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Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

True

False

Explanation

This is a theorem for well ordered sets and the proof is as follows.

First identify what is given in the statement.

1. The sets are onto order isomorphisms

$f:A_1\rightarrow A_2$ and $g:B_1\rightarrow B_2$

2. The goal is to make an onto order isomorphism. Let us call it .

$t:A_1\times B_1 \rightarrow A_2\times B_2$

Thus, can be defined as the function,

$\t:A_1\times B_1\rightarrow A_2\times B_2 \t(a,b)=(f(a),g(b))$

To show is a well defined, one-to-one, function on $A_1\times B_1$ since and are one-to-one, the following is performed.

$t(a,b)=t(u,v)\Rightarrow(f(a),g(b))=(f(u),g(v))$

$\\Rightarrow f(a)=f(u), g(b)=g(v) \\Rightarrow a=u, b=v \\Rightarrow (a,b)=(u,v)$

Thus, is one-to-one.

Now to prove in onto $A_2\times B_2$ if $(a,b)\ \epsilon\ A_2\times B_2$

for some

$u\ \epsilon\ A_1, v\ \epsilon\ B_1$

thus

proving is onto $A_2 \times B_2$ .

Lastly prove ordering.

$(a_1,b_1)\leq_{A\times B}(a_2,b_2)\Leftrightarrow \left{\begin{matrix} a_1<_aa_2\ a_1=a_2, b+1\leq_{r}b_2 \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} f(a_1)<_{A}f(a_2)\ f(a_1)=f(a_2), g(b_1\leq_{r}g(b_2)) \end{matrix}\right.$

$\Leftrightarrow\left{\begin{matrix} (f(a_1),g(b_1))<_{A\times B}(f(a_2),g(b_2))\ (f(a_1),g(b_1))\leq(f(a_2), g(b_2)) \end{matrix}\right.$

Thus proving is isomorphic. Therefore the statement is true.

For a bijective function from set to set defined by $f:A \rightarrow B$ , which of the following does NOT need to be true?

All of the conditions here must be true.

Every element of must map to one or more elements of .

No element of may map to multiple elements of .

Explanation

For a bijective function, every element in set must map to exactly one element of set , so that every element in set has exactly one corresponding element in set . All of the conditions presented must be true in order to satisfy this definition.

Determine if the following statement is true or false:

If be the set defined as,

$A=\begin{Bmatrix} x,\begin{Bmatrix} x,y,z \end{Bmatrix}, \begin{Bmatrix} y,z \end{Bmatrix} \end{Bmatrix}$

then

$x\subseteq A$ .

False

True

Explanation

Given is the set defined as,

$A=\begin{Bmatrix} x,\begin{Bmatrix} x,y,z \end{Bmatrix}, \begin{Bmatrix} y,z \end{Bmatrix} \end{Bmatrix}$

to state that $x\subseteq A$ , every element in must contain .

Looking at the elements in is is seen that the first two elements in fact do contain however, the third element in the set, $\begin{Bmatrix} y,z \end{Bmatrix}$ does not contain therefore $x\nsubseteq A$ .

Therefore, the answer is False.

Are the sets and equal?

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix} \S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

$T\neq S$

$T= S, x\geq0$

Explanation

To determine if two sets are equal to each other it must be proven that each set contains the same elements.

Recall the following terminology,

$\\mathbb{Z}=\text{Integers}{\begin{Bmatrix} ...,-2,-1,0,1,2,... \end{Bmatrix}} \\mathbb{R}=\text{Reals} \\mathbb{N}=\text{Naturals} {\begin{Bmatrix} 0,1,2,... \end{Bmatrix}}$

Now, identify the elements in each set.

$\T=\begin{Bmatrix} x|x\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

This means all integers are elements of .

Let ,

$\S=\begin{Bmatrix} 2n|n\ \epsilon\ \mathbb{Z} \end{Bmatrix}$

is the value for all integers.

Therefore if the element in is,

This means that all elements in will be divisible by two and thus be an even number therefore, 3 will never be an element in .

Thus, it is concluded that

$T\neq S$

For a bijective function from set to set defined by $f:A \rightarrow B$ , which of the following does NOT need to be true?

All of the conditions here must be true.

Every element of must map to one or more elements of .

No element of may map to multiple elements of .

Explanation

Given the set , calculate the cardinality.

$A=\begin{Bmatrix} a,a,b,c,d,e,e \end{Bmatrix}$

$|A|=\frac{a+e}{7}$

Explanation

Given a set , the cardinality is the number of elements in . This is also written as .

Looking at this particular problem,

$A=\begin{Bmatrix} a,a,b,c,d,e,e \end{Bmatrix}$

First, rewrite the set to depict a more accurate description of the space.

$A=\begin{Bmatrix} a,b,c,d,e \end{Bmatrix}$

Therefore, counting up the number of elements, results in the following cardinality,

Determine if the following statement is true or false:

If be the set defined as,

$A=\begin{Bmatrix} x,\begin{Bmatrix} x,y,z \end{Bmatrix}, \begin{Bmatrix} y,z \end{Bmatrix} \end{Bmatrix}$

then

$x\subseteq A$ .

False

True

Explanation

Given is the set defined as,

$A=\begin{Bmatrix} x,\begin{Bmatrix} x,y,z \end{Bmatrix}, \begin{Bmatrix} y,z \end{Bmatrix} \end{Bmatrix}$

to state that $x\subseteq A$ , every element in must contain .

Therefore, the answer is False.

Let denote all parabolas in the Cartesian plane. Does , , or both belong to ?

$\S\begin{Bmatrix} (x,y): y=2x^2\end{Bmatrix} \T\begin{Bmatrix} (x,y):y=-x+1 \end{Bmatrix}$

$S\subseteq A$

$S,T \subseteq A$

$T\subseteq A$

$\text{Neither belong to A}$

$A\subseteq T$

Explanation

is a set that contains all parabolas that live in the Cartesian plane, this is a vast set. To determine if , , or both belong to , identify if the elements of each set create a straight line, and if so, then that set will be a subset of . In other words, the set will belong to .

Identify the elements in first.

$\S\begin{Bmatrix} (x,y): y=2x^2 \end{Bmatrix}$

This statement reads, contains the coordinate pairs that live on the parabola .

Since

is a parabola that lives in the Cartesian plane, that means belongs to .

Now identify the elements in .

$T\begin{Bmatrix} (x,y):y=-x+1 \end{Bmatrix}$

This means that the elements of are those that live on the straight line . Thus does not create a parabola in the Cartesian plan therefore does not belong to .

Therefore, answering the question, belongs to .

$S\subseteq A$

Given the set , calculate the cardinality.

$A=\begin{Bmatrix} a,a,b,c,d,e,e \end{Bmatrix}$

$|A|=\frac{a+e}{7}$

Explanation

Given a set , the cardinality is the number of elements in . This is also written as .

Looking at this particular problem,

$A=\begin{Bmatrix} a,a,b,c,d,e,e \end{Bmatrix}$

First, rewrite the set to depict a more accurate description of the space.

$A=\begin{Bmatrix} a,b,c,d,e \end{Bmatrix}$

Therefore, counting up the number of elements, results in the following cardinality,

Determine if the following statement is true or false:

Let and be well ordered and order isomorphic sets. If and are also order isomorphic sets, then $A_1\times B_1$ and $A_2\times B_2$ are also order isomorphic.

True

False