Let $$ N=\{1,2,3,4, \ldots\} $$ be the set of natural numbers and $$ S=\{1,4,9,16, \ldots\} $$ be the set of squares of the natural numbers. Then $$ N \sim S $$, since we have the one-to-one correspondence $$ 1 \leftrightarrow 1,2 \leftrightarrow 4,3 \leftrightarrow 9,4 \leftrightarrow 16, \ldots n \leftrightarrow n^{2} $$. (This example is interesting, since it shows that an infinite set can be equivalent to a proper subset of itself.) Show that each of the following pairs of sets are equivalent by carefully describing a one-to-one correspondence between the sets. Complete parts (a) through (c) below. D. For each element in W, there is an element in N that is double that element. Write a general rule for the correspondence described in the previous step using $$ w $$ as an element in W. $$ \square $$ (b) The sets of odd and even numbers, $$ D=\{1,3,5,7, \ldots\} $$ and $$ E=\{2,4,6,8, \ldots\} $$ Which of the following describes a one-to-one correspondence between the two sets? A. For each element in $$ D $$, there is an element in $$ E $$ that is 3 greater than that element. B. For each element in D , there is an element in E that is 1 greater than that element. F. For each element in D, there is an element in E that is double that element. D. For each element in D , there is an element in E that is 2 less than double that element.

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**(a)** The correspondence from $$ W $$ to $$ N $$ is given by the rule

$$
w \mapsto 2w.
$$

That is, for every element $$ w $$ in $$ W $$, the associated element in $$ N $$ is double $$ w $$.

**(b)** We are given the sets of odd numbers

$$
D=\{1,3,5,7,\ldots\}
$$

and even numbers

$$
E=\{2,4,6,8,\ldots\}.
$$

We consider the following proposed bijections for elements $$ d \in D $$:

- Option A: $$ d \mapsto d+3 $$
- Option B: $$ d \mapsto d+1 $$
- Option F: $$ d \mapsto 2d $$
- Option D: $$ d \mapsto 2d-2 $$

For a valid one-to-one correspondence (bijection), the mapping must be both injective and surjective.

Consider Option B: $$ d \mapsto d+1 $$.

- For every odd number $$ d $$, $$ d+1 $$ is even.
- The function is clearly one-to-one since if $$ d_1+1 = d_2+1 $$, then $$ d_1 = d_2 $$.
- It is onto because every even number $$ e $$ can be written as $$ e = d+1 $$ where $$ d=e-1 $$ is odd.

Thus, the one-to-one correspondence between the sets $$ D $$ and $$ E $$ is given by

$$
d \mapsto d+1.
$$
**(a)** Each element $$ w $$ in $$ W $$ corresponds to $$ 2w $$ in $$ N $$. **(b)** The one-to-one correspondence between $$ D $$ and $$ E $$ is $$ d \mapsto d+1 $$.

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