Testing notation

master
Levi Pearson 2014-01-25 15:38:24 -07:00
parent c2017edf36
commit e61950fb7d
1 changed files with 33 additions and 9 deletions

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@ -159,7 +159,7 @@ $2^n$ total leaves of the tree.
**Exercises 1.3**
1. Subsets of \{1,2,3,4\}: $\emptyset, \{1\}, \{2\}, \{3\}, \{4\},
1. Subsets of $\{1,2,3,4\}$: $\emptyset, \{1\}, \{2\}, \{3\}, \{4\},
\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \{1,2,3\},
\{1,2,4\}, \{1,3,4\}, \{2,3,4\}, \{1,2,3,4\}$
@ -176,17 +176,24 @@ $A$. I.e. $\mathscr{P}(A) = \{X:X\subseteq A\}$.
* * * *
**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 *union* of $A$ and $B$ is the set $A\cup B = \{x : x \in A
\text{ or } x \in B\}$
- The *intersection* of $A$ and $B$ is the set $A\cap B = \{x : x \in
A and x \in B\}$
A \text{ and } x \in B\}$
- The *difference* of $A$ and $B$ is the set $A - B = \{x : x \in A
and x \notin B\}$
\text{ and } x \notin B\}$
The operations $\cup$ and $\cap$ obey the commutative law for sets,
but $-$ does not.
- If an expression involving sets uses only $\cap$ or $\cup$,
parentheses are optional.
- If it uses both $\cap$ and $\cup$, parentheses are required!
* * * *
We usually discuss sets in some context. Our sets in that context will
@ -200,9 +207,26 @@ 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 $\bar A$
is the set $\bar A = U - A$.
: $A$ is a set in the universe $U$. The *complement* of $A$ or $\overline A$
is the set $\overline A = U - A$.
E.g. if $P$ is the set of prime numbers, then $$
\bar P = \Bbb{N} - P = \{1,4,6,8,9,10,12,\dots\}
$$ so $\bar P$ is the set of composite numbers and 1.
\overline P = \Bbb{N} - P = \{1,4,6,8,9,10,12,\dots\}
$$ so $\overline P$ is the set of composite numbers and 1.
**Definition 1.7**
$A_1, A_2,\dots, A_n$ are sets.
$$
\align{
A_1 \cup A_2 \cup A_3 \cup \dots \cup A_n &=
\left{x : x \in A_i
\text{ for at least one set $A_i$, for }
1 \leq i \leq n\right}\\
A_1 \cap A_2 \cap A_3 \cap \dots \cap A-n &=
\left{x : x \in A_i
\text{ for every set $A_i$, for }
1 \leq i \leq n \right}\\
}
$$