Sets, Set Operations.

A set is understood as any collection of some objects. These objects are called elements of the set. Sets are denoted by capital letters, and elements are denoted by lowercase letters.

The fact that an element $a$ belongs to a set $A$ (i.e., is an element of set $A$) is written as $a\in A$, and the fact that an element $b$ does not belong to set $A$ (i.e., is not its element) is written as $b\notin A$.

A set that does not contain any elements is called empty and is denoted by the symbol $\emptyset$.

The notation $A\subset B$ ($A$ is contained in $B$) means that every element of set $A$ is an element of set $B$; in this case, set $A$ is called a subset of set $B$. Sets $A$ and $B$ are called equal $(A=B)$ if $A\subset B$ and $B\subset A$.

There are two main ways to define (describe) sets.

a) Set $A$ is defined by directly listing all its elements $a_1, a_2, ..., a_n$, that is, it is written in the form $$A=\{a_1, a_2, ..., a_n\}.$$

For example, the set of prime numbers from 10 to 20 can be written as: ${11, 13, 17, 19}$.

b) Set $A$ is defined as the set of those and only those elements from some basic set $T$ that possess a common property $\alpha$. In this case, the notation is used $$A=\{x\in T|\alpha(x)\},$$ where the notation $\alpha(x)$ means that the element $x$ possesses the property $\alpha$.

For example, $[0, 1)={x\in \mathbb{R}| 0\leq x<1}$.

The union of sets $A$ and $B$ is called the set $$A\cup B=\{x|(x\in A)\vee (x\in B)\}$$

The intersection of sets is called the set $$A\cap B=\{x|(x\in A)\wedge (x\in B)\}$$

The union and intersection operations possess the following properties:

1) Commutativity

$$A\cup B=B\cup A;\qquad A\cap B=B\cap A;$$

2) Associativity

$$A\cup(B\cup C)=(A\cup B)\cup C; (A\cap B)\cap C=A\cap(B\cap C);$$

3) Distributivity

$$(A\cup B)\cap C= (A\cap C)\cup(B\cap C);\quad (A\cap B)\cup C= (A\cup C)\cap(B\cup C);$$

4) Idempotence

$$A\cup A=A;\quad A\cap A=A.$$

The set consisting of all elements of set $A$ not belonging to set $B$ is called the difference of sets $A$ and $B$, and it is denoted by:

$$A\setminus B=\{x|(x\in A)\wedge (x\notin B)\}.$$

If $A\subset B$, then $B\setminus A$ is called the complement of set $A$ to set $B$, denoted by $A_B'$.

If, in particular, $A$ is a subset of some universal set $U$, then the difference $U\setminus A$ is denoted by the symbol $\overline{A}$ or $A'$ and is called the complement of set $A$ (to set $U$).

From the definition of the complement of a set, the following equalities follow:

$$A\cup A' =U;\quad A\cap A' =\emptyset,\quad (A')'=A.$$

The symmetric difference of sets $A$ and $B$ is denoted by $A \Delta B$ and consists of those elements that belong to exactly one of the sets $A$ or $B$. In other words, it consists of: $$A\Delta B =(A\setminus B)\cup (B\setminus A).$$

For any subsets $A$ and $B$ of the set $U$, the following equalities hold, which are called the laws of duality or de Morgan's laws: $$(A\cup B)'=A'\cap B';\quad (A\cap B)'=A'\cup B'.$$

Examples:

To prove the validity of the equalities:

1) $(A\cap B)'=A'\cup B'$

Proof:

$x\in(A\cap B)' \Leftrightarrow x\notin (A\cap B)\Leftrightarrow x\notin A\vee x\notin B \Leftrightarrow $ $x\in A'\vee x\in B' \Leftrightarrow$

$ x\in (A'\cup B')\Leftrightarrow (A\cap B)'= A'\cup B'. $

Which is what needed to be proven.

2) $(A\setminus B)\setminus C =A\setminus (B\cup C).$

Proof.

$x\in (A\setminus B)\setminus C \Leftrightarrow a\in A\setminus B \wedge a\notin C \Leftrightarrow (a\in A \wedge a\notin B)\wedge a\notin C\Leftrightarrow $

$\Leftrightarrow a\in A \wedge (a\notin B \wedge a\notin C)\Leftrightarrow a\in A \wedge a\notin (B\cup C)\Leftrightarrow a\in A\setminus(B\cup C).$

Which is what needed to be proven.

3) $A\setminus (A\setminus B)=A\cap B.$

Proof.

$a\in A\setminus(A\setminus B)\Leftrightarrow (a\in A)\wedge (a\notin A\setminus B)\Leftrightarrow $

$(a\in A) \wedge ((a\notin A)\vee (a\in B))\Leftrightarrow $

$((a\in A)\wedge(a\notin A))\vee((a\in A)\wedge(a\in B))\Leftrightarrow $

$(a\in A) \wedge (a\in B) \Leftrightarrow a\in A\cap B.$

Which is what needed to be proven.

4) $A\cup(B\setminus C)\supset (A\cup B)\setminus C.$

Proof.

$a\in (A\cup B)\setminus C\Leftrightarrow (a\in A \vee a\in B)\wedge a\notin C \Leftrightarrow$

$ (a\in A \wedge a\notin C)\vee(a\in B\wedge a\notin C)\Rightarrow (a\in A )\vee(a\in B\wedge a\notin C)\Leftrightarrow$

$ a\in A\cup(B\setminus C).$

Which is what needed to be proven.

1.28. (a)

To determine which of the two statements is correct:

${1, 2}\in{1, 2, {1, 2, 3}}$ and ${1, 2}\subset{1, 2, {1, 2, 3}}$.

Solution:

The statement ${1, 2}\in{1, 2, {1, 2, 3}}$ means that the element ${1, 2}$ is contained in the set consisting of three elements ${1},$ ${2},$ and ${{1, 2, 3}}.$ Obviously, the element ${1, 2}$ is not among these three elements. Therefore, this statement is incorrect.

The statement ${1, 2}\subset{1, 2, {1, 2, 3}}$ means that all elements of the set ${1, 2}$ (i.e., ${1}$ and ${2}$) are contained in the set ${1, 2, {1, 2, 3}}.$ The latter set consists of the elements ${1},$ ${2},$ and ${{1, 2, 3}},$ so this statement is correct.

Answer: ${1, 2}\subset{1, 2, {1, 2, 3}}$.

In problems 1.29 and 1.30, the given sets are to be defined by enumerating all their elements.

1.29. $A={x\in R| x^3-3x^2+2x=0}.$

Solution:

Let's find the set of real solutions to the equation $x^3-3x^2+2x=0:$

$x^3-3x^2+2x=x(x^2-3x+2)=0\Rightarrow$

$\Rightarrow x_1=0.$

We solve the quadratic equation $x^2-3x+2=0.$

$D=3^2-4\cdot 2=1.$

$$x_2=\frac{3+1}{2}=2,\qquad\quad x_3=\frac{3-1}{2}=1.$$

To determine which of the two statements is correct:

${1, 2}\in{1, 2, {1, 2, 3}}$ and ${1, 2}\subset{1, 2, {1, 2, 3}}$.

Solution:

Answer: ${1, 2}\subset{1, 2, {1, 2, 3}}$.

In problems 1.29 and 1.30, the given sets are to be defined by enumerating all their elements.

1.29. $A={x\in R| x^3-3x^2+2x=0}.$

Solution:

Let's find the set of real solutions to the equation $x^3-3x^2+2x=0:$

$x^3-3x^2+2x=x(x^2-3x+2)=0\Rightarrow$

$\Rightarrow x_1=0.$

We solve the quadratic equation $x^2-3x+2=0.$

$D=3^2-4\cdot 2=1.$

All roots of the equation are real, so we write the answer:

Answer: ${0, 1, 2}.$

1.30. $A={x\in R| x+\frac{1}{x}\leq 2,,,, x>0}.$

Solution:

Let's solve the inequality $x+\frac{1}{x}\leq 2:$

$\frac{x^2+1}{x}\leq\frac{2x}{x}.$

Since $x>0,$ then $x^2+1\leq 2x.$ We solve the quadratic equation

$x^2-2x+1=0\Rightarrow (x-1)^2=0\Rightarrow x_{1, 2}=1.$

Since $y=x^2-2x+1$ is a parabola opening upwards, it is positive for all other values of $x.$

Thus, the solution to the inequality $x+\frac{1}{x}\leq 2$ under the condition $x>0$ is:

$x=1.$

Answer: ${1}.$

Homework:

1.28 (b)

Determine which of the two statements is correct:

${1, 2}\in{1, 2, {1, 2}}$ and ${1, 2}\subset{1, 2, {1, 2}}.$

Answer: Both statements are correct.

In problems 1.31 and 1.32, the given sets are to be defined by enumerating all their elements.

1.31. $A={x\in N| x^2-3x-4\leq 0}.$

Answer: $A={1,, 2,, 3,, 4}.$

1.32. $A={x\in Z| \frac{1}{4}\leq 2^x<5}.$

Answer: $A={-2, , -1, , 0,, 1,, 2}.$

Draw the following sets on the coordinate plane:

1.35. ${(x, y)\in R^2 |, x+y-2=0}.$

1.36. ${(x, y)\in R^2 |, x^2-y^2>0}.$

1.37. ${(x, y)\in R^2 |, (x^2-1)(y+2)=0}.$

1.43. Describe all elements of the sets $A\cup B, ,, A\cap B, ,, A\setminus B, ,, B\setminus A.$

$$A=\{x\in R\,|\, x^2+x-20=0\},\qquad B=\{x\in R\, | \, x^2-x+12=0\}.$$

Answer: $A\cup B={-5, 3, 4},,, A\cap B ={4}, ,,, A\setminus B={5}, B\setminus A= {3}.$

In problems 1.50 and 1.53, taking the interval as the universal set $T=[0, 1]$, find and plot the complements of the following sets on the number line:

1.50. ${0, 1};$

Answer: $(0, , 1).$

1.53. ${1/4}\cup[3/4,, 1).$

Answer: $[0, , 1/4)\cup (1/4, 3/4)\cup {1}.$

1.55. Prove that the operations $\cup$ and $\cap$ are related by the distributive law:

$$(A\cup B)\cap C=(A\cap C)\cup(B\cap C);$$

$$(A\cap B)\cup C=(A\cup C)\cap(B\cup C);$$

Prove the equalities:

1.57. $A\setminus B=A\cap\overline B.$

1.58. $\overline{A\setminus B}=\overline A\cup B.$