Week 1: Functions, Sets, and Relations

Course: Jan 2026 - Mathematics I Difficulty: Foundational Focus: Set notation, relation properties, function behavior, and quick proof habits.

1. Set Theory Fundamentals

1.1 Core symbols

• Union $A \cup B$: in $A$ or $B$ or both.
• Intersection $A \cap B$: in both $A$ and $B$.
• Difference $A - B$: in $A$ but not in $B$.
• Complement $A'$: everything in the universe $U$ but not in $A$.

1.2 De Morgan's Laws

• $(A \cup B)' = A' \cap B'$
• $(A \cap B)' = A' \cup B'$

1.3 Common traps

• Do not confuse $\in$ with $\subseteq$.
• The empty set is a subset of every set.
• A set can contain another set as an element.

2. Cartesian Product and Relations

• Reflexive: $(a,a) \in R$ for every $a \in A$
• Symmetric: $(a,b) \in R \Rightarrow (b,a) \in R$
• Transitive: $(a,b) \in R$ and $(b,c) \in R \Rightarrow (a,c) \in R$

3. Functions

• Domain: where inputs come from
• Codomain: declared target set
• Range: actual outputs produced
• Injective: different inputs never share the same output
• Surjective: every codomain element is hit
• Bijective: both injective and surjective

2. Pattern A — Show a function is bijective

Abstract Solution (Strategy)

1. [Key concept]: To prove bijection, you must prove injectivity and surjectivity independently.
2. [Formula to use]: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ (Injectivity). For any $y$, $\exists x, f(x) = y$ (Surjectivity).
3. [Watch for]: Forgetting the horizontal line test or incorrectly handling $y = x^2$ style mapping in domains over all Reals.

Procedure

• Step 1: Write down the assumption $f(x_1) = f(x_2)$ and algebraically solve down to $x_1 = x_2$.
• Step 2: Write $y = f(x)$, solve for $x$ in terms of $y$. Check if the resulting $x$ is always within the given Domain.
• Step 3: Conclude bijection if both hold.

Question: Prove $f(x) = 2x + 3$ from $\mathbb{R}$ to $\mathbb{R}$ is bijective.

• Step 1: Injectivity. Let $2x_1 + 3 = 2x_2 + 3$, thus $2x_1 = 2x_2 \Rightarrow x_1 = x_2$.
• Step 2: Surjectivity. Let $y = 2x + 3 \Rightarrow x = \frac{y-3}{2}$. For any Real $y$, $x$ is Real.
• Answer: Yes, $f(x)$ is bijective.

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3. Pattern B — Evaluate relations on a Set

Abstract Solution (Strategy)

1. [Key concept]: Break the relation definition down into conditions on $x$ and $y$.
2. [Formula to use]: Reflexive ($xRx$), Symmetric ($xRy \Rightarrow yRx$), Transitive ($xRy, yRz \Rightarrow xRz$).
3. [Watch for]: Vacuous truth when testing properties or incorrectly mapping $y=x^2$ when symmetry is asked.

Procedure

• Step 1: Test $(x, x)$ for reflexivity.
• Step 2: Swap $x$ and $y$ and test symmetry constraint.
• Step 3: Plug into transitive property.
• Answer: Output the set of valid properties.

Question: Is $R = \{(1,1),(1,2),(3,1)\}$ a function on $\{1,2,3\}$?

• Step 1: An input can only produce one independent output.
• Step 2: Look at input $1$: It produces $1$ and $2$.
• Answer: No.

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4. Common Mistakes

5. Flashcards

<Flashcard front="What is the one-line test for a function?" back="Each input must have exactly one output." /> <Flashcard front="When is a function invertible?" back="When it is bijective." /> <Flashcard front="What does De Morgan's law do?" back="It swaps union/intersection under complement." />

6. Practice Matrix

• Prove or disprove bijection for a linear map.
• Classify a relation as reflexive, symmetric, transitive, or none.
• Expand a 2-set or 3-set expression using De Morgan's laws.

7. Quick Recall

• Relation = subset of a Cartesian product.
• Function = relation with one output per input.
• Bijection = perfect matching between domain and codomain.