Relations, Functions and Cartesian Product - Set Theory

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Question

Determine if the following statement is true or false:

If $D\subseteq E$ and $F\subseteq G$ then $D\times F\subseteq E\times G$ .

Answer

Assuming , , , and are classes where $D\subseteq E$ and $F\subseteq G$ .

Then by definition,

the product of and results in the ordered pair where is an element is the set and is an element in the set or in mathematical terms,

$D\times F=\begin{Bmatrix} (d,f):d\ \epsilon\ D \ and\ f\ \epsilon\ F \end{Bmatrix}$

and likewise

$E\times G=\begin{Bmatrix} (e,g):e\ \epsilon\ E \ and\ g\ \epsilon\ G \end{Bmatrix}$

Now,

$\(d,f)\ \epsilon\ D\timesF \\Rightarrow d\ \epsilon\ D\ and\ f\ \epsilon\ F \\Rightarrow d\ \epsilon\ E\ and\ f\ \epsilon\ G \\text{Because } D\subseteq E\ and\ F\subseteq G \\Rightarrow (d,f)\ \epsilon\ E\timesG$

therefore,

$D\times F\subseteq E \times G$ .

Thus by definition, this statement is true.

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Question

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

$x\subseteq A$ .

Answer

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 .

Therefore, the answer is False.

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Question

Determine if the following statement is true or false:

If be the set defined as,

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

then $x\subseteq A$ .

Answer

Given is the set defined as,

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

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

Looking at the elements in is is seen that all three elements in fact do contain therefore $x\subseteq A$ .

Thus, the answer is True.

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Question

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

Answer

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.

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Question

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

Answer

For an injective function, every element of must map to exactly one element of . Additionally, every element in must map to a different element in so that no element in has multiple pairings to elements in . It is not necessary for all elements of to be connected to an element in .

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Question

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

Answer

For a surjective function, every element of must map to exactly one element of . Additionally, all elements of to be paired with an element in , even if one or more elements of is connected to multiple elements in .

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Question

For which of the following pairs is the cardinality of the two sets equal?

Answer

The cardinality of $\mathbb{R}$ ( $\mathfrak{c}$ ) is greater than that of $\mathbb{N}$ ( $\aleph_{0}$ ,) as established by Cantor's first uncountability proof, which demonstrates that $\mathfrak{c}=2^{\aleph_{0}}>\aleph_{0}$ . The cardinality of the empty set is 0, while the cardinality of $\left { 0\right }$ is 1. $\left | \left { 1,2,3\right } \right |=3$ , while $\left | \left { 1,2,3,4\right } \right |=4$ . For sets and , where there exists an injective, non-surjective function $f:A\rightarrow B$ , must have more elements than , otherwise the function would be bijective (also called injective-surjective). Finally, for $\left { x|x\in \mathbb{N}\right },\left { 2y|y\in \mathbb{N}\right }$ , the cardinality of both sets is equal to the cardinality of $\mathbb{N}$ .

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Question

What type of function is $f:\mathbb{Z}\rightarrow \mathbb{N}$ where $x \mapsto \left |x \right |$ ?

Answer

Because multiple elements of $\mathbb{Z}$ can map to a single element of $\mathbb{N}$ (e.g. -2 and 2 map to 2), this function is surjective.

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Question

What type of function is $f:\mathbb{N}\rightarrow \mathbb{Z}$ where $x \mapsto x^2$ ?

Answer

Because every element of $\mathbb{N}$ maps to a single element of $\mathbb{Z}$ , but there are many elements of $\mathbb{Z}$ that do not pair with any element of $\mathbb{N}$ , this function is injective.

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Question

What type of function is $f:\mathbb{Z}\rightarrow \mathbb{R}$ where $x \mapsto \sqrt{x}$ ?

Answer

Because this operation is not defined for the negative elements of $\mathbb{Z}$ , it is either considered to be a partial function or not a function at all! Each element in the sub-domain of $\mathbb{Z}$ for which the operation is defined maps to exactly 1 element in $\mathbb{R}$ ; however, many elements of $\mathbb{R}$ are not paired to any element in this sub-domain of $\mathbb{Z}$ , thus the function could be considered a partial injective function.

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