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Week 1 · Day 2 Lesson 1.3–1.4 About 13 minutes Free

Sets: ∈ or ⊆, number sets, ∩ and ∪

By the end: Decide whether a symbol takes ∈\in or ⊆\subseteq, list the elements of a set, and find A∩BA\cap B and A∪BA\cup B for finite sets and for intervals, endpoints included.

The “element or subset” question was Question 1 in all five sittings we studied, and it is almost always the same test: ∈\in or ⊆\subseteq? This lesson secures the first question of the paper.

Understand it

A set is a collection of distinct objects, written between braces: A={1,2,3}A=\{1,2,3\}. Order does not matter and repeats do not count, so {1,2,2,3}\{1,2,2,3\} is the same set as {3,2,1}\{3,2,1\}. The objects are the elements.

Two different relations, and the exam tests the difference every time:

  • ∈\in links an element to a set: 2∈A2\in A reads “2 is an element of AA”. Its negation is ∉\notin.
  • ⊆\subseteq links a set to a set: B⊆AB\subseteq A means every element of BB is also in AA.

Read the left side first. A bare element (a number) takes ∈\in or ∉\notin. Anything in braces takes ⊆\subseteq. The empty set ∅\varnothing has no elements, and it is a subset of every set, so ∅⊆A\varnothing\subseteq A is always true while ∅∈A\varnothing\in A is false unless ∅\varnothing is itself listed in AA.

Sets are often described by a rule, called set-builder notation. A={x∣x2−9=0}A=\{x\mid x^2-9=0\} means “all xx such that x2−9=0x^2-9=0”. Solve the rule first and list the elements: A={−3,3}A=\{-3,3\}. Only then compare.

Number sets. N\mathbb{N} is the natural numbers 0,1,2,…0,1,2,\dots (zero is included); N∗\mathbb{N}^* (also written N+\mathbb{N}_+) is 1,2,3,…1,2,3,\dots without zero; Z\mathbb{Z} is all integers; Q\mathbb{Q} is every fraction of two integers with a nonzero denominator; R\mathbb{R} is all real numbers, which also include irrational numbers such as 2\sqrt2 and π\pi. Each set sits inside the next, so N⊆Z⊆Q⊆R\mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}.

Counting subsets. Each element is either in a subset or out of it, two choices per element, so a set with nn elements has 2n2^n subsets, counting ∅\varnothing and the set itself. Leave out the set itself and 2n−12^n-1 proper subsets remain; leave out ∅\varnothing as well and 2n−22^n-2 non-empty proper subsets remain.

Intersection and union. A∩BA\cap B keeps only what is in both sets (“and”). A∪BA\cup B collects everything that is in either set (“or”), writing shared elements once.

Intervals. Inequalities are written as intervals. A round bracket leaves the endpoint out and a square bracket keeps it: [a,b)[a,b) means a≤x<ba\le x<b. Infinity always gets a round bracket. On a number line, an open circle means “not included” and a filled circle means “included”. Draw both sets on one line, shade where they overlap for ∩\cap, and shade everything covered for ∪\cup.

Essential formulas

x∈AB⊆A∅⊆Ax\in A\qquad B\subseteq A\qquad \varnothing\subseteq A
x∈Ax\in A: the object xx is an element of AA. B⊆AB\subseteq A: every element of the set BB is in AA. The empty set is a subset of every set. Use braces whenever the object on the left is itself a set.
A∩B={x∣x∈A and x∈B}A∪B={x∣x∈A or x∈B}A\cap B=\{x\mid x\in A\text{ and }x\in B\}\qquad A\cup B=\{x\mid x\in A\text{ or }x\in B\}
∩\cap (intersection) needs both conditions; ∪\cup (union) needs at least one, so it includes everything in the intersection too.
2n subsets,2n−1 proper,2n−2 non-empty proper2^n\ \text{subsets},\qquad 2^n-1\ \text{proper},\qquad 2^n-2\ \text{non-empty proper}
For a set with nn elements. A proper subset is any subset except the set itself; a non-empty proper subset also leaves out ∅\varnothing. For {a,b,c}\{a,b,c\}: 23=82^3=8 subsets, 77 proper subsets and 66 non-empty proper subsets.
N⊆Z⊆Q⊆R\mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}
N={0,1,2,… }\mathbb{N}=\{0,1,2,\dots\} includes 00; N∗={1,2,3,… }\mathbb{N}^*=\{1,2,3,\dots\} does not. Z\mathbb{Z}: integers. Q\mathbb{Q}: fractions. R\mathbb{R}: all real numbers.
[a,b)={x∣a≤x<b}(a,+∞)={x∣x>a}[a,b)=\{x\mid a\le x<b\}\qquad (a,+\infty)=\{x\mid x>a\}
A square bracket includes the endpoint, a round bracket excludes it. ±∞\pm\infty are never included.

See it

AB 1, 234 5
A only: 1, 2. Overlap (A ∩ B): 3. B only: 4. Outside both: 5. A ∪ B is the two circles together: 1, 2, 3, 4.
ABA ∩ BA ∪ B−1135
A = [−1, 3), B = (1, 5]. A ∩ B = (1, 3). A ∪ B = [−1, 5].

Worked examples

Worked example 1

Let A={x∣x2−9=0}A=\{x\mid x^2-9=0\}. Which statements are true: 3∈A3\in A, {3}⊆A\{3\}\subseteq A, 3⊆A3\subseteq A, {3}∈A\{3\}\in A, ∅⊆A\varnothing\subseteq A, ∅∈A\varnothing\in A?

  1. Solve the rule first: x2=9x^2=9 gives x=−3x=-3 or x=3x=3, so A={−3,3}A=\{-3,3\}.
  2. 3∈A3\in A is true: 3 is a bare element and it is listed.
  3. {3}⊆A\{3\}\subseteq A is true: braces on the left take ⊆\subseteq, and the only element of {3}\{3\} is in AA.
  4. 3⊆A3\subseteq A and {3}∈A\{3\}\in A are false: they use the wrong symbol for the left side. A number is not a set, and AA does not contain the set {3}\{3\} as an element.
  5. ∅⊆A\varnothing\subseteq A is true for every set. ∅∈A\varnothing\in A is false because AA lists only −3-3 and 33.

Answer: True: 3∈A3\in A, {3}⊆A\{3\}\subseteq A, ∅⊆A\varnothing\subseteq A. False: the other three.

Worked example 2

Let A=[−1,3)A=[-1,3) and B=(1,5]B=(1,5]. Find A∩BA\cap B and A∪BA\cup B.

  1. Write the conditions: AA is −1≤x<3-1\le x<3 and BB is 1<x≤51<x\le5.
  2. For A∩BA\cap B both must hold: x>1x>1 and x<3x<3. The left end is open because 1∉B1\notin B, and the right end is open because 3∉A3\notin A. So A∩B=(1,3)A\cap B=(1,3).
  3. For A∪BA\cup B the sets overlap, so they join into one piece from the smallest left end to the largest right end. −1∈A-1\in A and 5∈B5\in B, so both ends are included: A∪B=[−1,5]A\cup B=[-1,5].

Answer: A∩B=(1,3)A\cap B=(1,3) and A∪B=[−1,5]A\cup B=[-1,5].

Worked example 3

Let A=(−∞,2]A=(-\infty,2] and B=(2,+∞)B=(2,+\infty). Find A∩BA\cap B and A∪BA\cup B.

  1. The point 22 belongs to AA but not to BB, and every other real number is in exactly one of the two sets.
  2. Nothing is in both, so A∩B=∅A\cap B=\varnothing.
  3. Together they cover every real number, so A∪B=RA\cup B=\mathbb{R}. The shared endpoint 2 is covered by AA, so the union has no gap.

Answer: A∩B=∅A\cap B=\varnothing and A∪B=RA\cup B=\mathbb{R}.

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

Let M={x∈N∣x<3}M=\{x\in\mathbb{N}\mid x<3\} and N=[1,4)N=[1,4). List MM, then find M∩NM\cap N and M∪NM\cup N.

Hint 1 of 3

N\mathbb{N} includes 00. List every natural number below 3.

Hint 2 of 3

For M∩NM\cap N, test each element of MM against N={x∣1≤x<4}N=\{x\mid 1\le x<4\}.

Hint 3 of 3

The union keeps MM’s element that falls outside NN as a separate point.

Show the step-by-step solution
  1. M={0,1,2}M=\{0,1,2\} because 00 is a natural number and x<3x<3.
  2. 0∉N0\notin N (it is below 1), while 1∈N1\in N and 2∈N2\in N. So M∩N={1,2}M\cap N=\{1,2\}.
  3. The union is everything in MM or in NN: [1,4)[1,4) already contains 11 and 22, and 00 is extra. So M∪N={0}∪[1,4)M\cup N=\{0\}\cup[1,4).

Answer: M={0,1,2}M=\{0,1,2\}, M∩N={1,2}M\cap N=\{1,2\}, M∪N={0}∪[1,4)M\cup N=\{0\}\cup[1,4).

Common mistakes

  • Writing 3⊆A3\subseteq A or {3}∈A\{3\}\in A. Look at the left side: a bare element takes ∈\in, braces take ⊆\subseteq.
  • Leaving 00 out of N\mathbb{N}. Zero is a natural number; only N∗\mathbb{N}^* and N+\mathbb{N}_+ start at 1.
  • Saying ∅∈A\varnothing\in A is true. The empty set is a subset of every set, but it is an element of AA only if AA lists it.
  • Merging intervals with a gap: (1,3)∪(3,5)(1,3)\cup(3,5) is not (1,5)(1,5), because 33 is in neither piece.
  • Swapping the symbols: ∩\cap is “and” (the overlap), ∪\cup is “or” (everything covered).
  • Counting 2n−12^n-1 when the question says non-empty proper subsets. Leave out both ∅\varnothing and the set itself: 2n−22^n-2.

Exam tips

  • Read the left side first. Bare element: ∈\in or ∉\notin. Braces: ⊆\subseteq. This single habit answers most of Question 1.
  • Solve any rule such as {x∣x2−4=0}\{x\mid x^2-4=0\} and list the elements before you compare anything.
  • Draw a number line for every interval question, with open and filled circles. If the options differ only at an endpoint, check just that endpoint.
  • To count subsets of a set with nn elements, start from 2n2^n: subtract 1 for “proper” and 2 for “non-empty proper”.

In short

  • ∈\in is for elements, ⊆\subseteq is for sets; ∅⊆\varnothing\subseteq every set.
  • N\mathbb{N} includes 00; N∗\mathbb{N}^* does not.
  • An nn-element set has 2n2^n subsets, 2n−12^n-1 proper and 2n−22^n-2 non-empty proper.
  • A∩BA\cap B is the overlap (and); A∪BA\cup B is everything covered (or).
  • Square bracket includes an endpoint, round bracket excludes it; infinity is always round.

Knowledge check

4 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

All of Week 1 is free: the eight lessons, the daily questions and the Week 1 test.

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