2. In each item below, you are given sets A and B, and a formula for a (non?) function f: A → B. In each case, determine whether or not the formula gives a well-defined function. Fully justify your answers. (a) A = Q = B, f(r) = a root in B of the polynomial X²-r (b) A=C=B, f(2)= a root in B of the polynomial X² - z (c) A = P(R) - {0}¹, B = R, ƒ(E)= the smallest element of E (d) A = P(R), B = R, f(E)= the smallest element of E if E admits a smallest ele- ment, and f(E)= 0 otherwise A function f A → B is injective (or one-to-one) if it satisfies the following condition: for all a₁, a₂ € A, if f(a₁) = f(a₂), then a₁ = a2; f is surjective (or onto) if f(A) = B, or, more explicitly, if for each b E B there exists a € A such that f(a) = b; f is bijective if f is both injective and surjective. Here are some examples. • The squaring function f: RR given by f(x) = x² is not injective because f(-1) = 1 = f(1); f also fails to be surjective because -1 is not in the range of f. The function g: R→ [0,00) given by g(x) = x² is not injective for the same reason that the function f of the previous example is not injective, since g(-1) = 1 = g(1); but g is surjective, because every nonnegative real number admits a square root (this is a result in introductory real analysis). • The sine function sin: R R is very far from injective: it is 27-periodic, and assumes all of its values infinitely often. It is not surjective either, because its range is [-1,1]. • If we define S: [-π/2, π/2] → [-1, 1] by S(x) = sin(x), then methods from calculus show that S is both injective and surjective, i.e., S is bijective. Restricting the domain and codomain of sin in this way is how one obtains the arcsin function, which is the inverse of the function we have called S (see below for more on inverse functions). • The swapping function from Rx R to itself sending (x, y) to (y, z) is bijective. (Verify this!) • The exponential function exp: R→ (0, ∞) is bijective. (I write exp(r) for what is typically written e in calculus textbooks. This notation is pretty standard outside

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### Well-defined Functions in Set Theory

#### Problem Statement

2. In each item below, you are given sets \( A \) and \( B \), and a formula for a (non?) function \( f: A \to B \). In each case, determine whether or not the formula gives a well-defined function. Fully justify your answers.

(a) \( A = \mathbb{Q} \), \( B = \mathbb{Q} \), \( f(r) \) = a root in \( B \) of the polynomial \( X^2 - r \)

(b) \( A = \mathbb{C} \), \( B = \mathbb{C} \), \( f(z) \) = a root in \( B \) of the polynomial \( X^2 - z \)

(c) \( A = \mathcal{P}(\mathbb{R} - \{0\})^1 \), \( B = \mathbb{R} \), \( f(E) \) = the smallest element of \( E \)

(d) \( A = \mathcal{P}(\mathbb{R}) \), \( B = \mathbb{R} \), \( f(E) \) = the smallest element of \( E \) if \( E \) admits a smallest element, and \( f(E) = 0 \) otherwise


#### Definitions and Concepts Explained

A function \( f: A \to B \) is **injective** (or one-to-one) if it satisfies the following condition: for all \( a_1, a_2 \in A \), if \( f(a_1) = f(a_2) \), then \( a_1 = a_2 \). It is **surjective** (or onto) if \( f(A) = B \), or more explicitly, if for each \( b \in B \) there exists \( a \in A \) such that \( f(a) = b \); \( f \) is **bijective** if it is both injective and surjective. Here are some examples:

- The squaring function \( f: \mathbb{R} \to \mathbb{R} \) given by \( f(x) = x^2 \) is not injective because \( f(-1) = 1 = f(1) \); \( f \
Transcribed Image Text:### Well-defined Functions in Set Theory #### Problem Statement 2. In each item below, you are given sets \( A \) and \( B \), and a formula for a (non?) function \( f: A \to B \). In each case, determine whether or not the formula gives a well-defined function. Fully justify your answers. (a) \( A = \mathbb{Q} \), \( B = \mathbb{Q} \), \( f(r) \) = a root in \( B \) of the polynomial \( X^2 - r \) (b) \( A = \mathbb{C} \), \( B = \mathbb{C} \), \( f(z) \) = a root in \( B \) of the polynomial \( X^2 - z \) (c) \( A = \mathcal{P}(\mathbb{R} - \{0\})^1 \), \( B = \mathbb{R} \), \( f(E) \) = the smallest element of \( E \) (d) \( A = \mathcal{P}(\mathbb{R}) \), \( B = \mathbb{R} \), \( f(E) \) = the smallest element of \( E \) if \( E \) admits a smallest element, and \( f(E) = 0 \) otherwise #### Definitions and Concepts Explained A function \( f: A \to B \) is **injective** (or one-to-one) if it satisfies the following condition: for all \( a_1, a_2 \in A \), if \( f(a_1) = f(a_2) \), then \( a_1 = a_2 \). It is **surjective** (or onto) if \( f(A) = B \), or more explicitly, if for each \( b \in B \) there exists \( a \in A \) such that \( f(a) = b \); \( f \) is **bijective** if it is both injective and surjective. Here are some examples: - The squaring function \( f: \mathbb{R} \to \mathbb{R} \) given by \( f(x) = x^2 \) is not injective because \( f(-1) = 1 = f(1) \); \( f \
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