Notes I took during studying MIT OCW Real Analysis. The class taught by Professor Casey Rodriguez, he also taught Functional analysis.
Resources (Useful link)
Lecture notes
Goal of the course - Gain experience with proofs - Prove statements about the real numbers, function and limits
Lecture 1: Sets, Set operations, and Mathematical Induction
Definition (Sets) A sets is a collection of objects called elements/members.
Definition (Empty set) A set with no elements, denoted as \(\emptyset\)
Notation
- \(a \in S\): \(a\) is a element of \(S\)
- \(a \notin S\): \(a\) is not a element of \(S\)
- \(\forall\): for all
- \(\exists\): there exists
- \(\implies\): implies
- \(\iff\): if and only if
Definition
- (subset) A set \(A\) is a subset of \(B\), denoted as \(A \subset B\) if: \(a \in A \implies a \in B\)
- (equal) Two sets are equal if \(A \subset B \land B \subset A\)
- (proper subset) \(A \subsetneqq B \iff A \subset B \land A \neq B\)
Set building notation
$$ \{ x \in A : P(x) \} $$
Examples
- \(\mathbb{N} = \{1, 2, 3 \cdots\}\)
- \(\mathbb{Z} = \{\cdots,-2, -1, 0, 1, 2, \cdots\}\)
- \(\mathbb{Q} = \{\frac{m}{n}: m, n \in \mathbb{Z}\}\)
- Real number set \(\mathbb{R}\)
Remark: \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\)
Goal: describe the real number set $\mathbb{R}$
Definition (union) The union of A and B is the set $$ A \cup B := {x: x\in A \lor x\in B} $$ Definition (intersection) The intersection of A and B is the set $$ A \cap B := {x: x\in A \land x \in B} $$ Definition (different) The set different between A w.r.t B is the set $$ A\backslash B = {x\in A: x\notin B} $$ Definition (complement) A complement of set A is the set $$ A^c = {x: x\notin A} $$ Definition (disjoint) Two sets A and B are disjoint if \(A \cap B = \emptyset\).
Theorem (De-Morgan) If A, B, C are sets then
- \((B \cup C)^c = B^c \cap C^c\)
- \((B \cap C)^c = B^c \cup C^c\)
- \(A\backslash (B\cup C) = (A\backslash B) \cap (A\backslash C)\)
- \(A\backslash (B\cap C) = (A\backslash B) \cup(A\backslash C)\)
Induction A way to prove theorem about natural number.
\(\mathbb{N} = \{1, 2, 3, \cdots \}\) has an ordering \(1 < 2 < 3< 4 < \cdots\)
Axiom (Well ordering of natural numbers) if \(S\subset \mathbb{N}\) and \(S\neq \emptyset\) has a least element \(\exists x\in S\) st \(\forall y \in S: x\leq y\).
Theorem (Induction) Let \(P(n)\) be a statement depending on \(n\in \mathbb{N}\). Assume:
- (Base case) \(P(1)\) is true
- (Inductive step) If \(P(m)\) is true, then \(P(m+1)\) is true. Then \(P(n)\) is true for all \(n\in \mathbb{N}\).
Proof: Let \(S = {n\in\mathbb{N}: P(n) \text{ is not true}}\). Want to show \(S=\emptyset\)
Suppose \(S\neq \emptyset\). By WOP.\(\mathbb{N}\), \(S\) has a least element \(x\in S\). Since \(P(1)\) is true, \(1\notin S\), so \(x>1\).
Since \(x\) is the least element in \(S\) \(\implies x-1 \notin S\).
By the definition of \(S\), \(P(x-1)\) is true, by 2) \(\implies P(x)\) is true \(\implies x \notin S\).
\(\therefore S = \emptyset\)
Using induction We want to prove some statement \(\forall n\in\mathbb{N}:P(n)\) is true, we have to do two things:
- Prove \(P(1)\).
- Prove \(P(m) \implies P(m+1)\)
Example For all \(c\neq 1, \forall n\in\mathbb{N}\):
$$ 1 + c + c^2 + \cdots +c^n=\frac{1-c^{n+1}}{1-c} $$
Proof
- (Base case):
$$ 1 + c^1 = \frac{1-c^{1+1}}{1-c}=\frac{(1-c)(1+c)}{1-c} = {1+c}; \forall c\neq1 $$
- (Inductive step) Assume: $$ 1+c+c^2+\cdots+c^m=\frac{1-c^{m+1}}{1-c}\quad(*) $$ We want to show $$ 1+c+c^2+\cdots+c^n=\frac{1-c^{n+1}}{1-c}\quad(**) $$ for \(n = m+1\).
We have:
$$ \begin{aligned} 1+c+c^2+\cdots+c^m+c^{m+1} &= \frac{1-c^{m+1}}{1-c}+c^{m+1} \\ & = \frac{1-c^{m+1}+c^{m+1}-c^{m+2}}{1-c}\\ & = \frac{1-c^{(m+1)+1}}{1-c} \end{aligned} $$ So (*) hold for \(n=m+1\). By induction, \(P(n)\) is true \(\forall n\in\mathbb{N}\).