Created: January 14, 2025
Modified: January 14, 2025
Notes and exercises for my independent study of the MIT course Calculus with Theory 18.014.
The notes follow the structure given on the readings page. Recitations, problem sets, and exams are grouped inside or near the sessions with which they are associated.
The course makes heavy use of Tom Apostol's excellent book, Calculus, Volume 1: One-Variable Calculus, with An Introduction to Linear Algebra. Additional course notes can be viewed on the course website.
Most of this section addresses basic ideas of sets, membership, union, intersection, etc. Since I'm familiar with most of these ideas, I'll only include things that seemed worth reviewing for me, specifically.
Two sets A and B are said to be disjoint if A \cap B = \varnothing.
Union and intersection are commutative and associative.
Union and intersection can be extended to finite or infinite collections of sets. Let \mathscr{F} be a class of sets. Then,
\bigcup_{A \in \mathscr{F}} A
is the set of all elements in any set within the class, and
\bigcap_{A \in \mathscr{F}} A
is the set of all elements common to every member of the class.
Apostol's Calculus, p. 16: 10, 15, 18
Problem 10: Prove the distributive laws: A \cap (B \cup C) = (A \cap B) \cup (A \cap C) and A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
a. A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
Let X = A \cap (B \cup C) and Y = (A \cap B) \cup (A \cap C). We will prove that X \subseteq Y and Y \subseteq X.
X \subseteq Y: Suppose that x \in X. Then x \in A and x \in B \cup C. x \in B \cup C means that x must be a member of B or C (or both). If x is a member of B, then x is a member of both A and B, so x \in A \cap B. If x is a member of C, then x is a member of both A and C, so x \in A \cap C. Thus, x must be a member of (A \cap B) \cup (A \cap C) = Y.
Y \subseteq X: Suppose that x \in Y. Then x \in A \cap B or x \in A \cap C (or both). In either case, x must be a member of A to be a member of X. Thus, x \in B or x \in C (or both) which means that x \in B \cup C. Since x \in A and x \in B \cup C, then x must be a member of A \cap (B \cup C) = X. \square
b. A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
Let X = A \cup (B \cap C) and Y = (A \cup B) \cap (A \cup C). We will prove that X \subseteq Y and Y \subseteq X.
X \subseteq Y: Suppose that x \in X. Then x \in A or x \in B \cap C (or both). If x \in A, then x is certainly a member of A \cup B and A \cup C. So, x is also a member of (A \cup B) \cap (A \cup C) = Y. If x \in B \cap C, then x is a member of both B and C. x must then be a member of A \cup B and A \cup C and is therefore also a member of (A \cup B) \cap (A \cup C) = Y.
Y \subseteq X: Suppose that x \in Y. Then x \in A \cup B and x \in A \cup C. If x is a member of A, then clearly x \in A \cup (B \cap C), so consider x \notin A. Then, x must be a member of both B and C to satisfy x \in Y. Therefore, x \in B \cap C, and so x is a member of A \cup (B \cap C) = X. \square
Problem 15: If A \subseteq C and B \subseteq C, then prove that A \cup B \subseteq C.
Suppose that x \in A \cup B. Then x is a member of A or B (or both). If x \in A, then we know x \in C since A \subseteq C. If x \in B, we know that x \in C since B \subseteq C. Thus, A \cup B \subseteq C. \square
Problem 18: Prove that A - (B \cap C) = (A - B) \cup (A - C).
Let X = A - (B \cap C) and Y = (A - B) \cup (A - C). We'll prove that X \subseteq Y and Y \subseteq X.
X \subseteq Y: Suppose that x \in X. Then x is some member of A that is not common to both B and C. If x \in B and not in C, then A - C must still contain x. So, x \in (A - C) and thus x \in (A - B) \cup (A - C). We can make an analogous argument for the case where x \in C but x \notin B. If x \notin B \cup C, then x remains in both A - B and A - C. Therefore, in all cases x is a member of (A - B) \cup (A - C) = Y.
Y \subseteq X: Suppose that x \in Y. Then x \in A - B or x \in A - C. Clearly, x \in A, and for x to be a member of either A - B or A - C, then x must not be in both B and C. So, x \notin B \cap C, and thus A - (B \cap C) still contains x. \square