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Let \(A=\\{1,2,3, \ldots, 10\\} .\) Consider the function \(f: \mathcal{P}(A) \rightarrow \mathbb{N}\) given by \(f(B)=|B|\). That is, \(f\) takes a subset of \(A\) as an input and outputs the cardinality of that set. (a) Is \(f\) injective? Prove your answer. (b) Is \(f\) surjective? Prove your answer. (c) Find \(f^{-1}(1)\). (d) Find \(f^{-1}(0)\). (e) Find \(f^{-1}(12)\)

Short Answer

Expert verified
f is not injective because different subsets can have the same cardinality. f is not surjective onto \(\mathbb{N}\), but it is onto \{0, 1, ..., 10\}. \(f^{-1}(1)\) is the set of all single-element subsets, \(f^{-1}(0)\) is the empty set, and \(f^{-1}(12)\) does not exist.

Step by step solution

01

Understanding injective (one-to-one) functions

A function is injective if different elements in the domain map to different elements in the codomain. To prove injectivity for this function, we must show that if two subsets of A, say B and C, have the same cardinality, then they must be the same subset.
02

Proving/disproving the injectivity of f

Assume we have two subsets, B and C, such that B is not equal to C. It is possible for subsets to have the same number of elements without being identical, e.g. \(\{1,2\}\) and \(\{3,4\}\) both have a cardinality of 2 but are different sets. Therefore, \(f\) is not injective.
03

Understanding surjective (onto) functions

A function is surjective if every element in the codomain has a preimage in the domain. To prove surjectivity for this function, we must show that for every natural number n in the codomain, there is a subset of A with cardinality n.
04

Proving/disproving the surjectivity of f

Since the set A has 10 elements, \(f\) maps to the natural numbers from 0 to 10, as these are the possible cardinalities of subsets of A. Hence, \(f\) is surjective onto the set \(\{0, 1, 2, ..., 10\}\) but not onto all of \(\mathbb{N}\) because there is no subset of A with more than 10 elements.
05

Finding \(f^{-1}(1)\)

The preimage of 1 under function \(f\) will be the set of all subsets of A that contain exactly one element. These subsets are \(\{1\}\), \(\{2\}\), ..., \(\{10\}\).
06

Finding \(f^{-1}(0)\)

The preimage of 0 under \(f\) is the set of all subsets of A with no elements. There is only one subset that satisfies this condition, which is the empty set \(\emptyset\).
07

Finding \(f^{-1}(12)\)

Since there are no subsets of A with a cardinality of 12 (because A has only 10 elements), \(f^{-1}(12)\) is the empty set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Functions
Understanding the nature of injective functions, also known as one-to-one functions, is essential in discrete mathematics. An injective function guarantees that each element in the domain of the function is mapped to a unique element in its codomain. In other words, if you have two different elements in the domain, they will never map to the same element in the codomain.

To determine if the given function f is injective, consider two different subsets of the set A. If it is found that these subsets, despite being distinct, can have the same cardinality, the function cannot be considered injective. For example, different pairs of elements like \(\{1, 2\}\) and \(\{3, 4\}\) both result in the same output when applying the function f, proving it’s not injective.
Surjective Functions
A surjective function, also termed an onto function, covers the entirety of its codomain. This means that for every element in the codomain, you can find a preimage in the function’s domain. This concept is a crucial part of understanding how functions connect two sets.

For the function f in the exercise, which maps subsets and their cardinality, you might examine if every natural number is achieved by the cardinality of some subset of A. Given that the largest subset has at most 10 elements, any number beyond 10 does not have a corresponding subset, and therefore, the function is not surjective onto all natural numbers. It is only surjective onto \(\{0, 1, 2, ..., 10\}\).
Preimage of a Function
The preimage of a function involves finding all the elements in the domain that map to a particular element in the codomain. It is often denoted as f-1 for a given function f. Grasping the idea of preimages is vital for solving problems related to the behavior of functions.

In the context of the exercise, finding \(f^{-1}(1)\) means identifying all subsets of A with exactly one element. Similarly, finding \(f^{-1}(0)\) involves identifying the subset with no elements, which is inherently the empty set. However, for \(f^{-1}(12)\), there are no subsets with 12 elements, and so the preimage is the empty set—a key point illustrating the concept of domain and codomain restriction.
Discrete Mathematics
In discrete mathematics, we study mathematical structures that are fundamentally discrete rather than continuous. It includes a wide range of topics such as logic, set theory, combinatorics, graph theory, and functions, which are pertinent to computer science and algorithm analysis.

The exercise with function f demonstrates a typical discrete mathematics problem—analyzing properties of functions related to set theory. Through this lens, students learn about the relationships and mappings between sets, which are the foundation of more complex discrete systems. This domain of mathematics encourages a deep understanding of how discrete structures interact—a key skill for addressing computational and theoretical problems.

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Most popular questions from this chapter

Consider the statement "If Oscar eats Chinese food, then he drinks milk." (a) Write the converse of the statement. (b) Write the contrapositive of the statement. (c) Is it possible for the contrapositive to be false? If it was, what would that tell you? (d) Suppose the original statement is true, and that Oscar drinks milk. Can you conclude anything (about his eating Chinese food)? Explain. (e) Suppose the original statement is true, and that Oscar does not drink milk. Can you conclude anything (about his eating Chinese food)? Explain.

Find the least element of each of the following sets, if there is one. (a) \(\left\\{n \in \mathbb{N}: n^{2}-3 \geq 2\right\\}\). (b) \(\left\\{n \in \mathbb{N}: n^{2}-5 \in \mathbb{N}\right\\}\). (c) \(\left\\{n^{2}+1: n \in \mathbb{N}\right\\}\) (d) \(\left\\{n \in \mathbb{N}: n=k^{2}+1\right.\) for some \(\left.k \in \mathbb{N}\right\\}\).

Consider the statement, "For all natural numbers \(n,\) if \(n\) is prime, then \(n\) is solitary." You do not need to know what solitary means for this problem, just that it is a property that some numbers have and others do not. (a) Write the converse and the contrapositive of the statement, saying which is which. Note: the original statement claims that an implication is true for all \(n,\) and it is that implication that we are taking the converse and contrapositive of. (b) Write the negation of the original statement. What would you need to show to prove that the statement is false? (c) Even though you don't know whether 10 is solitary (in fact, nobody knows this), is the statement "if 10 is prime, then 10 is solitary" true or false? Explain. (d) It turns out that 8 is solitary. Does this tell you anything about the truth or falsity of the original statement, its converse or its contrapositive? Explain. (e) Assuming that the original statement is true, what can you say about the relationship between the set \(P\) of prime numbers and the set \(S\) of solitary numbers. Explain.

Let \(A_{2}\) be the set of all multiples of 2 except for \(2 .\) Let \(A_{3}\) be the set of all multiples of 3 except for \(3 .\) And so on, so that \(A_{n}\) is the set of all multiples of \(n\) except for \(n,\) for any \(n \geq 2 .\) Describe (in words) the set \(\overline{A_{2} \cup A_{3} \cup A_{4} \cup \cdots}\)

Find the following cardinalities: (a) \(|A|\) when \(A=\\{4,5,6, \ldots, 37\\}\). (b) \(|A|\) when \(A=\\{x \in \mathbb{Z}:-2 \leq x \leq 100\\}\). (c) \(|A \cap B|\) when \(A=\\{x \in \mathbb{N}: x \leq 20\\}\) and \(B=\\{x \in \mathbb{N}: x\) is prime \(\\}\).

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