2 Sets

Homework 2 BCM.pdf

Yue Shi 999027873

Exercise 1. (20 points) Write the power set of \(A = \{ \emptyset, 5, \{ 2, \{ 3 \} \} \}\)
Decide if the following statements are true or false.

The elements of \(A\) are \(\emptyset, 5, \{2, \{3\}\}\)

Then \(\wp(A)=\{\emptyset,\left\lbrace\emptyset\right\rbrace,\{5\},\{\{2,\{3\}\}\},\{\emptyset,5\},\{\emptyset,\{2,\{3\}\}\},\{5,\{2,\{3\}\}\},\{\emptyset,5,\{2,\{3\}\}\}\}\)

(a) \(\{\emptyset\} \subseteq \mathcal{P}(A)\).

True, \(\{\emptyset\}\) is an subset of \(\mathcal{P}(A)\)

(b) \(\{\emptyset\} \in \mathcal{P}(A)\).

True, \(\{\emptyset\}\) is an element of \(\mathcal{P}(A)\)

(c) \(\{2, \{3\}\} \in A\)

True, \(\{2, \{3\}\}\) is an element of \(A\)

Exercise 2. (20 points)

(a) Give an example of sets \(A, B, C\) such that \(A \subseteq B\), \(B \not\subseteq C\) and \(A \subseteq C\).

Let \(A=[1,2],B=[0,3],C=[1,4]\), \(A \subseteq B\), \(B \not\subseteq C\) and \(A \subseteq C\).

(b) Give an example of sets \(A, B, C\) such that \(A \subseteq B\), \(B \subseteq C\), and \(C \subseteq A\).

\(A=[1,2],B=[1,2],C=[1,2]\) then \(A \subseteq B\), \(B \subseteq C\), and \(C \subseteq A\).

Exercise 3. (20 points) Let the universe be the set of integers.

Let \(E, D, \mathbb{Z}^+\) and \(\mathbb{Z}^-\) be the sets of all even, odd, positive and negative integers, respectively (0 is not positive nor negative).

Find

(a) \(E - \mathbb{Z}^+\).

Since we know \(A-B=A\cap B^c\) and \(A\cap\left(B\cup C)=\left(A\cap B)\cup\left(A\cap C\right.\right)\right.\), then

\(E - \mathbb{Z}^{+} = E \cap (\mathbb{Z}^{+})^{c} = E \cap (\mathbb{Z}^{-} \cup \{0\}) = (E \cap \mathbb{Z}^{-}) \cup (E \cap \{0\}) = E \cap \mathbb{Z}^{-} \cup \{0\}\)

Thus the result is all non-positive even number

(b) \((\mathbb{Z}^-)^c\).

\((\mathbb{Z}^{-})^{c}=\mathbb{Z}^{+}\cup\{0\}\), thus the result is all non-negative integer

(c) \((D \cup \mathbb{Z}^+)^c\).

Since we know \((A\cup B)^c=A^c\cap B^c\) and \(A\cap\left(B\cup C)=\left(A\cap B)\cup\left(A\cap C\right.\right)\right.\)

Then \((D \cup \mathbb{Z}^{+})^{c} = D^{c} \cap (\mathbb{Z}^{+})^{c} = E \cap (\mathbb{Z}^{-} \cup \{0\}) = E \cap \mathbb{Z}^{-} \cup (E \cap \{0\}) = E \cap \mathbb{Z}^{-} \cup \{0\}\)

Thus the result is all non-positive even number

Exercise 4. (20 points) Decide if the following sentences are true for every sets \(A, B, C, D\).

Write a proof when a sentence is true, or give a counterexample when it is false.

(a) \((A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)\).

By definition of cartesian product

\((A \times B) = \{(a, b) : a \in A, b \in B\}\)

\((C \times D) = \{(c, d) : c \in C, d \in D\}\)

\((A \cap C) \times (B \cap D) = \{(v, w) : v \in A \cap C, w \in B \cap D\}\)

Then \((A \times B) \cap (C \times D) = \{(a, b) : a \in A, b \in B\} \cap \{(c, d) : c \in C, d \in D\}\)

\(\subseteq\)) Take \((x,y)\in(A\times B)\cap(C\times D)\Rightarrow(x,y)\in(A\times B)\text{ and }(x,y)\in(C\times D)\\\Rightarrow x\in A\text{ and }y\in B\text{ and }x\in C\text{ and }y\in D\Rightarrow x\in A\text{ and }x\in C\text{ and }y\in B\text{ and }y\in D\\\Rightarrow x\in A\cap C\text{ and }y\in B\cap D\Rightarrow(x,y)\in(A\cap C)\times(B\cap D)\)

\(\supseteq\)) Take \((x, y) \in (A \cap C) \times (B \cap D) \Rightarrow (x, y) \in \{(v, w) : v \in A \cap C, w \in B \cap D\}\)

\(\Rightarrow x\in A\cap C\text{ and }y\in B\cap D\Rightarrow x\in A\text{ and }x\in C\text{ and }y\in B\text{ and }y\in D\)

\(\Rightarrow x\in A\text{ and }y\in B\text{ and }x\in C\text{ and }y\in D\)

\(\Rightarrow (x, y) \in (A \times B) \text{ and } (x, y) \in (C \times D)\)

\(\Rightarrow (x, y) \in (A \times B) \cap (C \times D)\)

Thus \((A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)\)

(b) \((A \times B) \cup (C \times D) = (A \cup C) \times (B \cup D)\).

By definition of cartesian product

\((A \times B) = \{(a, b) : a \in A, b \in B\}\)

\((C \times D) = \{(c, d) : c \in C, d \in D\}\)

\((A\cup C)\times(B\cup D)=\{(v,w):v\in A\cup C,w\in B\cup D\}\)

Then \((A\times B)\cup(C\times D)=\{(a,b):a\in A,b\in B\}\cup\{(c,d):c\in C,d\in D\}\)

\(\subseteq\)) Take \((x,y)\in(A\times B)\cup(C\times D)\Rightarrow(x,y)\in(A\times B)\text{ or }(x,y)\in(C\times D)\\\Rightarrow x\in A\text{ and }y\in B\text{ or }x\in C\text{ and }y\in D\)

However, from this we cannot conclude that \(x\in A\) or \(x\in C\) and \(y\in B\) or \(y\in D\)

Because \((A\cup B)\cap\left(C\cup D\right)=\left\lbrack(A\cup B)\cap C\right\rbrack\cup\left\lbrack(A\cup B)\cap D\right\rbrack=\left\lbrack\left(A\cap C\right)\cup\left(B\cap C\right)\right\rbrack\cup\left\lbrack\left(A\cap D\right)\cup\left(B\cap D\right)\right\rbrack\)

Which cannot be \((A\cup B)\cup(C\cup D)\)​

Thus \((A\times B)\cup(C\times D)\neq(A\cup C)\times(B\cup D)\)

Counter example: \(A=\left\lbrace1\right\rbrace,B=\left\lbrace2\right\rbrace,C=\left\lbrace3\right\rbrace,D=\left\lbrace4\right\rbrace\)

\((A\times B)\cup(C\times D)=\left\lbrace\left(1,2\right)\right\rbrace\cup\left\lbrace\left(3,4\right)\right\rbrace=\left\lbrace\left(1,2\right),\left(3,4\right)\right\rbrace\)

\((A\cup C)\times(B\cup D)=\left\lbrace\left(1,2\right),\left(1,4\right),\left(3,2\right),\left(3,4\right)\right\rbrace\)

Thus \((A\times B)\cup(C\times D)\neq(A\cup C)\times(B\cup D)\)

Exercise 5. (20 points) Compute the following sets.

(a) \(\bigcup_{n \in \mathbb{N}} [-n, n]\).

Since we know \(\N\) is not bounded, thus \(n\) can goes close to infinity, but \(n\) cannot reach.

Also, by definition, we know if \(n\) is in the set for some set in the union of family set, then \(n\) is also in the union of family set.

Thus \(\bigcup_{n\in\mathbb{N}}[-n,n]=(-\infty,+\infty)=\mathbb{R}\)

(b) \(\bigcup_{n\in\mathbb{N}}\left[\frac{1}{n},n\right)\).

We know \(-\frac1n\) is monotonic decreasing when \(n\in\N\), thus when \(n=1\), \(\frac{1}{n}=1\) is the maximum element, when n increases, \(\frac1n\) goes to \(0\).

Also \(n\) is monotonic increasing when \(n\in\N\), thus \(n\) goes close to infinity.

Also, by definition, we know if \(n\) is in the set for some set in the union of family set, then \(n\) is also in the union of family set.

Thus \(\bigcup_{n\in\mathbb{N}}\left[\frac{1}{n},n\right)=\left(0,+\infty\right)\).

(c) \(\bigcap_{r > 0} \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq r^2 \}\).

By definition, we know if \(n\) is in the set for any set in the intersection of family set, then \(n\) is also in the intersection of family set.

Since we know \(r^2\geq x^2+y^2\geq0\), when \(r\) gets close to \(0\), \(r^2\) gets close to \(0\).

Thus \(x^2+y^2\text{ has to be }\)​\(0\Rightarrow x=0,y=0\)

Thus \(\bigcap_{r > 0} \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq r^2 \}=\{(0,0)\}\)

(d) \(\bigcap_{n \in \mathbb{N}} \{ n, n + 1, n + 2, \dots \}\).

By definition, we know if \(n\) is in the set for any set in the intersection of family set, then \(n\) is also in the intersection of family set.

Since we know \(\N\) is not bounded, thus we can always choose a element that is not included by the previous set.

Thus \(\bigcap_{n \in \mathbb{N}} \{ n, n + 1, n + 2, \dots \}=\empty\)