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 $x\nsubseteq A$ .

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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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 $x\nsubseteq A$ .

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.

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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 .

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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