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

Math 308 · September 4, 2026

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 N,Z,Q,R\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 {Aα}α∈I\{A_\alpha\}_{\alpha \in I}{Aα​}α∈I​ and compute ⋃α∈IAα\bigcup_{\alpha \in I} A_\alpha⋃α∈I​Aα​ and ⋂α∈IAα\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∈Ax \in Ax∈A for “xxx is an element of AAA” and x∉Ax \notin Ax∈/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\}A={1,2,3,4,5}.

  • Set-builder notation: A={ n∈N  :  1≤n≤5 }A = \left\{\, n \in \mathbb{N} \;:\; 1 \le n \le 5 \,\right\}A={n∈N:1≤n≤5}, read “the set of nnn in N\mathbb{N}N such that 1≤n≤51 \le n \le 51≤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\}{1,2,3}={3,1,2}={1,1,2,3}.

Standard number systems.

N={1,2,3,… },Z={…,−2,−1,0,1,2,… },Q={ ab  :  a,b∈Z, b≠0 },R=the real numbers.\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∈N0 \in \mathbb{N}0∈N is a convention; in this course N\mathbb{N}N starts at 111.)

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

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

Subsets (§1.2)

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

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

A⊆B⟺for every x,  x∈A⇒x∈B.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 AAA:  (i) ∅⊆A\varnothing \subseteq A∅⊆A,  and  (ii) A⊆AA \subseteq AA⊆A.

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

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

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

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

P(A)={ ∅, {1}, {2}, {1,2} },∣P(A)∣=4.\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∣A∣=n then ∣P(A)∣=2n|\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,BA, BA,B (thought of as living inside a universal set UUU when we take complements).

Definition 5.

A∪B={ x  :  x∈A or x∈B }(union)A∩B={ x  :  x∈A and x∈B }(intersection)A−B={ x  :  x∈A and x∉B }(difference)Ac=U−A={ x∈U  :  x∉A }(complement)\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,BA, BA,B are disjoint if A∩B=∅A \cap B = \varnothingA∩B=∅.

Notation. The textbook writes A‾\overline{A}A for the complement of AAA. We will write AcA^{c}Ac instead, and reserve the overline A‾\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\}A={1,2,3,4} and B={3,4,5}B = \{3,4,5\}B={3,4,5} with universe U={1,…,6}U = \{1,\dots,6\}U={1,…,6}. Then

A∪B={1,2,3,4,5},A∩B={3,4},A−B={1,2},Ac={5,6}.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}.

figure figure figure figure

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

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

(A∪B)c=Ac∩Bc(one of De Morgan’s laws).(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 III. We write {Aα}α∈I\{A_\alpha\}_{\alpha \in I}{Aα​}α∈I​ (read “the sets AαA_\alphaAα​, as α\alphaα ranges over III”). Union and intersection extend naturally:

⋃α∈IAα={ x  :  x∈Aα for some α∈I },⋂α∈IAα={ x  :  x∈Aα for every α∈I }.\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 An=[0,1n]A_n = [0, \tfrac{1}{n}]An​=[0,n1​] for each n∈Nn \in \mathbb{N}n∈N. Every AnA_nAn​ contains 000, and as nnn grows the intervals shrink toward {0}\{0\}{0}:

⋃n∈NAn=[0,1],⋂n∈NAn={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: 000 lies in every AnA_nAn​; but no positive number does, since 1n→0\tfrac1n \to 0n1​→0.

The nested intervals An=(0,1n) (top to bottom: A1,A2,A3,A4). Only 0 survives every one: An=0.

The nested intervals An=[0,1n]A_n=[0,\tfrac1n]An​=[0,n1​] (top to bottom: A1,A2,A3,A4A_1,A_2,A_3,A_4A1​,A2​,A3​,A4​). Only 000 survives every one: ⋂nAn={0}\bigcap_n A_n=\{0\}⋂n​An​={0}.

Homework problems

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

    1. Show that

      {x∈R:0≤x≤3}∪{x∈R:−1<x<1}={x∈R:0≤x<1}.\{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}∪{x∈R:x2−4x+3=0}∩{x∈R:−1≤x<3}.\{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∈R:−2≤x≤0}∪{x∈R:2<x<4}∩{x∈R:0≤x≤3}={x∈R:2<x≤3}∪{0}.\{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 AAA is a subset of the set SSS, show that

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

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

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

  3. Let AAA, BBB, CCC be subsets of a set SSS. Prove the following statements and illustrate them with Venn diagrams.

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

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

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

  4. If AAA, BBB, CCC are sets, show that

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

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

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

  5. Let III be a nonempty set and for each i∈Ii \in Ii∈I let XiX_iXi​ be a set. Prove that

    1. for any set BBB we have

      B∩⋃i∈IXi=⋃i∈I(B∩Xi)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 XiX_iXi​ is a subset of a given set SSS, then

      (⋃i∈IXi)C=⋂i∈IXiC.\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.

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