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Levi Pearson 2014-01-25 15:21:00 -07:00
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@ -111,3 +111,98 @@ The idea extends to a 3-list, or *ordered triple*. In general:
$$ A_1 \times A_2 \times \dots \times A_n = \{ $$ A_1 \times A_2 \times \dots \times A_n = \{
(x_1,x_2,\dots,x_n) : x_i \in A_i, i \in \Bbb{N} \} (x_1,x_2,\dots,x_n) : x_i \in A_i, i \in \Bbb{N} \}
$$ $$
We can also take *Cartesian powers* of sets. For a set $A$ and
positive integer $n$, $A^n$ is the Cartesian product of $A$ with
itself $n$ times: $$
A^n = A \times A \times \dots \times A = \{(x_1,x_2,\dots,x_n):x_i \in
A,i \in \{1,\dots,n\}\} $$
* * * *
**Exercises 1.2**
1. $A = \{1,2,3,4\}, B = \{a,c\}$
a. $A \times B =
\{(1,a),(2,a),(3,a),(4,a),(1,c),(2,c),(3,c),(4,c)\}$
b. $B \times A =
\{(a,1),(c,1),(a,2),(c,2),(a,3),(c,3),(a,4),(c,4)\}$
d. $B \times B = \{(a,a),(a,c),(c,a),(c,c)\}$
e. $\emptyset \times B = \emptyset$
f. $(A \times B) \times B = \{((1,a),a),((2,a),a),
((3,a),a),((4,a),a),((1,c),a),((2,c),a),((3,c),a),
((4,c),a),((1,a),a),((2,a),c),((3,a),c),((4,a),c),
((1,c),c),((2,c),c),((3,c),c),((4,c),c)\}$
* * * *
**Definition 1.3**
: $A$ and $B$ are sets. If every element of $A$ is also an element of
$B$, then $A$ is a *subset* of $B$ and we write $A \subseteq B$. If
this is not the case, we write $A \subsetneq B$, which means there is
at least one element of $A$ that is not in $B$.
**Fact 1.2**
: It follows from **1.3** that for any set $B$, $\emptyset \subseteq
B$. I.e., the empty set is a subset of every set.
**Fact 1.3**
: If a finite set has $n$ elements, it has $2^n$ subsets.
This can be shown by drawing a decision tree starting with the empty
set, with each fork representing a choice of whether to insert the
next element of the set in question. Since there are two possibilities
at each fork and $n$ elements to consider for insertion, that gives
$2^n$ total leaves of the tree.
* * * *
**Exercises 1.3**
1. Subsets of \{1,2,3,4\}: $\emptyset, \{1\}, \{2\}, \{3\}, \{4\},
\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \{1,2,3\},
\{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}
* * * *
**Definition 1.4**
: $A$ is a set. The *power set* of $A$ is another set,
$\mathscr{P}(A)$, defined to be the set of all subsets of
$A$. I.e. $\mathscr{P}(A) = \{X:X\subseteq A\}$.
**Fact 1.4**
: If $A$ is a finite set, $|\mathscr{P}(A)| = 2^{|A|}$.
* * * *
**Definition 1.5**
$A$ and $B$ are sets.
- The *union* of $A$ and $B$ is the set $A\cup B = \{x : x \in A or x
\in B\}$
- The *intersection* of $A$ and $B$ is the set $A\cap B = \{x : x \in
A and x \in B\}$
- The *difference* of $A$ and $B$ is the set $A - B = \{x : x \in A
and x \notin B\}$
The operations $\cup$ and $\cap$ obey the commutative law for sets,
but $-$ does not.
* * * *
We usually discuss sets in some context. Our sets in that context will
naturally be subsets of some other set, which we call the *universal
set* or just *universe*. If we don't know specifically which set it
is, we call it $U$.
For example, when discussing the set of prime numbers $P$, the
*universal set* is $\Bbb{N}$. When we discuss geometric figures such
as the set of points in a circle $C$, the universe would be
$\Bbb{R}^2$.
**Definition 1.6**
: $A$ is a set in the universe $U$. The *complement* of $A$ or $A\bar$
is the set $A\bar = U - A$.
E.g. if $P$ is the set of prime numbers, then $$
P\bar = \Bbb{N} - P = \{1,4,6,8,9,10,12,\dots\}
$$ so $P\bar$ is the set of composite numbers and 1.