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2. Methodical guidelines / Методичні вказівки

This section is self-contained: it collects all the theory needed for the theme — finite sets, their operations, diagrams, and algebra — together with the techniques used to solve the tasks in 3classroom.md. No outside material is required.

2.1 Sets and how to write them

A set is an unordered collection of distinct objects, called its elements. We write xAx \in A if xx is an element of AA, and xAx \notin A otherwise. Two properties define what a set is:

  • Distinctness — repeating an element changes nothing: {1,2,2,3}={1,2,3}\{1,2,2,3\} = \{1,2,3\};
  • No order — the elements have no positions: {1,2,3}={3,1,2}\{1,2,3\} = \{3,1,2\}.

There are two standard ways to describe a set:

  • Roster form — list the elements between braces: A={2,4,6,8}A = \{2,4,6,8\};
  • Set-builder form — state a defining property: A={xxN and x is even}A = \{\, x \mid x \in \mathbb{N} \ \text{and}\ x \ \text{is even} \,\}, read “the set of all xx such that …”.

Useful named sets: the natural numbers N={1,2,3,}\mathbb{N} = \{1,2,3,\dots\} and the empty set ={}\varnothing = \{\,\}, which has no elements. The cardinality A|A| is the number of distinct elements of AA; =0|\varnothing| = 0.

Subset and equality. ABA \subseteq B (“AA is a subset of BB”) means every element of AA is also in BB. Two sets are equal, A=BA = B, exactly when they have the same elements — equivalently, when ABA \subseteq B and BAB \subseteq A. This “prove both inclusions” idea is one way to establish a set identity (§2.5).

Universal set. Many problems fix in advance a set UU of all elements under consideration — the universe. It is the reference domain against which a complement is taken (§2.2). On a diagram the universe is the bounding rectangle.

2.2 The operations on sets

Let AA and BB be sets inside a universe UU.

Operation Notation Definition Example
Union ABA \cup B {x:xA or xB}\{\, x : x \in A \ \text{or}\ x \in B \,\} {1,2,3}{3,4}={1,2,3,4}\{1,2,3\}\cup\{3,4\}=\{1,2,3,4\}
Intersection ABA \cap B {x:xA and xB}\{\, x : x \in A \ \text{and}\ x \in B \,\} {1,2,3}{3,4}={3}\{1,2,3\}\cap\{3,4\}=\{3\}
Difference ABA \setminus B {x:xA and xB}\{\, x : x \in A \ \text{and}\ x \notin B \,\} {1,2,3}{3,4}={1,2}\{1,2,3\}\setminus\{3,4\}=\{1,2\}
Complement A\overline{A} UA={xU:xA}U \setminus A = \{\, x \in U : x \notin A \,\} {1,2}={3,4,5}\overline{\{1,2\}}=\{3,4,5\} if U={1,,5}U=\{1,\dots,5\}

Two facts used constantly:

  • Complement is relative. A\overline{A} has no meaning without a universe; always state UU before taking a complement.
  • Difference is directed. In general ABBAA \setminus B \neq B \setminus A. Sets with AB=A \cap B = \varnothing are disjoint.

The single most useful bridge between the operations is

AB=AB,A \setminus B = A \cap \overline{B},

i.e. “everything in AA that is not in BB” equals “AA intersected with the complement of BB”. It converts differences into intersections and complements, which is what makes the algebraic proofs in §2.5 work.

2.3 Euler and Venn diagrams

A Venn diagram draws each set as a circle inside the rectangle UU, arranging the circles so that every possible overlap is shown. For two sets there are four regions; for three sets, eight:

  • Two sets A,BA,B: only AA (ABA\setminus B), the overlap (ABA\cap B), only BB (BAB\setminus A), and outside both (AB\overline{A\cup B}).
  • Three sets A,B,CA,B,C: the three “petals” A,B,CA,B,C, the three pairwise lenses (ABA\cap B, ACA\cap C, BCB\cap C), the central region ABCA\cap B\cap C, and the outside region.

Two-set Venn diagram inside the universe U, showing its four regions

An Euler diagram is the same idea but shows only the relationships that actually occur — for example, if ABA \subseteq B it draws circle AA inside circle BB with no spurious overlap. Venn diagrams fix the layout and shade; Euler diagrams tailor the layout to the situation.

How to shade an expression (work inside-out). To draw the region a formula denotes, evaluate it the way you would arithmetic — innermost operation first:

  1. shade each parenthesised sub-expression on a scratch copy;
  2. combine: union shades the union of the two regions; intersection keeps only the common part; difference XYX\setminus Y keeps the part of XX outside YY; complement shades everything outside the region.

For example, to shade A(BC)A \setminus (B \cup C): first shade BCB \cup C (both of those circles), then keep only the part of AA that lies outside that shaded area.

How to read a formula off a diagram. Name the shaded region by the simplest combination of circles that produces it — e.g. “the parts of AA and BB but not CC” reads as (AB)C(A \cup B)\setminus C.

2.4 The algebra of sets

The operations obey algebraic laws. They let you simplify expressions and prove identities by symbol manipulation, without drawing.

Law Statement
Commutativity AB=BA,AB=BAA \cup B = B \cup A,\quad A \cap B = B \cap A
Associativity (AB)C=A(BC)(A \cup B) \cup C = A \cup (B \cup C), and likewise for \cap
Distributivity A(BC)=(AB)(AC)A \cap (B \cup C) = (A \cap B) \cup (A \cap C), and dually A(BC)=(AB)(AC)A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
Identity A=A,AU=AA \cup \varnothing = A,\quad A \cap U = A
Domination AU=U,A=A \cup U = U,\quad A \cap \varnothing = \varnothing
Complement AA=U,AA=,A=AA \cup \overline{A} = U,\quad A \cap \overline{A} = \varnothing,\quad \overline{\overline{A}} = A
De Morgan AB=AB,AB=AB\overline{A \cup B} = \overline{A} \cap \overline{B},\quad \overline{A \cap B} = \overline{A} \cup \overline{B}
Difference AB=ABA \setminus B = A \cap \overline{B}

Note especially De Morgan (a complement of a union/intersection flips the operation and complements each part) and the difference law — these two do most of the work in the proof tasks.

2.5 Proving a set identity

To show two expressions denote the same set, use either method:

  • Diagram / membership method. Shade the region for each side on its own Venn diagram. If the two shaded regions coincide, the identity holds. (Formally, this is checking that an arbitrary element lies in the left side exactly when it lies in the right side — a membership argument.)
  • Algebraic method. Rewrite one side step by step using the laws of §2.4 until it becomes the other side, citing one law per step. Rewriting ABA \setminus B as ABA \cap \overline{B} is usually the first move, after which De Morgan and distributivity finish the job.

Both methods are legitimate; the diagram makes the result visible, the algebra makes it rigorous. Where practical, do both and check that they agree.

2.6 The power set

The power set P(A)\mathcal{P}(A) is the set of all subsets of AA, including the empty set \varnothing and AA itself. For example,

P({1,2})={, {1}, {2}, {1,2}}.\mathcal{P}(\{1,2\}) = \{\, \varnothing,\ \{1\},\ \{2\},\ \{1,2\} \,\}.

Each element of AA is independently in or out of a given subset, so an nn-element set has

P(A)=2A|\mathcal{P}(A)| = 2^{|A|}

subsets. To list a power set without omissions, enumerate subsets by size: first \varnothing, then all singletons, then all pairs, and so on up to AA itself.

2.7 Working checklist

  • Write sets in a consistent order (e.g. increasing) so equal sets are easy to compare, even though order does not matter formally.
  • Before any complement, write down UU.
  • For a diagram task, shade inside-out and label the final region.
  • For a proof task, prefer converting \setminus to     \cap\overline{\;\cdot\;} first, then apply De Morgan and distributivity.
  • For a listing task, translate the set-builder condition into a test applied to each candidate, then collect those that pass.

2.8 Symmetric difference, Cartesian product, and inclusion–exclusion

Three further tools appear in the counting and product tasks.

  • Symmetric difference AB=(AB)(BA)A \triangle B = (A \setminus B) \cup (B \setminus A): the elements in exactly one of AA and BB — the “exclusive or” of sets. Equivalently AB=(AB)(AB)A \triangle B = (A \cup B) \setminus (A \cap B).
  • Cartesian product A×B={(a,b):aA, bB}A \times B = \{\,(a,b) : a \in A,\ b \in B\,\}: the set of all ordered pairs whose first component lies in AA and second in BB. Its size is A×B=AB|A \times B| = |A|\cdot|B|.
  • Inclusion–exclusion (two sets). AB=A+BAB|A \cup B| = |A| + |B| - |A \cap B|: adding A|A| and B|B| counts the overlap ABA \cap B twice, so it is subtracted once. For three sets, ABC=A+B+CABACBC+ABC|A \cup B \cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|.

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