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Chapter 2: Problem 9

Let \(S=\\{a, b, c, d, e\\}\) and define \(f: S \rightarrow g(S)\) by \(f(a)=\\{a, e\\}, f(b)=\) \(\\{a, c, d\\}, f(c)=\\{b, d\\}, f(d)=\varnothing\), and \(f(e)=\\{c, d, e\\}\). (a) Find the set \(T=\\{x \in S: x \notin f(x)\\}\). (b) Note that \(T \notin \mathrm{mg} f\). Is it possible to find some function \(g: S \rightarrow \mathscr{( S )}\) such that \(T \in \mathrm{rng} g\), where \(T=\\{x \in S: x \notin g(x)\\}\) ? t

Short Answer

Expert verified
The set \(T = \{b, c, d\}\), as these elements are not in their respective output sets after applying the function \(f\). There does not exist a function \(g: S \rightarrow \mathscr{P}(S)\) such that \(T\) is in the range of \(g\), as it would create a contradiction due to Russell's paradox.

Step by step solution

01

To find the set T, we need to check for each element x in set S, if x is not in the set f(x). We are given the function f and its outputs for each input in set S. Let's verify if each x is not in f(x): 1. \(x = a\) : Since f(a) = {a, e}, a is in f(a). So, a 鈭� T. 2. \(x = b\) : Since f(b) = {a, c, d}, b is not in f(b). So, b 鈭� T. 3. \(x = c\) : Since f(c) = {b, d}, c is not in f(c). So, c 鈭� T. 4. \(x = d\) : Since f(d) = 鈭�, d is not in f(d). So, d 鈭� T. 5. \(x = e\) : Since f(e) = {c, d, e}, e is in f(e). So, e 鈭� T. After evaluating each element in S for the given condition, we find that T = {b, c, d} as these are the elements not in their respective sets after applying the function f. #b. Determine if function g exists#

We need to check if there exists a function g: S 鈫� 鈩�(S) such that T is in rng g and follows the condition: T={x 鈭� S: x 鈭� g(x)}. It seems like this question is asking if we can find a function g that satisfies the conditions similar to part (a). The current function f does not satisfy this condition as T is not included in the range of f. The range of f includes the following sets: rng f = {f(a), f(b), f(c), f(d), f(e)} = {{a, e}, {a, c, d}, {b, d}, 鈭�, {c, d, e}} Since T is not in rng f, there is no way to apply f to obtain T. Thus, we need to analyze the possibility of modifying the given function f or creating a new function g that fulfills the criteria for T. However, finding such a function g is where the theorem of Russell's paradox can be applied, which is a fundamental paradox in set theory. It implies that for set S and function g: S 鈫� 鈩�(S), there will always be some set T that cannot belong to the range of g, as this would create a contradiction. Therefore, it is not possible to find a function g: S 鈫� 鈩�(S) that satisfies the given conditions for T.

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Key Concepts

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

Functions
Functions are fundamental elements in mathematics. They create a mapping between two sets where each input from the first set corresponds to one unique output in the second set. For example, in the given exercise, the function \( f \) maps elements from set \( S \) to the power set of \( S \), denoted as \( \mathscr{P}(S) \), which includes all possible subsets of \( S \). When defining a function like \( f \), it鈥檚 crucial to note that:
  • Each element in the domain (set \( S \) here) must be paired with exactly one element in the codomain (power set of \( S \)).
  • The function鈥檚 domain in the exercise is the set \( \{a, b, c, d, e\} \) and the function provides specific subsets of \( S \) as outputs.
This one-to-one relationship is what underlies set theory-based function mapping, ensuring each input is well-defined to a single output subset.
Range
The range of a function, also known as the image, is the set of all possible outputs the function can produce. In the exercise, the range of the function \( f \) consists of the subsets of \( S \) that \( f \) maps to:\[ \text{rng } f = \{ \{a, e\}, \{a, c, d\}, \{b, d\}, \emptyset, \{c, d, e\} \} \]Here, you see that \( f \) creates specific subsets as its output for each element in \( S \). Each subset forms a part of the range as it reflects possible outcomes the function \( f \) might yield.To determine if a particular subset, such as \( T = \{b, c, d\} \), belongs to the range, it鈥檚 crucial to inspect each output individually. If the specific output set does not appear in the function outputs, as is the case with \( T \), it means \( T \) is not in the range of \( f \). This understanding helps in exploring whether new functions can be defined to include \( T \) within their range.
Russell's paradox
Russell's paradox is a famous concept in set theory that reveals a fundamental problem with naive set definitions, particularly when defining sets that contain themselves. The paradox highlights limitations in defining functions such as \( g: S \rightarrow \mathscr{P}(S) \) that aspire to produce a set \( T \), where \( T = \{ x \in S : x otin g(x) \}\). This paradox arises when we try to construct \( T \) such that \( T \) must precisely avoid being in \( g(x) \); doing so creates a contradiction: if \( T \) is part of \( g(x) \), then by definition, \( T \) shouldn't be, and vice versa. This contradiction suggests it's impossible to find a function \( g \) where \( T \) is part of its range without contradicting itself. Thus, Russell's Paradox elegantly demonstrates the challenges within set theory, particularly when self-referential sets come into play, and why certain function mappings like \( g \) cannot be defined to include sets like \( T \) in their range.

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Most popular questions from this chapter

Let \(S\) and \(T\) be nonempty sets. Prove that \(|T| \leq|S|\) iff there exists a surjection \(f: S \rightarrow T\). ??

Use the axiom of regularity to show that there cannot exist three sets \(w, x\), and \(y\) such that \(w \in x, x \in y\), and \(y \in w\).

Mark each statement True or False. Justify each answer. (a) The Zermelo-Fraenkel axioms are widely accepted as a foundation for set theory. (b) Russell's paradox conflicts with the axiom of separation. \({ }^{\dagger}\) What we are presenting here is only a special case of their more general theorem, but this is sufficient to make our point. For a good discussion of this and related paradoxes, see the article by Blumethal (1940) and the book by Wagon (1993).

Mark each statement True or False. Justify each answer. (a) The axiom of regularity rules out the possibility that a set is a member of itself. (b) The Banach-Tarski paradox uses the axiom of choice.

Use the axiom for regularity to show that for any set \(x, x \cup\\{x\\} \neq x\).
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