A chapter companion covering membership, subsets, union, intersection and complement with finite examples. Infinite sets, intervals and the full textbook exercise map still need further lessons.
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LESSON 1
Describe membership precisely
Represent a finite set and distinguish membership from inclusion.
A mathematical set needs a clear rule deciding whether each candidate belongs.
Use a definite rule
Let A contain the positive even integers below 9. The rule gives A={2,4,6,8}.
Ignore order and repetition
{8,2,4,6,2} describes the same set: repeated writing does not create another element.
Distinguish two statements
2∈A means that the number 2 belongs to A. {2}⊆A means that every member of the one-element set {2} belongs to A.
Test inclusion
B={2,8} is a subset of A because both its elements occur in A. C={2,9} is not, because 9 is a counterexample.
Worked example: If D={letters in MATH}, then D={M,A,T,H} and |D|=4. The sets {A,M} and the empty set are subsets of D; {A,S} is not because S∉D.
Watch for this: The number 2 and the set {2} are different objects. Use membership for an element and inclusion for a set.
Discuss: Find a vague collection, rewrite its rule to remove opinion, and list its members.
LESSON 2
Combine sets without double counting
Calculate union, intersection and the size of a union.
Union uses either set; intersection requires membership in both.
Place each member once
Take A={1,2,3} and B={3,4,5}. The shared member 3 sits in the overlap of the diagram.
Collect the union
A∪B={1,2,3,4,5}. It includes both circles but lists the shared member only once.
Collect the intersection
A∩B={3}. Only membership in both sets qualifies; 1 and 4 do not.
Check the count
For finite sets, |A∪B|=|A|+|B|−|A∩B|. Here 3+3−1=5. Subtract the overlap once because adding both sizes counted it twice.
Worked example: In a class, 12 pupils join chess, 9 join art and 4 join both. The number joining at least one is 12+9−4=17. This alone does not tell us how many join neither; the class size is also needed.
Watch for this: Union does not mean add the cardinalities without checking the overlap.
Discuss: Create two sets with four members each whose union has six members. Explain the overlap.
LESSON 3
Use the universe when taking a complement
Explain why a complement depends on the chosen universal set.
A universal set specifies the objects currently under consideration.
State the universe
Let U={1,2,3,4,5,6} and A={1,2,3}. Without U, a complement is not fully specified.
Remove the members of A
The elements of U not in A are 4,5,6. Thus A′={4,5,6}.
Check the partition
A and A′ do not overlap, and their union is U. Their sizes add to |U|.
Change the universe carefully
If U is enlarged to {1,2,3,4,5,6,7}, A stays the same but its complement becomes {4,5,6,7}.
Worked example: Let U={a,b,c,d,e}, A={a,c}. Then A′={b,d,e}. The complement of A′ is A again. If B={c,d}, A−B={a}; subtraction of sets removes membership, not numerical values.
Watch for this: A complement is not every imaginable object. It is relative to the stated universe.
Discuss: Choose a universe of six symbols. Make a set and its complement, then check both partition rules.
