HOW TO DEFINE A SET ...
Ex: Let M be the set of all positive integers whose square is
strictly between 10 and 101. Define M :
M = {x ∈ N | 10 < x2 < 101}.
We could describe the same set by
M = {4, 5, 6, 7, 8, 9, 10}
because these are the only positive integers satisfying the condition 10 < x2 < 101.
Ex: Let S be the set of all integers that are squares:
S = {x ∈ Z | ∃y ∈ Z : x = y 2}.
Generally: The set of all elements of a set A that have property
P is described by
{x ∈ A | x has property P }.
If a set is described in this way, then all we know about the set
is that an element is in the set if and only if it has property P .
Ex: The unit disc in the real plane, centred at (1, 1) is the set
{(x, y) ∈ R × R | (x − 1)2 + (y − 1)2 ≤ 1}.
Remark: Sometimes a set is defined by substituting elements:
Ex: The set S can also be defined by
S = {x2 | x ∈ Z}.
That means to obtain the set S we have to substitute all x ∈ Z
in x2. So since Z has the elements 0, 1, −1, 2, −2, 3, −3, . . ., S has
the elements 0, 1, 1, 4, 4, 9, 9, . . ..
HOW TO SHOW A SUBSET ...
Recall: If A and B are sets, then A ⊆ B means:
every element of A is also in B.
Ex: {1, 2, 4} ⊆ {1, 2, 3, 4}.
Q: Given sets A and B, how do we show A ⊆ B?
A: We have to show that x ∈ A implies x ∈ B, (that is x ∈ A ⇒
x ∈ B).
Usually we start with ”Assume that x ∈ A”, and then somehow
conclude ”x ∈ B”.
Ex: Let A = {z ∈ C | z 4 = −1} and B = {z ∈ C | z 8 ∈ Z}. We
want to show
A ⊆ B.
Proof: Assume that x ∈ A. Then x4 = −1. Hence (x4)2 =
(−1)2, and so x8 = 1. Since 1 ∈ Z, we have x8 ∈ Z. Hence x ∈ B.
General example: Let A1 be the set of all numbers that have
property P1, and A2 the set of all numbers that have property
P2. If we want to show that
A 1 ⊆ A2 ,
then this would look as follows:
Assume that x ∈ A1. Then x has property P1.....................and
so x has property P2, hence x ∈ A2.
Ex: Let
A = {(x, y) ∈ R2 | (x − 1)2 + (y − 1)2 ≤ 1},
B = {(x, y) ∈ R2 | x ≥ 0 and y ≥ 0}.
Show that A ⊆ B:
Let (x, y) ∈ A. Then (x − 1)2 + (y − 1)2 ≤ 1. Hence (x − 1)2 ≤ 1
and (y − 1)2 ≤ 1. Then |x − 1| ≤ 1 and |y − 1| ≤ 1. Hence
−1 ≤ x − 1 ≤ 1 and −1 ≤ y − 1 ≤ 1. This implies x ≥ 0 and y ≥ 0.
Thus (x, y) ∈ B.
HOW TO SHOW ”NOT A SUBSET ...”
Q: If A is not a subset of B: How do we show that A * B?
A: We have to show that ” x ∈ A implies x ∈ B” does not hold.
A good way to do that is to present an element x with x ∈ A
but x ∈
/ B.
Ex: Let A = {x ∈ C | x4 = 81} and B = {3, −3}. We want to
show
A * B.
Proof: We have 3i ∈ A since
(3i)4 = 34 · i4 = 81 · 1 = 81.
But 3i ∈
/ B. Hence A * B.
Ex (homework): Let
A = {x ∈ R | x2 + x > 0},
B = {x ∈ R | x > 0}.
Is A ⊆ B? Prove your answer, either by showing that x ∈ A
implies x ∈ B, or by exhibiting an element of A that is not in B.
Similarly decide (with proof) if B ⊆ A.
HOW TO SHOW ”... EQUALS...”
Q: If A and B are sets: How do we show that A = B?
A: Usually the best way is to show two statements:
A⊆B
and
B ⊆ A.
Ex: Let S be the set of squares of integers, i.e., S = {0, 1, 4, 9, . . .}.
Let
A = {x ∈ S | x is a multiple of 4},
B = {x ∈ S | x is a multiple of 2}.
We want to show that A = B.
Proof: Step 1: Prove that A ⊆ B:
Let x ∈ A. Then x is a multiple of 4. Since 4 is a multiple of 2,
x is a multiple of 2, so x ∈ B.
Step 2: Prove that B ⊆ A:
Let x ∈ B. Then x is a multiple of 2. Then 2 is a prime factor
of x. Since x is a square, at least two prime factors of x equal
2. Then x is a multiple of 4. Hence x ∈ A.
HOW TO DO INTERSECTION
Def: If A and B are sets, then the intersection of A and B is
the set containing those elements that are in A and in B. It is
denoted by A ∩ B.
Ex: Let A = {1, 2, 3}, B = {2, 3, 4}, then A ∩ B = {2, 3}.
Rem: If we want to show that an element x is in A ∩ B, we need
to show that x ∈ A and x ∈ B.
Ex: Let M2 be the set of integers that are multiples of 2, similarly
M3 and M6. Show that
M2 ∩ M3 = M6.
Proof: Step 1: Prove that M2 ∩ M3 ⊆ M6:
Let x ∈ M2 ∩ M3. Then x ∈ M2 and x ∈ M3. Hence there is a 2
and a 3 among the prime factors of x. Therefore, x is a multiple
of 2 · 3 (which equals 6), and so x ∈ M6.
Step 2: Prove that M6 ⊆ M2 ∩ M3:
Let x ∈ M6. Then x is a multiple of 6. Since 6 is a multiple of
2, x is also a multiple of 2. Hence x ∈ M2. Similarly we get that
x ∈ M3. Hence x ∈ M2 and x ∈ M3, and so x ∈ M2 ∩ M3.
HOW TO DO UNION
Def: If A and B are sets, then the union of A and B is the set
containing those elements that are in A or in B. It is denoted by
A ∪ B.
Rem: An element of A ∪ B may be in both sets, A and B.
Ex: Let A = {1, 2, 3}, B = {2, 3, 4}, then A ∪ B = {1, 2, 3, 4}.
Rem: If we want to show that an element x is in A ∪ B, we need
to show that x ∈ A or x ∈ B.
Ex (homework): Let
f1(x) = x12 − 2x7 + 3x,
f2(x) = x5 +
f3(x) = (x12 − 2x7 + 3x)(x5 +
√
√
2x3 + 2,
2x3 + 2).
For i = 1, 2, 3 let Ri be the set of roots of fi. Show that R1 ∪R2 =
R3 .
Ex (homework): Let M2, M3 and M6 be as above. Show that
M2 ∪ M3 * M6.
EXERCISES
Q1 Define the following sets in at least two of the different ways
shown in the lecture:
a) The set of all even positive integers whose square is less than
100.
b) The set of all integers strictly between −100 and 100 that
are multiples of 3.
c) The set of all points in the real plane that are on the line
passing through (0, 0) and (1, 1).
Q2 Let A and B be two sets with A ⊆ B ⊆ R. Let f : R → R
be a real function. Give a proof of the following statement that
follows the format of the proofs in the lecture:
{y ∈ R | ∃x ∈ A : f (x) = y} ⊆ {y ∈ R | ∃x ∈ B : f (x) = y}.