$0 lies in every $A_n$An​; but no positive number does, since $\tfrac1n \to 0$n1​→0.

The nested intervals $A_n=[0,\tfrac1n]$An​=[0,n1​] (top to bottom: $A_1,A_2,A_3,A_4$A1​,A2​,A3​,A4​). Only $0$0 survives every one: $\bigcap_n A_n=\{0\}$⋂n​An​={0}.

Homework problems

1. Let $\mathbb{R}$R be the set of real numbers and let the symbols $<$<, $\leq$≤ have their conventional meanings.

   1. Show that

      $\{x \in \mathbb{R}: 0 \leq x \leq 3\} \cup \{x \in \mathbb{R}: -1 < x < 1\} = \{x \in \mathbb{R}: 0 \leq x < 1\}.${x∈R:0≤x≤3}∪{x∈R:−1<x<1}={x∈R:0≤x<1}.

   2. List the elements of

      $\{2, 3, 4\} \cup \{x \in \mathbb{R}: x^2 - 4x + 3 = 0\} \cap \{x \in \mathbb{R}: -1 \leq x < 3\}.${2,3,4}∪{x∈R:x2−4x+3=0}∩{x∈R:−1≤x<3}.

   3. Show that

      $\{x \in \mathbb{R}: -2 \leq x \leq 0\} \cup \{x \in \mathbb{R}: 2 < x < 4\} \cap \{x \in \mathbb{R}: 0 \leq x \leq 3\} = \{x \in \mathbb{R}: 2 < x \leq 3\} \cup \{0\}.${x∈R:−2≤x≤0}∪{x∈R:2<x<4}∩{x∈R:0≤x≤3}={x∈R:2<x≤3}∪{0}.

2. If $A$A is a subset of the set $S$S, show that

   1. $(A^C)^C = A$(AC)C=A

   2. $A \cup A = A \cap A = A \cup \varnothing = A$A∪A=A∩A=A∪∅=A

   3. $A \cap \varnothing = \varnothing$A∩∅=∅

3. Let $A$A, $B$B, $C$C be subsets of a set $S$S. Prove the following statements and illustrate them with Venn diagrams.

   1. $(A \cup B)^C = A^C \cap B^C$(A∪B)C=AC∩BC

   2. $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$A∩(B∪C)=(A∩B)∪(A∩C)

   3. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$A∪(B∩C)=(A∪B)∩(A∪C).

4. If $A$A, $B$B, $C$C are sets, show that

   1. $(A - B) \cap C = (A \cap C) - B$(A−B)∩C=(A∩C)−B

   2. $(A \cup B) - (A \cap B) = (A - B) \cup (B - A)$(A∪B)−(A∩B)=(A−B)∪(B−A)

   3. $A - (B - C) = (A - B) \cup (A \cap B \cap C)$A−(B−C)=(A−B)∪(A∩B∩C)

   4. $(A - B) \times C = (A \times C) - (B \times C)$(A−B)×C=(A×C)−(B×C).

5. Let $I$I be a nonempty set and for each $i \in I$i∈I let $X_i$Xi​ be a set. Prove that

   1. for any set $B$B we have

      $B \cap \bigcup_{i \in I} X_i = \bigcup_{i \in I} (B \cap X_i)$B∩i∈I⋃​Xi​=i∈I⋃​(B∩Xi​)

   2. if each $X_i$Xi​ is a subset of a given set $S$S, then

      $\left(\bigcup_{i \in I} X_i\right)^C = \bigcap_{i \in I} X_i^C.$(i∈I⋃​Xi​)C=i∈I⋂​XiC​.

Suggested textbook exercises (Chartrand §1.1--1.3): 1.2, 1.6, 1.14, 1.22, 1.30, 1.36.