Definition 1..1. A set A is a subset of a set B, written as A ⊆ B, if every element of A is also an element of B. The relation ⊆ is called the inclusion relation.

So, A ⊆ B whenever x ∈ A implies x ∈ B. Observe carefully the difference between ⊆ and ∈:

If B = {1, 2, 3}, then 1 is an element of B, but 1 is not a subset of B. The set A = {1}, which has 1 as its only element, however, is a subset of B, since it fulfills the definition: Whenever x ∈ A, then x ∈ B. Note that there is only one possibility for x, namely, x = 1. It is very important that you learn to distinguish between an object (which may of course be a set itself), and the set formed from objects, e.g.

• 1 is different from {1},

• {1} is different from {{1}}.

Observe that the unique element of the last set is the set {1}.

It is conceivable that a set contains no elements at all; this set is called the empty set, and it is denoted by the symbol ∅. It can also be described by a property, e.g.

                   ∅ = {x : x ≠x}.

Here are some elementary properties of the ⊆ relation:

Lemma 1.1. ∅ is a subset of every set.

Proof. Recall that by definition, ∅ is a subset of a set A, if every element of ∅ is also an element of A. Since ∅ has no elements, this is trivially true.

Lemma 1..2. For any set A, A ⊆ A.

Proof. By our definition of ⊆ , A is a subset of A, if every element of A is an element of A, and this is of course true.

Lemma 1.3. If A is a subset of B, and B is a subset of C, then A is a subset of C.

Proof. We want to show that A ⊆ C, and we have as hypothesis that A ⊆ B and B ⊆ C. Using our definition of ⊆ , we have to show that if x ∈ A, then x is an element of C. Thus, let x be an arbitrary element of A. Our hypothesis A ⊆ B tells us that x ∈ B, and, since B ⊆ C, we also have x ∈ C. This is what we wanted to prove.

The next definition allows us to form a new set from a given one:

Definition 1..2. If A is a set, then P(A) = {X : X ⊆ A} is called the power set of A. It is the set of all subsets of A.

Let us look at some power sets:

1. A = ∅: Since ∅ is a subset of every set, we must have ∅ ⊆ ∅, i.e. ∅ ∈

P(∅). But ∅ has no elements, and therefore no other subsets; hence, we have

P(∅) = {∅}. Observe that ∅ is different from {∅}: While ∅ has no elements,  {∅} has exactly one element, namely the empty set ∅.

2. A = {x}: By Lemma 1. 1 and Lemma 1. 2, A has at least two subsets,  namely the empty set ∅, and A itself. Since A has only one element, there can be no other subsets,thus, P(A) = {∅, {x}}.

3. A = {x, y}: Using the previous lines of reasoning we obtain that

                          (1.1)                                P(A) = {∅, {x}, {y}, {x,y}}.

So, we see that P(A) has four elements.

4. A = {x, y, z}: Except for the subsets ∅ and A, we can pick one or two elements of A at a time to form a subset. Hence,

P(A) = {∅, {x}, {y}, {z}, {x,y}, {x,z}, {y,z}, {x,y, z}}.

Observe that P(A) has eight elements.

When shall we call two sets equal? An intuitive definition is to say that two sets are equal if they contain the same elements. We shall use our already defined notion ⊆ to define equality of sets:

Definition 1.3. Two sets A and B are equal, if A ⊆ B and B ⊆ A. If A and Bare equal, we write A = B.

Observe that this definition expresses just what we would intuitively mean by equality of sets: If every element of A is an element of B,and if every element of B is an element of A, then they must contain the same elements. To prove that two sets A and B are equal, we must show that A is a subset of B, and that B is a subset of A. The procedure shall be illustrated by an easy

Example 1. 1. Let A = {x ∈ R : x² = 1}, and B = {1, −1}; we want to show that A = B:

“A ⊆ B”: Let x ∈ A; then, by the definition of A, x solves the equation x² = 1,  hence, x = 1 or x = −1. In either case, x ∈ B.

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مواضيع ذات صلة

Subset

Set Class

Set

Second Category

Proper Subset

Power Set

Matroid

Inductive Set

Independent Vertex Set

Independent Set

Graphoid

G_delta Set