Question

Verify each of the properties of sets listed on page 511.

Short answer

Thus, the properties of sets listed are verified.

Step-by-step solution

Show that B∪B=B

As given that the statement to verify the properties of the set are provided below:

Now, the property to be shown is$B\cup B=B$ . Let solve as follows:

$B=\left\{x:x\in B\right\}$then, it gives:

$$\begin{gathered} B=\left\{x:x\in Borx\in B\right\} \\ =\left\{x:x\in B\right\} \\ =B \end{gathered}$$

Hence, the statement is proved

Show that B∩B=B

Now, the property to be shown is$B\cap B=B.$

Let $B=\left\{x:x\in B\right\},$, then,

$$\begin{gathered} B\cap B=\left\{x:x\in Bandx\in B\right\}, \\ =\left\{x:x\in B\right\} \\ =B \end{gathered}$$

Hence, the statement is proved.

Show that B∪ϕ=B

Now, the property to be shown is $B\cap \varphi =B$.

Let role="math" localid="1659158968119" $B=\left\{x:x\in B\right\}$, then,

role="math" localid="1659159098595" $$\begin{gathered} B\cap \varphi =\left\{x:x\in Borx\in \varphi \right\} \\ =\left\{x:x\in B\right\} \\ =B \end{gathered}$$

Hence, the statement is proved

Show that B∩ϕ=B

Now, the property to be shown is$B\cap \varphi =B$

Let $B=\left\{x:x\in B\right\}$, then,

$$\begin{gathered} B\cap \varphi =\left\{x:x\in Bandx\in \varphi \right\} \\ =\left\{x:x\in B\right\} \\ =\varphi \end{gathered}$$

Hence, the statement is proved

Show that A∪B=B∪A and A∩B=B∩A

Now, it is considered as $x\in A\cup Bthenx\in Aorx\in B.$

It implies that, $x\in Borx\in Athus,x\in B\cup A.$

Then,$A\cup B\subseteq B\cup A$

Similarly, solved as follows:

$B\cup A\subseteq A\cup B$

Therefore, the statement $A\cup B=B\cup A$is proved.

Show that $A\cap B=B\cap A$

It is given that$x\in A\cap Bthen,x\in Aorx\in B.Itimpliesthat,x\in Borx\in A$

Thus,$x\in B\cap A$

So, $x\in A\cap B$

It implies that, $x\in A\cap B$

Therefore, $(A\cap B)\subset (B\cap A)$

Similarly, solve as follows;

$(B\cap A)\subset (A\cap B)$

Therefore, the given statement$(A\cap B)=(B\cap A)$ is proved

Show that B⊆B∪C and B∩C⊆B 

It is given that therefore, it is true when at least one of $x\in Borx\in C$is true.

Since, from the definition of union $x\in Borx\in C$is true.

Thus, $x\in B\cup C$

Therefore, the given statement is proved.

Show that $B\cap C\subseteq B$

It is given that $x\in B\cap C$thus, by the definition of intersection$x\in Borx\in C.$

Thus, in particular $x\in B$.

Therefore, the given statement is proved.

Show that  B∪(C∪D)=(B∪C)∪D and B∩(C∩D)=(B∩C)∩D 

It is given and solve as follows:

$$\begin{gathered} X\in B\cup \left(C\cup D\right) \\ =\left\{X\in Bor\left(X\in CorX\in D\right)\right\} \\ =\left\{X\in BorX\in CorX\in D\right\} \\ =\left\{B\cup C\right\}\cup D \end{gathered}$$

Hence, the given statement is proved.

Show that$B\cap (C\cap D)=(B\cap C)\cap D$

It is given and solve as follows:

$$\begin{gathered} X\in B\cup \left(C\cup D\right) \\ =\left\{X\in Band\left(X\in C\cup \right)\right\} \\ =\left\{\left(X\in BandX\in C\right)andX\in D\right\} \\ =\left\{B\cup C\right\}\cup D \end{gathered}$$

Hence, the given statement is proved.

Show that B∩(C∪D)=(B∩C)∪(B∩D) and B∪(C∩D)=(B∪C)∩(B∪D) 

It is given and solve as follows:

$$\begin{gathered} X\in B\cup \left(C\cup D\right) \\ =\left\{X\in Band\left(X\in C\cup D\right)\right\} \\ =\left\{X\in Band\left(X\in CorX\in D\right)\right\} \\ =\left\{\left(X\in BandX\in C\right)or\left(X\in BandX\in D\right)\right\} \\ =\left(B\cup C\right)\cup \left(B\cup D\right) \end{gathered}$$

Hence, the given statement is proved.

Show that$\mathbf{}\mathit{B}\mathbf{\cup }\mathbf{(}\mathit{C}\mathbf{\cap }\mathit{D}\mathbf{)}\mathbf{=}\mathbf{(}\mathit{B}\mathbf{\cup }\mathit{C}\mathbf{)}\mathbf{\cap }\mathbf{(}\mathit{B}\mathbf{\cup }\mathit{D}\mathbf{)}$

It is given and solve as follows:

$$\begin{gathered} X\in B\cup \left(C\cup D\right) \\ =\left\{X\in Bor\left(X\in C\cup D\right)\right\} \\ =\left\{X\in Bor\left(X\in CandX\in D\right)\right\} \\ =\left\{\left(X\in BorX\in C\right)and\left(X\in BorX\in D\right)\right\} \\ =\left(B\cup C\right)\cup \left(B\cup D\right) \end{gathered}$$

Hence, the given statement is proved.