Lecture Notes
A set is a well-defined collection of distinct objects, called elements.
The phrase well-defined means that for any object, it must be unambiguous whether the object belongs to the set or not.
We write \(x \in A\) to mean "\(x\) is an element of \(A\)" and \(x \notin A\) to mean "\(x\) is not an element of \(A\)."
Not well-defined: "The set of all colors" — unclear whether certain shades should be included.
Well-defined: The set of natural numbers \(\mathbb{N} = \{1, 2, 3, \ldots\}\).
In this course: \(\mathbb{N} = \{1, 2, 3, \ldots\}\) (does not include 0). Some textbooks include 0 — always check!
Let \(A\) and \(B\) be sets, and let \(U\) denote the universal set. Click each operation below to see its Venn diagram:
\(A \setminus B = A \cap B^c\)
Both expressions describe "elements in \(A\) but not in \(B\)." This identity is used frequently in proofs.
Two sets \(A\) and \(B\) are equal if and only if they contain exactly the same elements:
$$A = B \iff (A \subseteq B) \text{ and } (B \subseteq A)$$To prove \(A = B\), use a double containment argument:
Step 1 (⊆): Show that if \(x \in A\), then \(x \in B\).
Step 2 (⊇): Show that if \(x \in B\), then \(x \in A\).
Intuition: Negation "flips" \(\cup\) to \(\cap\) and \(\cap\) to \(\cup\).
(⊆) Let \(x \in (A \cup B)^c\). Then \(x \notin A \cup B\), which means \(x \notin A\) and \(x \notin B\). Hence \(x \in A^c\) and \(x \in B^c\), so \(x \in A^c \cap B^c\). ✓
(⊇) Let \(x \in A^c \cap B^c\). Then \(x \notin A\) and \(x \notin B\), so \(x \notin A \cup B\), giving \(x \in (A \cup B)^c\). ✓
Since both containments hold, \((A \cup B)^c = A^c \cap B^c\). □
Click any row to highlight it. The blue columns are identical, proving the equivalence!
| \(p\) | \(q\) | \(p \lor q\) | \(\neg(p \lor q)\) | \(\neg p\) | \(\neg q\) | \((\neg p) \land (\neg q)\) |
|---|
(N1) \(1 \in \mathbb{N}\). — There is a starting element
(N2) Every \(n \in \mathbb{N}\) has a successor. — We can always go to \(n+1\)
(N3) 1 is not the successor of any natural number. — Nothing before 1
(N4) If two natural numbers have the same successor, they are equal. — No "merging"
(N5) (Induction Axiom) If \(A \subseteq \mathbb{N}\) contains 1 and contains \(n+1\) whenever it contains \(n\), then \(A = \mathbb{N}\).
Let \(P(n)\) be a statement for \(n \in \mathbb{N}\). To prove \(P(n)\) for all \(n\):
1. Base Step: Prove \(P(1)\).
2. Inductive Step: Assume \(P(k)\) (inductive hypothesis), prove \(P(k+1)\).
Click Start to watch the dominoes fall. The base step knocks over the first, and the inductive step ensures each knocks over the next.
Prove: \(\displaystyle 1 + 2 + \cdots + n = \frac{n(n+1)}{2}\) for all \(n \in \mathbb{N}\).
Base Step (\(n = 1\)): Left = 1. Right = \(\frac{1 \cdot 2}{2} = 1\). ✓
Inductive Hypothesis: Assume for some \(k\):
\(\displaystyle 1 + 2 + \cdots + k = \frac{k(k+1)}{2}\)
Inductive Step:
$$1 + 2 + \cdots + k + (k+1) = \underbrace{\frac{k(k+1)}{2}}_{\text{by I.H.}} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2} \quad\checkmark$$By induction, the formula holds for all \(n \in \mathbb{N}\). □
Choose \(n\) and watch the sum build up:
The sum \(1 + 2 + \cdots + n\) forms a staircase. Complete it to an \(n \times (n{+}1)\) rectangle — the sum is half the rectangle!
Prove: \(5^n - 4n - 1\) is divisible by 16 for all \(n \in \mathbb{N}\).
Base Step (\(n=1\)): \(5 - 4 - 1 = 0 = 16 \cdot 0\). ✓
Inductive Hypothesis: Assume \(5^k - 4k - 1 = 16m\) for some integer \(m\).
Inductive Step:
$$5^{k+1} - 4(k+1) - 1 = 5 \cdot 5^k - 4k - 5 = 5\underbrace{(5^k - 4k - 1)}_{=16m} + 16k = 16(5m + k) \quad\checkmark$$By induction, the result holds for all \(n \in \mathbb{N}\). □
Click values of \(n\) to check that \(5^n - 4n - 1\) is divisible by 16:
Three pegs, \(n\) disks of different sizes. Rules: move one disk at a time; never place a larger disk on a smaller one. Let \(T(n)\) = minimum moves to transfer all disks.
Recurrence: \(T(n) = 2T(n-1) + 1\). Solution by induction: \(T(n) = 2^n - 1\).
Union ↔ "there exists" ↔ ∃
Intersection ↔ "for all" ↔ ∀
Let \(A_n = \{k \in \mathbb{N} : k \geq n\}\). Use the slider to see how the sets shrink:
Union: \(\bigcup A_n = \mathbb{N}\) — every number appears in at least one \(A_n\).
Intersection: \(\bigcap A_n = \varnothing\) — no number is \(\geq n\) for every \(n\).
\(\bigcup\) (Union) ↔ \(\exists\) (there exists) ↔ "for some"
\(\bigcap\) (Intersection) ↔ \(\forall\) (for all) ↔ "for every"
Exercise 1. Prove that \(A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)\).
Exercise 2. Show that \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\). (Distributive law)
Exercise 3. Prove that \((A \cap B)^c = A^c \cup B^c\). (The other De Morgan's Law)
Exercise 4. Prove \(1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}\).
Exercise 5. Prove that \(3^n - 1\) is divisible by 2 for all \(n \in \mathbb{N}\).
Exercise 6. Prove by induction that the Tower of Hanoi requires \(2^n - 1\) moves.
Exercise 7 (Binomial Theorem). (a) Verify \((a+b)^n\) for \(n=1,2,3\). (b) Show Pascal's identity. (c) Prove by induction.
Math 205: Real Analysis — Lecture 01