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Chapter 2: Problem 42

Let \(f\) be a function from the set \(A\) to the set \(B .\) Let \(S\) and \(T\) be subsets of \(A .\) Show that a) \(f(S \cup T)=f(S) \cup f(T)\) b) \(f(S \cap T) \subseteq f(S) \cap f(T)\)

Short Answer

Expert verified
a) \( f(S \cup T) = f(S) \cup f(T) \); b) \( f(S \cap T) \subseteq f(S) \cap f(T) \).

Step by step solution

01

Define the problem

Given a function \( f: A \rightarrow B \) and subsets \( S, T \subseteq A \), we need to show: a) \( f(S \cup T) = f(S) \cup f(T) \) and b) \( f(S \cap T) \subseteq f(S) \cap f(T) \).
02

Prove part (a)

To show \( f(S \cup T) = f(S) \cup f(T) \):- Let \( y \in f(S \cup T) \). By definition, there exists some \( x \in S \cup T \) such that \( f(x) = y \). According to the definition of the union, \( x \in S \) or \( x \in T \). This means \( y \in f(S) \cup f(T) \). Hence, \( f(S \cup T) \subseteq f(S) \cup f(T) \).- Conversely, let \( y \in f(S) \cup f(T) \). Then there exists some \( x_S \in S \) such that \( f(x_S) = y \), or \( x_T \in T \) such that \( f(x_T) = y \). In either case, \( x \in S \cup T \), hence \( y \in f(S \cup T) \). So, \( f(S) \cup f(T) \subseteq f(S \cup T) \).Since both inclusions hold, \( f(S \cup T) = f(S) \cup f(T) \).
03

Prove part (b)

To show \( f(S \cap T) \subseteq f(S) \cap f(T) \):- Let \( y \in f(S \cap T) \). By definition, there exists some \( x \in S \cap T \) such that \( f(x) = y \). Since \( x \in S \cap T \), it means \( x \in S \) and \( x \in T \). Thus, \( y \in f(S) \) and \( y \in f(T) \); hence, \( y \in f(S) \cap f(T) \).Thus, \( f(S \cap T) \subseteq f(S) \cap f(T) \).

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Key Concepts

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

Set Theory
Set theory is a branch of mathematical logic that studies sets, which are collections of objects. These objects can be anything: numbers, letters, symbols, and even other sets. The basic operations of set theory include:
  • Union: Combining all elements from two or more sets.
  • Intersection: Finding common elements between sets.
  • Subset: Determining if all elements of one set are contained within another set.
Understanding these operations is crucial for working with more complex concepts like function mappings and subset relations. In set theory, we often use different notations to represent these operations, such as \(\cup\) for union and \(\cap\) for intersection.
Function Mapping
Function mapping involves creating a relationship between elements of two sets. If we have two sets, say \(A\) and \(B\), and a function \(f\) that maps elements from \(A\) to \(B\), we denote this as \(f: A \rightarrow B\).
  • Domain: The set from which elements are taken, here it is \(A\).
  • Codomain: The set to which elements are mapped, here it is \(B\).
The function assigns each element in \(A\) to exactly one element in \(B\). This function mapping forms the foundation for understanding how subsets of \(A\) relate to sets in \(B\) through \(f\). When dealing with subsets \(S\) and \(T\) of \(A\), functions can map their unions and intersections into corresponding unions and intersections in \(B\) through specific properties.
Subset Relations
A set \(S\) is a subset of another set \(T\) if every element in \(S\) is also in \(T\). In notation, this is expressed as \(S \subseteq T\). If \(S\) and \(T\) are subsets of \(A\), we can explore their relations through function mappings.

For example, the intersection of two subsets, \(S \cap T\), consists of elements that are in both \(S\) and \(T\). When mapped through a function \(f\), the intersection \(S \cap T\) gets a set of elements in \(B\) which is the subset of the intersection of mapped sets \(f(S) \cap f(T)\). This ensures that the relationship between the sets while mapping remains consistent.
Union and Intersection
The union and intersection operations in set theory are fundamental for combining and comparing sets. With function mapping:
  • Union Operation: For subsets \(S\) and \(T\) of \(A\), the union \(S \cup T\) represents all elements that are in either \(S\) or \(T\). Using a function \(f\), we find that \(f(S \cup T) = f(S) \cup f(T)\), indicating that the union operation distributes over function mapping.
  • Intersection Operation: The intersection \(S \cap T\) includes elements that are in both \(S\) and \(T\). Through the function \(f\), we get \(f(S \cap T) \subseteq f(S) \cap f(T)\), meaning the mapped intersection is a subset of the intersection of the mapped sets.
These relationships help in understanding how set operations translate under function mappings, maintaining logical consistency in more complex scenarios.

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Most popular questions from this chapter

Let \(a_{n}\) be the \(n\) th term of the sequence \(1,2,2,3,3,3,\) \(4,4,4,4,5,5,5,5,5,6,6,6,6,6,6, \ldots,\) constructed by including the integer \(k\) exactly \(k\) times. Show that \(a_{n}=\left\lfloor\sqrt{2 n}+\frac{1}{2}\right\rfloor\)

Let \(A\) be a \(3 \times 4\) matrix, \(B\) be a \(4 \times 5\) matrix, and \(C\) be a \(4 \times 4\) matrix. Determine which of the following products are defined and find the size of those that are defined.

Show that if \(S\) is a set, then there does not exist an onto function \(f\) from \(S\) to \(\mathcal{P}(S),\) the power set of \(S .\) Conclude that \(|S|<|\mathcal{P}(S)| .\) This result is known as Cantor's theorem. [Hint: Suppose such a function \(f\) existed. Let \(T=\\{s \in S | s \notin f(s)\\}\) and show that no element \(s\) can exist for which \(f(s)=T . ]\)

Show that if \(x\) is a real number and \(n\) is an integer, then a) \(x < n\) if and only if \(\lfloor x\rfloor < n .\) b) \(n < x\) if and only if \(n < \lceil x\rceil\)

In this exercise, we prove the Schr枚der-Bernstein theorem. Suppose that \(A\) and \(B\) are sets where \(|A| \leq|B|\) and \(|B| \leq|A| .\) This means that there are injections \(f : A \rightarrow B\) and \(g : B \rightarrow A\) . To prove the theorem, we must show that there is a bijection \(h : A \rightarrow B,\) implying that \(|A|=|B|\) To build \(h,\) we construct the chain of an element \(a \in A .\) This chain contains the elements \(a, f(a), g(f(a)), f(g(f(a))), g(f(g(f(a)))), \ldots\) It also may contain more elements that precede \(a,\) extending the chain backwards. So, if there is a \(b \in B\) with \(g(b)=a\) , then \(b\) will be the term of the chain just before \(a .\) Because \(g\) may not be a surjection, there may not be any such \(b,\) so that \(a\) is the first element of the chain. If such a \(b\) exists, because \(g\) is an injection, it is the unique element of \(B\) mapped by \(g\) to \(a ;\) we denote it by \(g^{-1}(a) .\) (Note that this defines \(g^{-1}\) as a partial function from \(B\) to \(A .\) . We extend the chain backwards as long as possible in the same way, adding \(f^{-1}\left(g^{-1}(a)\right), g^{-1}\left(f^{-1}\left(g^{-1}\left(g^{-1}(a)\right)\right), \ldots \text { To construct }\right.\) the proof, complete these five parts. a) Show that every element in \(A\) or in \(B\) belongs to exactly one chain. b) Show that there are four types of chains: chains thaform a loop, that is, carrying them forward from every element in the chain will eventually return to this element (type 1\()\) , chains that go backwards without stopping (type 2\()\) , chains that go backwards and end nin the set \(A\) (type \(3 ),\) and chains that go backwards and end in the set \(B\) (type 4\()\) . c) We now define a function \(h : A \rightarrow B .\) We set \(h(a)=\) \(f(a)\) when \(a\) belongs to a chain of type \(1,2,\) or \(3 .\) Show that we can define \(h(a)\) when \(a\) is in a chain of type \(4,\) by taking \(h(a)=g^{-1}(a)\) . In parts \((\mathrm{d})\) and (e), we show that this function is a bijection from \(A\) to \(B,\) proving the theorem. d) Show that \(h\) is one-to-one. (You can consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.) e) Show that \(h\) is onto. (You need to consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.)
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