More notes

math/proof/notes.page

@@ -111,3 +111,98 @@ The idea extends to a 3-list, or *ordered triple*. In general:
 $$ A_1 \times A_2 \times \dots \times A_n = \{
  (x_1,x_2,\dots,x_n) : x_i \in A_i, i \in \Bbb{N} \}
 $$
+
+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.