집합과 명제 전체 문제

Let $A = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 9 } \right\}$ and let $B = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 3 } \right\}$.

Is the set $A$ a subset of $B$? Justify your conclusion.

Let $A = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 9 } \right\}$ and let $B = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 3 } \right\}$.

Is the set $A$ equal to the set $B$? Justify your conclusion.

Prove the following proposition. To do so, prove each set is a subset of the other set by using the choose-an-element method.

Let $A$ and $B$ be subsets of some universal set. Then $A - B = A \cap B^c$.

Let $A = \left\{ {\left. {x \in \mathbb{R}} \right| x^2 \lt 4} \right\}$ and let $B = \left\{ {x \in \mathbb{R}\left. \right| x \lt 2} \right\}$.

Is $A \subseteq B$? Justify your conclusion with a proof or a counterexample.

Let $A = \left\{ {\left. {x \in \mathbb{R}} \right| x^2 \lt 4} \right\}$ and let $B = \left\{ {x \in \mathbb{R}\left. \right| x \lt 2} \right\}$.

Is $B \subseteq A$? Justify your conclusion with a proof or a counterexample.

Let $A$, $B$, and $C$ be subsets of a universal set $U$.

Prove the following proposition:

If

A \subseteq B

and

B \subseteq C

, then

A \subseteq C

.

Note: This may seem like an obvious result. However, one of the reasons for this exercise is to provide practice at properly writing a proof that one set is a subset of another set. So we should start the proof by assuming that $A \subseteq B$ and $B \subseteq C$. Then we should choose an arbitrary element of $A$.

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

• The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if $x \left( x - 2 \right) \lt 0$, then we can conclude that either (1) $x \lt 0$ and $x - 2 > 0$ or (2) $x > 0$ and $x - 2 \lt 0$. However, in the first case, we must have $x \lt 0$ and $x > 2$, and this is impossible. Therefore, we conclude that $x > 0$ and $x - 2 \lt 0$, which means that $0 \lt x \lt 2$.

Use the choose-an-element method to prove each of the following:

$\left\{x \in \mathbb{R} \mid x^2 - 3x - 10 \lt 0 \right\} = \left\{ x \in \mathbb{R} \mid -2 \lt x \lt 5 \right\}$

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

• The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if $x \left( x - 2 \right) \lt 0$, then we can conclude that either (1) $x \lt 0$ and $x - 2 > 0$ or (2) $x > 0$ and $x - 2 \lt 0$. However, in the first case, we must have $x \lt 0$ and $x > 2$, and this is impossible. Therefore, we conclude that $x > 0$ and $x - 2 \lt 0$, which means that $0 \lt x \lt 2$.

Use the choose-an-element method to prove each of the following:

$\left\{x \in \mathbb{R} \mid x^2 - 5x + 6 \lt 0 \right\} = \left\{ x \in \mathbb{R} \mid 2 \lt x \lt 3 \right\}$

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

• The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if $x \left( x - 2 \right) \lt 0$, then we can conclude that either (1) $x \lt 0$ and $x - 2 > 0$ or (2) $x > 0$ and $x - 2 \lt 0$. However, in the first case, we must have $x \lt 0$ and $x > 2$, and this is impossible. Therefore, we conclude that $x > 0$ and $x - 2 \lt 0$, which means that $0 \lt x \lt 2$.

Use the choose-an-element method to prove each of the following:

$\left\{ x \in \mathbb{R} \mid x^2 \geq 4 \right\} = \left\{ x \in \mathbb{R} \mid x \leq -2 \right\} \cup \left\{ x \in \mathbb{R} \mid x \geq 2 \right\}$

Let $A$ and $B$ be subsets of some universal set $U$. Prove each of the following:

$A \cap B \subseteq A$

Let $A$ and $B$ be subsets of some universal set $U$. Prove each of the following:

$A \subseteq A \cup B$

Let $A$ and $B$ be subsets of some universal set $U$. Prove each of the following:

$A \cap A = A$

Let $A$ and $B$ be subsets of some universal set $U$. Prove each of the following:

$A \cup A = A$

Let $A$ and $B$ be subsets of some universal set $U$. Prove each of the following:

$A \cup \emptyset = A$

Let $A$, $B$, and $C$ be subsets of a universal set $U$. Prove:

If $A \subseteq B$, then $A \cap C \subseteq B \cap C$.

Let $A$, $B$, and $C$ be subsets of a universal set $U$. Prove:

If $A \subseteq B$, then $A \cup C \subseteq B \cup C$.

Prove the following proposition:

For all sets

A

,

B

, and

C

that are subsets of some universal set, if

A \cap B = A \cap C

and

A^c \cap B = A^c \cap C

, then

B = C

.

Let $S$, $T$, $X$, and $Y$ be subsets of some universal set. Assume that

• (i) $S \cup T \subseteq X \cup Y$; - (ii) $S \cap T = \emptyset$; and - (iii) $X \subseteq S$.

Using all three assumptions, either prove that $T \subseteq Y$ or explain why it is not possible to do so.

Let $A$, $B$, and $C$ be subsets of some universal set $U$. Prove or disprove each of the following:

$\left( A \cap B \right) - C = \left( A - C \right) \cap \left( B - C \right)$

Let $A$, $B$, and $C$ be subsets of some universal set $U$. Prove or disprove each of the following:

$\left( A \cup B \right) - \left( A \cap B \right) = \left( A - B \right) \cup \left(B - A \right)$