blog/content/Unit 1 Sets and Logic.md

tags, Slides, Topic, Unit

tags	Slides	Topic	Unit
Math-301 Math School notes Spring-25 public	Wed 1-7.pdf	Sets and Logic	1.1 - 1.6

1.1 Basic definitions

A set is a collection of objects. The objects in a set are called its elements or members.

Let A={a,b,c}. a\in A -- means a is an element of A d \notin A -- means d is not an element of A

Def. the Cardinality of a finite set S, demoted |S|. is the number of elements in S. In the example, |A| = 3.

Notation for some sets of numbers

Natural numbers: \mathbb{N} = {1,2,3,...} Whole Numbers: \mathbb{W} = {0,1,2,3,...} The set of integers: \mathbb{Z} = {...-3,-2,-1,0,1,2,3,} = {0,1,-1,2,-2,3,-3...} Rational Numbers: \mathbb{Q} = \{\frac{a}{b}\vert a,b\in\mathbb{Z},b\neq0\} Set of real numbers: \mathbb{R}

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2 subsets

• Let A and B be sets. We say that B is a subset of A, if every element of B is also an element of A, denoted B\subseteq A
• Two sets are equal, denoted A = B, if A\subseteq B and B\subseteq A

Ex. Let A = \{x\vert x\in\mathbb{Z}\hspace{7px}and\hspace{7px}0\lt x\lt6\}. We see that A=\{1,2,3,4,5\}. Note: 1\in A, 4\in A, but 6\notin A. \{1,3,5\}\subseteq A,\{2\}\subseteq A, but \{2\}\notin A {2,4,6}\subsetneq A since 6 \notin A |A| = 5. Ex. \mathbb{N}\subseteq \mathbb{W} \subseteq \mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R}

Proper subset: B\subset A if B\subseteq A and B\neq A

Intervals of \mathbb{R}

(a,b) = {x$\in \mathbb{R}$ | a < x < b}, where -\infty\le a \le b \le\infty [a,b] = {x$\in \mathbb{R}$ | a \le x \le b}, where -\infty\lt a \le b \lt\infty [a,b) = {x$\in \mathbb{R}$ | a \le x < b}, where -\infty\lt a \lt b \le\infty (a,b] = {x$\in \mathbb{R}$ | a \lt x \le b}, where -\infty\le a \lt b \lt\infty

3. Collections of Sets

The elements of a set may themselves be sets, and so is is a collections of sets

Ex. \mathcal{C} = { {1}, {1, 2}, {1, 2, 3} }. Note that {1} \in \mathcal{C}, {1,2} \notin\mathcal{C}, {1,2,3} \in\mathcal{C} ?? {1} \subsetneq\mathcal{C} since 1\notin\mathcal{C} {{1}, {1,2}} \subseteq\mathcal{C} {{1}, {1,2}} \notin\mathcal{C}.

Indexed Collection of Sets

Let I be a set. Suppose S_i is a set for each i \in I. Then we say that \{S_i\}_{i\in I} = \{S_i|i\in I\} Is called a collection of sets indexed by I.

Ex. Let S_n = (n-1, n) for each n\in\mathbb{N}. Then \{S_n\}_{n=1}^{3} = {$S_1,S_2,S_3$} = \{(0,1), (1,2), (2,3)\} \{S_n\}_{n\in\mathbb{N}} = \{(0,1), (1,2), (2,3),...\}

4. The Empty Set

Let E be a set with no elements. Then for any Set A, we have E$\subseteq$A (Vacuously True). If E' is another set with no elements, then E$\subseteqE'` `E'\subseteq$E
So E=E' Therefore, there is a unique set with no elements. We call it the empty set, denoted by \emptyset = {}. Property: For every set A, $\emptyset\subseteq$A.

5. The Power Set of a Set

the power set of a set S is the collection of all subsets of S and is denoted \wp(S).

\wp(S) = \{A | A \subseteq S\}.

Ex. Let A = \{a,b,c\}. Then \wp(A)\{\emptyset\,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\} Note: |\wp(A) = 8 = 2^3 = 2^{|A|}.

6. Summarizing example

Consider the set:

S = \{1,2,\emptyset,\{a,b\}\}. Then 2\in S, 2\subsetneq S {2}\notin S, 2\subseteq S

\emptyset, 2\subsetneq S

1.2 Set Operations

1 Intersections and Unions

Let A and B be sets. The Intersection of A and B is the set. A\cap B {X | x \in A and x \in B} !

The Union of A and B is the set

A \cup B = {x | \in a A or x \in B } ## "Or" is the "Inclusive or" !

Ex. Let A={0,2,4,6,8}, B={0,3,6,9}, and C={1,2,3,4,5,6,8,9}. Then (a) A\cap B = {0,6} (b) A\cup B = \{0,2,4,6,8,3,9\}=\{0,2,3,4,6,8,9\} (c) B\cap C = {3} (d) A\cup(B\cap C) = \{0,2,3,4,6,8\} (e) (A\cup B)\cap C = \{2,3,4\} Note: A\cup(B\cap C)\neq(A\cup B)\cap C

Def. We say that two sets A and B and disjoint if $A\cap B=\emptyset$ !

Theorem. Let A and B be finite sets. (a) |A\cup B|=|A|+|B|-|A\cup B| ! (b) if A and B and disjoint then |A\cup B| = |A| + |B| ! (c) If A\subseteq B, then |A|\leq|B|

2 Arbitrary Collections

Let \mathcal{C}=\{S_i|i\in I\} be a collection of sets indexed by a set I (Assume I\neq0.) Then the Intersection of \mathcal{C} is defined as

\cap\mathcal{C}=\cap_{i\in I} S_i=\{x|x\in S_i for all i\in I\}.

The Union of the collection \mathcal{C} is the set

\cup\mathcal{C}= \cup_{i\in I}S_i=\{x|x\in S_i for at least one i\in I\}

For a finite collection of sets indeed by I=\{1,2 ... , n\} we often write the intersection and unions of \mathcal{C}=\{S_1,S_2, ... ,S_n\} as

\cap_{i=1}^n S_i = S_1\cap S_2\cap S_3\cap ... \cap S_n \cup_{i\in I}^n S_i=S_1\cup S_2\cup ... \cup S_n

Def. Let \{\S_i|i\in I\} be an indexed collection of sets.

(a) The collection is Mutually disjoint if for all i,j\in I, if S_i\neq S_j, then S_I =\cap S_j\space\emptyset Equivalently: for all i,j\in I, S_i=S_j\space or\space S_i\cap S_j=\emptyset (b) The collection is nested if for all i,j\in I, S_i\subseteq S_j or S_j\subseteq S_i

S=\{x\in\mathbb{W}|x\notin M_{5}, x\in M_3\}

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B_n = \{x\in\mathbb{R}\space\vert\space |x| < n\}

B_1 = (-1,1) B_2 = (-2,2) B_3 =(-3, 3)

[!note] incluse vs exclusion [-1,1]\neq (-1,1) where n=\mathbb{R} let A = (-1,1) let B = [-1,1] A\subseteq B A\neq B

Q	P	A
Mutually disjoint?	B_n\subseteq B_{n+1}	False
Nested?	B_n\subseteq B_{n+1}	True
Intersect	B\cap B_n	B_1
Union	B\cup B_n	(-\infty,\infty)