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Set Theory : Relations, Functions and Cartesian Product

Study concepts, example questions & explanations for Set Theory

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Example Questions

Example Question #1 : Relations, Functions And Cartesian Product

Determine if the following statement is true or false:

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

Possible Answers:

False

True

Correct answer:

True

Explanation:

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

Then by definition,

the product of and results in the ordered pair (d,f) 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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Example Question #2 : Relations, Functions And Cartesian Product

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

$x\subseteq A$ .

Possible Answers:

False

True

Correct answer:

False

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.

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Example Question #3 : Relations, Functions And Cartesian Product

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 $x\subseteq A$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

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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Example Question #4 : Relations, Functions And Cartesian Product

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

Possible Answers:

All of the conditions here must be true.

No element of may map to multiple elements of .

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

No element of may map to multiple elements of .

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

Correct answer:

All of the conditions here must be true.

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.

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Example Question #5 : Relations, Functions And Cartesian Product

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

Possible Answers:

No element of may map to multiple elements of .

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

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 .

Correct answer:

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

Explanation:

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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Example Question #6 : Relations, Functions And Cartesian Product

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

Possible Answers:

No element of may map to multiple elements of .

All of the conditions here must be true.

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

Correct answer:

No element of may map to multiple elements of .

Explanation:

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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Example Question #7 : Relations, Functions And Cartesian Product

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

Possible Answers:

$\left \{ 0\right \},\varnothing$

$\mathbb{R},\mathbb{N}$

$\left \{ x|x\in \mathbb{N}\right \},\left \{ 2y|y\in \mathbb{N}\right \}$

and , where there exists an injective, non-surjective function $f:A\rightarrow B$ .

$\left \{ 1,2,3\right \},\left \{ 0,1,2,3\right \}$

Correct answer:

$\left \{ x|x\in \mathbb{N}\right \},\left \{ 2y|y\in \mathbb{N}\right \}$

Explanation:

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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Example Question #1 : Relations, Functions And Cartesian Product

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

Possible Answers:

Surjective

Injective

None of these answers is correct.

Bijective

Correct answer:

Surjective

Explanation:

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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Example Question #2 : Relations, Functions And Cartesian Product

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

Possible Answers:

None of these answers is correct.

Surjective

Bijective

Injective

Correct answer:

Injective

Explanation:

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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Example Question #3 : Relations, Functions And Cartesian Product

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

Possible Answers:

Injective

Bijective

None of these answers is correct.

Surjective

Correct answer:

None of these answers is correct.

Explanation:

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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Set Theory : Relations, Functions and Cartesian Product

Study concepts, example questions & explanations for Set Theory

All Set Theory Resources

Example Questions

Example Question #1 : Relations, Functions And Cartesian Product

Example Question #2 : Relations, Functions And Cartesian Product

Example Question #3 : Relations, Functions And Cartesian Product

Example Question #4 : Relations, Functions And Cartesian Product

Example Question #5 : Relations, Functions And Cartesian Product

Example Question #6 : Relations, Functions And Cartesian Product

Example Question #7 : Relations, Functions And Cartesian Product

Example Question #1 : Relations, Functions And Cartesian Product

Example Question #2 : Relations, Functions And Cartesian Product

Example Question #3 : Relations, Functions And Cartesian Product

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