Sec 3 E-Math Set Language & Notation notes

A set is a collection of distinct objects, written inside curly brackets: {1, 2, 3}. Each object is an element: we write 2 ∈ {1, 2, 3} ("2 is an element of") and 5 ∉ {1, 2, 3} ("5 is not an element of"). The number of elements in a set A is written n(A). Order never matters and a repeat adds nothing: {3, 7, 7, 11} is really {3, 7, 11}, so its size is 3.

• Universal set ξ (sometimes written U): everything under discussion for the current problem. If ξ = {1, 2, …, 10}, then no other numbers exist in that problem.
• Empty set ∅ (also { }): the set with nothing in it, so n(∅) = 0. Note that {0} is not empty — it contains the number 0.
• Set-builder notation: {x : x is an integer and 3 < x < 8} reads "all x such that…", which here lists as {4, 5, 6, 7}.

A is a subset of B (written A ⊆ B) when every element of A also belongs to B — equality is allowed. A is a proper subset of B (written A ⊂ B) when every element of A is in B and B has at least one extra element. So {1, 2} ⊆ {1, 2, 3} and {1, 2} ⊂ {1, 2, 3}, but {1, 2, 3} is a subset of itself (⊆ true) while not a proper subset of itself (⊂ false).

• Every set is a subset of itself: A ⊆ A always.
• The empty set sits inside everything: ∅ ⊆ B for every set B.
• If A ⊆ B and B ⊆ A, the two sets are equal — they hold exactly the same elements.

The intersection A ∩ B collects elements in both A and B at once. The union A ∪ B collects elements in at least one of them. A helpful memory hook: ∩ looks like "AND" (narrower result), ∪ looks like a cup that holds everything (broader result). Two sets with empty intersection (A ∩ B = ∅) are called disjoint.

The complement of A, written A′ (or Ā), is the set of elements of the universal set ξ that are not in A. The universal set is part of the definition: A′ changes when ξ changes. Two identities always hold: A ∪ A′ = ξ (together the pair covers everything) and A ∩ A′ = ∅ (nothing can be both in and out). Taking the complement twice returns the original set: (A′)′ = A.

1. 1Fix ξ first

   Let ξ = {1, 2, 3, …, 10} and A = {2, 4, 6, 8, 10}.

2. 2Sweep ξ, keep outsiders

   The elements of ξ missing from A are 1, 3, 5, 7, 9.

3. 3Write the complement

   A′ = {1, 3, 5, 7, 9}, and indeed A ∪ A′ = ξ ✓ and n(A) + n(A′) = 5 + 5 = 10 = n(ξ).

A Venn diagram draws the universal set as a rectangle and each set as a circle inside it. With two sets there are four regions: elements in A only, elements in B only, elements in both (the overlap), and elements outside both. Shading questions translate notation into pictures: A ∪ B shades everything in the two circles, A ∩ B shades just the lens-shaped overlap, and A′ shades the rectangle outside circle A.

1. 1Start at the overlap

   In a class of 35, 20 play football (F), 15 play basketball (B), 8 play both. Write 8 in the overlap first.

2. 2Subtract for the "only" regions

   F only: 20 − 8 = 12. B only: 15 − 8 = 7.

3. 3Account for the rectangle

   Inside the circles: 12 + 8 + 7 = 27, so outside both: 35 − 27 = 8.

4. 4Answer the actual question

   "Exactly one sport" = 12 + 7 = 19; "at least one sport" = 27; "neither" = 8.

Always enter the overlap region first when filling a diagram from counts — “only” regions are found by subtracting the overlap. Writing the raw totals 20 and 15 into the circles directly double counts the 8 shared elements and sinks the rest of the question.