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Sets, Relations, and Functions

Basic operations:

  • Union: AB=x:xAorxBA \cup B = \\{x : x \in A \mathrm{ or} x \in B\\}
  • Intersection: AB=x:xAandxBA \cap B = \\{x : x \in A \mathrm{ and} x \in B\\}
  • Difference: AB=x:xAandxBA \setminus B = \\{x : x \in A \mathrm{ and} x \notin B\\}
  • Complement: Ac=UAA^c = U \setminus A (where UU is the universal set)

De Morgan”s Laws:

(AB)c=AcBc,(AB)c=AcBc(A \cup B)^c = A^c \cap B^c, \quad (A \cap B)^c = A^c \cup B^c

Power set: P(A)=B:BA\mathcal{P}(A) = \\{B : B \subseteq A\\}. If A=n|A| = n Then P(A)=2n|\mathcal{P}(A)| = 2^n.

A binary relation RR from set AA to set BB is a subset of A×BA \times B.

A relation RR on AA is:

  • Reflexive: aA\forall a \in A, (a,a)R(a,a) \in R.
  • Symmetric: (a,b)R    (b,a)R(a,b) \in R \implies (b,a) \in R.
  • Antisymmetric: (a,b)R(a,b) \in R and (b,a)R    a=b(b,a) \in R \implies a = b.
  • Transitive: (a,b)R(a,b) \in R and (b,c)R    (b,c) \in R \implies(a,c) \in R$.

Equivalence relation: reflexive, symmetric, transitive. Partitions the set into equivalence classes.

Partial order: reflexive, antisymmetric, transitive. Written (A,)(A, \preceq).

A Hasse diagram is a graphical representation of a finite poset (A,)(A, \preceq): an element aa Is drawn below bb whenever aba \prec b (i.e., aba \preceq b and aba \neq b), and an edge is drawn From aa to bb whenever bb covers aa (there is no cc with acba \prec c \prec b).

Worked Example. Show that RR on Z\mathbb{Z} defined by aRba\,R\,b iff ab(mod5)a \equiv b \pmod{5} is An equivalence relation. Describe the equivalence classes.

Solution

Reflexive: aa=0=50a - a = 0 = 5 \cdot 0 So aa(mod5)a \equiv a \pmod{5} for all aa.

Symmetric: If ab(mod5)a \equiv b \pmod{5} Then 5(ab)5 \mid (a - b) So 5(ba)5 \mid (b - a)Giving ba(mod5)b \equiv a \pmod{5}.

Transitive: If 5(ab)5 \mid (a - b) and 5(bc)5 \mid (b - c) Then 5(ab)+(bc)=ac5 \mid (a - b) + (b - c) = a - c So ac(mod5)a \equiv c \pmod{5}.

The equivalence classes are [0]=5k:kZ[0] = \\{5k : k \in \mathbb{Z}\\} [1]=5k+1:kZ[1] = \\{5k+1 : k \in \mathbb{Z}\\} [2]=5k+2:kZ[2] = \\{5k+2 : k \in \mathbb{Z}\\} [3]=5k+3:kZ[3] = \\{5k+3 : k \in \mathbb{Z}\\} [4]=5k+4:kZ[4] = \\{5k+4 : k \in \mathbb{Z}\\}. There are exactly 5 equivalence classes, forming the quotient Z/5Z\mathbb{Z}/5\mathbb{Z}.

Worked Example. Let \preceq be divisibility on A=1,2,3,4,6,12A = \\{1, 2, 3, 4, 6, 12\\}: aba \preceq b iff aba \mid b. Verify this is a partial order and identify the cover relations.

Solution

Reflexive: aaa \mid a for all aAa \in A. ✓

Antisymmetric: If aba \mid b and bab \mid a Then b=kab = ka and a=lba = lb for positive k,lk, l So a=lkaa = lkaGiving lk=1lk = 1 and l=k=1l = k = 1Hence a=ba = b. ✓

Transitive: If aba \mid b and bcb \mid c Then c=lb=l(ka)=(lk)ac = lb = l(ka) = (lk)a So aca \mid c. ✓

Cover relations (bb covers aa when aba \mid b and no element lies strictly between):

  • 12 covers 6, 4
  • 6 covers 2, 3
  • 4 covers 2
  • 3 covers 1
  • 2 covers 1

Reading from bottom to top the Hasse diagram is: 11 at the bottom with edges to 22 and 33; 22 connects up to 44 and 66; 33 connects up to 66; 44 and 66 connect up to 1212 at the top.

A function f:ABf : A \to B is a relation where each aAa \in A appears exactly once as a first element.

  • Injective (one-to-one): f(a1)=f(a2)    a1=a2f(a_1) = f(a_2) \implies a_1 = a_2.
  • Surjective (onto): for every bBb \in BThere exists aAa \in A with f(a)=bf(a) = b.
  • Bijective: both injective and surjective.

Theorem 2.1. If AA and BB are finite sets, f:ABf : A \to B is:

  • Injective if and only if AB|A| \leq |B|.
  • Surjective if and only if AB|A| \geq |B|.
  • Bijective if and only if A=B|A| = |B|.

Theorem 2.2 (Pigeonhole Principle). If A>B|A| \gt{} |B| Then no function f:ABf : A \to B is injective. Equivalently, placing nn items into mm boxes with n>mn \gt{} m forces at least one box to contain at least n/m\lceil n/m \rceil items.

Function composition. Given f:ABf : A \to B and g:BCg : B \to CThe composition gf:ACg \circ f : A \to C is defined by (gf)(a)=g(f(a))(g \circ f)(a) = g(f(a)) for all aAa \in A.

Theorem 2.3. If f:ABf : A \to B and g:BCg : B \to C are both injective, then gfg \circ f is injective.

Proof. Suppose (gf)(a1)=(gf)(a2)(g \circ f)(a_1) = (g \circ f)(a_2). Then g(f(a1))=g(f(a2))g(f(a_1)) = g(f(a_2)). Since gg is Injective, f(a1)=f(a2)f(a_1) = f(a_2). Since ff is injective, a1=a2a_1 = a_2. Hence gfg \circ f is injective. \blacksquare

Theorem 2.4. If f:ABf : A \to B and g:BCg : B \to C are both surjective, then gfg \circ f is surjective.

Proof. Let cCc \in C. Since gg is surjective, bB\exists\, b \in B with g(b)=cg(b) = c. Since ff is Surjective, aA\exists\, a \in A with f(a)=bf(a) = b. Then (gf)(a)=g(f(a))=g(b)=c(g \circ f)(a) = g(f(a)) = g(b) = c. Hence gfg \circ f is surjective. \blacksquare

Corollary 2.5. The composition of two bijections is a bijection.

A function f:ABf : A \to B is invertible if there exists f1:BAf^{-1} : B \to A such that f^{-1} \circ f = \mathrm{id{}_A and f \circ f^{-1} = \mathrm{id{}_B. A function is invertible If and only if it is bijective.

Definition. A set SS is countable if it is finite or countably infinite. A set is countably infinite if there exists a bijection NS\mathbb{N} \to S. A set that is not countable Is uncountable.

Theorem 2.6. Z\mathbb{Z} is countably infinite.

Proof. The function f:NZf : \mathbb{N} \to \mathbb{Z} defined by

f(n)={n/2if  n  is  even(n+1)/2if  n  is  oddf(n) = \begin{cases} n/2 & \mathrm{if}\; n\; \mathrm{is}\; even \\ -(n+1)/2 & \mathrm{if}\; n\; \mathrm{is}\; odd \end{cases}

Is a bijection, enumerating 0,1,1,2,2,3,3,0, -1, 1, -2, 2, -3, 3, \ldots \blacksquare

Theorem 2.7. Q\mathbb{Q} is countably infinite.

Proof. Every positive rational can be written as p/qp/q with p,qN+p, q \in \mathbb{N}^+. Arrange the Pairs (p,q)(p, q) in an infinite grid and traverse them diagonally:

1/1,  1/2,  2/1,  3/1,  1/3,  1/4,  2/3,  3/2,  4/1,1/1,\; 1/2,\; 2/1,\; 3/1,\; 1/3,\; 1/4,\; 2/3,\; 3/2,\; 4/1, \ldots

Skipping duplicates (where p/q=p/qp/q = p'/q' in reduced form) yields an enumeration of Q+\mathbb{Q}^+. Extending with negatives and zero gives an enumeration of Q\mathbb{Q}. \blacksquare

Theorem 2.8 (Cantor, 1891). R\mathbb{R} is uncountable.

Proof (Diagonal argument). Suppose for contradiction that R\mathbb{R} is countable. Then the Interval [0,1)[0, 1) can be listed as r1,r2,r3,r_1, r_2, r_3, \ldots where each rir_i has a unique decimal Expansion ri=0.di1di2di3r_i = 0.d_{i1}d_{i2}d_{i3}\ldots with each dij0,1,,9d_{ij} \in \\{0, 1, \ldots, 9\\} (choosing the expansion that does not end in all 9s to avoid dual representations).

Define s=0.s1s2s3s = 0.s_1 s_2 s_3 \ldots by

si={5if  dii56if  dii=5s_i = \begin{cases} 5 & \mathrm{if}\; d_{ii} \neq 5 \\ 6 & \mathrm{if}\; d_{ii} = 5 \end{cases}

Then s[0,1)s \in [0, 1) and ss differs from rir_i in the ii-th decimal place for every ii So sr1,r2,s \notin \\{r_1, r_2, \ldots\\}Contradicting the assumption that the list was complete. Therefore R\mathbb{R} is uncountable. \blacksquare

flowchart TD
A[2_Sets Relations And Functions] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Sets are unordered collections of distinct elements. Relations connect elements from different sets, with properties like reflexivity, symmetry, and transitivity. Functions are special relations where each input maps to exactly one output. These concepts underpin database theory (relations), type systems (functions), and abstract algebra (groups, rings). Understanding set operations (union, intersection, complement) is essential for query languages and logic.

Confusing reflexivity with symmetry. A reflexive relation requires (a,a)R(a, a) \in R for every aAa \in A, while symmetry requires that if (a,b)R(a, b) \in R then (b,a)R(b, a) \in R. These are independent properties: a relation can be reflexive without being symmetric, and vice versa. Always verify each property separately.

Misapplying the Pigeonhole Principle. Students often state that placing nn items into mm boxes guarantees one box has at least n/mn/m items, but forget the ceiling function. The correct bound is n/m\lceil n/m \rceil, not n/mn/m. For example, 5 items into 3 boxes guarantees at least 2 in one box, not 1.67.

Assuming injectivity from surjectivity (or vice versa). For finite sets, a function f:ABf: A \to B is injective if and only if AB|A| \leq |B|, and surjective if and only if AB|A| \geq |B|. These are not interchangeable: an injective function from a smaller set to a larger set cannot be surjective.

  • [[1-discrete-mathematics/1_propositional-and-predicate-logic]] - Logic foundations
  • [[1-discrete-mathematics/3_proof-techniques]] - Proof techniques
  • [[2-systems/databases]] - Relational database theory
  • [[1-discrete-mathematics/4_combinatorics]] - Counting and sets