Fubction From a Set to Cartesian Product of Itself is Continuous

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Determine if the following statement is true or false:

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

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.

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

$x\subseteq A$ .

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.

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 $x\subseteq A$ .

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.

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:

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:

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.

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.

All of the conditions here must be true.

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

Correct answer:

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

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:

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

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.

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

Possible Answers:

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

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

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

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

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

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 $\mathbb{N}$ .

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

Possible Answers:

None of these answers is correct.

Bijective

Injective

Surjective

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.

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

Possible Answers:

None of these answers is correct.

Injective

Bijective

Surjective

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.

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

Possible Answers:

Bijective

Injective

Surjective

None of these answers is correct.

Correct answer:

None of these answers is correct.

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Fubction From a Set to Cartesian Product of Itself is Continuous

All Set Theory Resources

All Set Theory Resources

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