Are {1, 2, 3, 4} and {1, 2, 2, 3, 4, 4, 4} two different sets? And does the set with nothing in it sit quietly inside every other set? This note answers both, and shows you the trap hiding between ⊆ and ⊂.
1. Subsets
If A and B are sets and every element of A also belongs to B, then A is a subset of B, written A ⊆ B. For example, A = {2, 3} is a subset of B = {2, 3, 5, 7}. One element outside is enough to break it: {1, 3} ⊈ {2, 3, 5, 7}, because 1 ∉ B. Many books and papers write A ⊄ B with the same meaning.
Example 1: let A = {x : x is a positive integer}, B = {0} and X = {x : x is an integer}. Then A ⊆ X and B ⊆ X, but B ⊈ A, because zero is not positive.
2. The empty set
A set that has no element is called the empty set. It is written ∅ or { }, and its number of elements is 0.
An empty set often hides behind a condition. {x : x is a real number and x² < 0} is empty, because the square of a real number is never negative. {x : x is an African country that won the FIFA World Cup up to 2014} is empty too, because by the book's count no African country had won it up to 2014. And in {x ∈ N : x² + 9x + 20 = 0}, (x + 4)(x + 5) = 0 gives x = −4, −5. Neither is a natural number, so the set is ∅. Walk through them step by step.
step by steppress next
Spotting an empty set
{x ∈ R : x² < 0} A square is never negative, so no real number meets the condition. Nothing is left, so the set is ∅.
{x ∈ N : x² + 9x + 20 = 0} Find the candidates first: (x + 4)(x + 5) = 0, so x = −4 or x = −5.
−4, −5 ∉ N Now test the condition: neither is a natural number. None survives, so the set is ∅ with 0 elements.
African World Cup winners up to 2014 By the book's count there is none, so ∅. The fact holds up to 2014 only.
The same move all three times: find the candidates, then test the condition. If none survives, the set is empty.
3. Equal sets
If A and B have the same elements they are the same set, A = B. An element written again and again still counts once: {1, 2, 3, 4} = {1, 2, 2, 3, 4, 4, 4}. So the opening question has its answer: not two sets, one.
Here is the working rule for proofs: A = B if and only if A ⊆ B and B ⊆ A. Show the subset in both directions and equality is proved.
equal sets?flip to check
Equal or not
Equal?{1, 2, 3, 4} and {1, 2, 2, 3, 4, 4, 4}
AnswerEqual: a repeat counts once.
Equal?{1, 2} and {2, 1}
AnswerEqual: order does not matter.
Equal?∅ and {∅}
AnswerNot equal: {∅} has one element.
Equal?{0} and ∅
AnswerNot equal: {0} contains 0.
4. Proper subsets
If A ⊆ B and also A ≠ B, then A is a proper subset of B, written A ⊂ B. That is, B has at least one element that A does not. For example, {1, 2} ⊂ {1, 2, 3}.
For any set A, A ⊆ A, because x ∈ A gives x ∈ A. But A is not a proper subset of itself, because A ≠ A is false.
The empty set plays too: for any set A, ∅ ⊆ A. Otherwise ∅ would hold an element that A lacks, which is impossible. As the book states it, ∅ is a proper subset of any set. Two more facts: if A ⊆ ∅ then A = ∅ (since ∅ ⊆ A holds anyway), and if A ⊆ B and B ⊆ C then A ⊆ C.
Where each sign goes: ∈ sits between an element and a set, ⊆ or ⊂ between two sets. With P = {2, 4} and Q = {2, 4, 6}, P ⊂ Q and 2 ∈ Q, but P ∈ Q is false, because the elements of Q are only 2, 4, 6.
To pick a proper subset, list the main set first, then look for a set whose elements all belong to it and which is not the whole set. In U = {3, 4, 5, 6}, S = {x : x + 2 ≤ 7} = {3, 4, 5}. Here {3, 5} ⊂ S, but {3, 4, 5} is S itself, and {3, 6} has 6 ∉ S.
Build the sets yourself. Add elements to A and B or take them out, and watch which signs stay true.
set playgroundtap the numbers
⊆, ⊂ and =
Now sort eight pairs into three boxes. Each time, judge the first set against the second.
sort thempair, then box
⊆, ⊂ or ⊈
Pick up a pair, then drop it in a box. First check that every element of the first set is in the second.
Right. Next one.
Not yet. First see whether any element of the first set is missing from the second; if so, ⊈. If not, see whether the second has something extra.
All in place. 1 ∉ {2, 3, 5, 7} and 6 ∉ {3, 4, 5}, so those two are ⊈; zero is not positive, so {0} is ⊈ as well. −1 ∈ Z but is not positive, so the positive integers are ⊂ Z. By the book, ∅ is a proper subset of any set. And two equal sets are ⊆ but not ⊂.
5. Check yourself
The first question is in the exam's style.
check yourselfpick one answer
Question 1
If A ∩ B = B and A ≠ B, which relation between A and B is true?
A ∩ B = B means every element of B is in A, so B ⊆ A, not the other way round.
Right. A ∩ B = B gives B ⊆ A, and A ≠ B makes it proper: B ⊂ A.
With B ⊆ A, A ∪ B equals A, not B, because A ≠ B.
A ∩ B = B says all of B lies in A, so B ⊄ A is false.
In the second, hold the definitions up against each option.
check yourselfpick one answer
Question 2
If A = {x : x ∈ ℕ, x² + 9x + 20 = 0}, then n(A) = ?
There are not 4 elements; x² + 9x + 20 = (x + 4)(x + 5) has roots −4 and −5, which are not natural, so n(A) = 0.
The two roots are −4 and −5, but x ∈ ℕ rules out both, so n(A) = 0.
Neither −4 nor −5 is a natural number, so A has no element at all.
Right. The roots −4 and −5 are not natural numbers, so A = ∅ and n(A) = 0.
In the last one, go through the elements one by one.
check yourselfpick one answer
Question 3
If P = {2, 4} and Q = {2, 4, 6}, which of these statements are correct?
i. P ∈ Q ii. P ⊂ Q iii. x ∈ P ⟹ x ∈ Q
i is false: the elements of Q are 2, 4 and 6; the set P = {2, 4} is not one of them.
i is false: P is not an element of Q but a subset, so the true pair is ii and iii.
Right. Every element of P is in Q and 6 is extra, so P ⊂ Q; P ∈ Q is false.
i is false: the set P is not an element of Q; only ii and iii are true.
The whole note on seven cards.
cut from the notetap to turn over
Seven cards
QuestionWhen is A called a subset of B?
AnswerWhen every element of A belongs to B; it is written A ⊆ B.
QuestionHow many elements does the empty set have, and how is it written?