Class 11 Maths - Chapter 01 - Comprehensive Interactive Notes
Start here: A set is a well-defined collection of distinct objects. For any object, membership must have a clear yes-or-no answer.
Memory cue: a set records membership, not order or repetition. \( \{a,e,a\}=\{e,a\} \).
"Collection of all vowels in English alphabet"
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We know exactly what they are: {a, e, i, o, u}. This IS a set.
"Collection of best cricket players"
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"Best" is subjective. My list might differ from yours. This is NOT a set.
One set, two descriptions: list its elements, or write the rule that selects exactly those elements.
| Form | Description | Example |
|---|---|---|
| Roster (Tabular) | Elements listed within braces { }, separated by commas. | \( A = \{2, 4, 6, 8\} \) |
| Set-Builder | State the number system and the condition an element must satisfy. | \( A = \{x \in \mathbb{N} : x \text{ is even and } x < 10\} \) |
Classify by size: first ask how many distinct elements the set contains. Use \( n(A) \) for the number of elements of a finite set A.
Choose the universe first: the universal set U contains every element allowed in the problem.
Then cross out A: the complement \( A' = U-A \) contains exactly the elements of U that are not in A. Changing U can change the complement.
Example:
If \( U = \{1,2,3,4,5\} \) and \( A = \{1,3\} \), then
\( A' = \{2,4,5\} \)
No overlap: A and B are disjoint when they share no element. Their intersection is the empty set.
\( A \cap B = \phi \)
Example:
\( A = \{1,3,5\}, \; B = \{2,4,6\} \)
Check every element: \( A \subseteq B \) means that every element of A also belongs to B. Equality is allowed. For a proper subset, also require \( A \ne B \).
Notation note: some school texts use \( \subset \) for “is a subset of”, including equality. Here \( \subseteq \) makes that meaning explicit.
The power set \( P(A) \) is the set of all subsets of A, including \( \varnothing \) and A itself.
If \( n(A)=m \), then \( n(P(A))=2^m \). Each element has two choices: included or left out.
Worked example: for \( A=\{a,b\} \), write \( P(A)=\{\varnothing,\{a\},\{b\},\{a,b\}\} \). Check: \( 2^2=4 \) subsets.
List up to 4 distinct elements, separated by commas. Predict the number of subsets before generating them.
Read the endpoints: an interval describes a stretch of the real line. Assume \( a < b \); a round bracket leaves an endpoint out, and a square bracket includes it.
| Interval Type | Notation | Set-Builder Form | Description |
|---|---|---|---|
| Open Interval | \( (a, b) \) | \( \{x : a < x < b\} \) | Excludes endpoints \( a \) and \( b \). |
| Closed Interval | \( [a, b] \) | \( \{x : a \le x \le b\} \) | Includes endpoints \( a \) and \( b \). |
| Semi-Open (L) | \( (a, b] \) | \( \{x : a < x \le b\} \) | Excludes \( a \), includes \( b \). |
| Semi-Open (R) | \( [a, b) \) | \( \{x : a \le x < b\} \) | Includes \( a \), excludes \( b \). |
Length of any interval \( (a,b), [a,b], (a,b] \) or \( [a,b) \) is \( b - a \).
Shade the meaning: the rectangle represents U; each circle represents a set. Decide which region the operation selects before listing its elements.
Work it once: take \( A=\{1,2,3\} \) and \( B=\{3,4,5\} \). Then \( A\cup B=\{1,2,3,4,5\} \), \( A\cap B=\{3\} \), \( A-B=\{1,2\} \), and \( A\triangle B=\{1,2,4,5\} \).
Enter elements for Set A and Set B.
Use laws to simplify: commutative changes order, associative changes grouping, and distributive opens a bracket. All complements use the same universal set.
| Law | Expression |
|---|---|
| Commutative |
\( A \cup B = B \cup A \) \( A \cap B = B \cap A \) |
| Associative |
\( A \cup (B \cup C) = (A \cup B) \cup C \) \( A \cap (B \cap C) = (A \cap B) \cap C \) |
| Distributive |
\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \) \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \) |
| Identity | \( A \cup \phi = A \), \( A \cap U = A \) |
Complement the whole expression: swap union with intersection, then complement each set. These two rules connect all three operations.
\( (A \cup B)' = A' \cap B' \)
\( (A \cap B)' = A' \cup B' \)
Diagram check: shade each side separately in the same universe. The shaded regions must match. Outside both sets means \( A'\cap B' \); outside their overlap means \( A'\cup B' \).
Count the overlap only once: for finite sets A and B, adding their sizes counts shared elements twice. Subtract one copy of the intersection.
Suppose \( n(A)=20 \), \( n(B)=15 \), and \( n(A\cap B)=5 \). Then \( n(A\cup B)=20+15-5=30 \).
Read the regions: only A has \( 20-5=15 \) elements; only B has \( 15-5=10 \). Check: \( 15+5+10=30 \). If \( n(U)=40 \), neither set contains \( 40-30=10 \) elements.
Let \( A - B \), \( A \cap B \), and \( B - A \) be mutually disjoint sets whose union is \( A \cup B \).
Therefore, \( n(A \cup B) = n(A - B) + n(A \cap B) + n(B - A) \).
Substituting \( n(A - B) = n(A) - n(A \cap B) \) and \( n(B - A) = n(B) - n(A \cap B) \), we get:
\( n(A \cup B) = [n(A) - n(A \cap B)] + n(A \cap B) + [n(B) - n(A \cap B)] \)
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Add, subtract, add: for three finite sets, add the individual sizes, subtract all three pairwise overlaps, then add back the triple overlap.
Fill the diagram from the centre: enter the triple intersection first. Subtract it from each pairwise total, then find the “only A”, “only B”, and “only C” regions. Pairwise intersection totals include the triple intersection unless the question says “only”.
Keep this symbol key nearby: check the number system before deciding which values belong to a set.
Before you finish: run these four checks over your answer.
Choose your first step: use this revision checklist to connect each question type to a method.
Close your notes and try: choose an answer, then explain the rule behind it in one line. Revisit the matching notebook page if you are unsure.
1. If \( A = \{1, 2\} \), how many subsets does it have?
2. \( A \cup A' \) is equal to:
3. Which set is infinite?
4. If \( A \subset B \), then \( A \cap B \) is:
5. The set of "intelligent students" is: