Problem 9 - Analysis with an introduction to proof

Question

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

Step-by-Step Solution

Answer

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 1: Find T = {x ∈ S : x ∉ f(x)}

Check each element:

• \(a\): \(f(a) = \{a,e\}\), \(a \in f(a)\), so \(a \notin T\)
• \(b\): \(f(b) = \{a,c,d\}\), \(b \notin f(b)\), so \(b \in T\)
• \(c\): \(f(c) = \{b,d\}\), \(c \notin f(c)\), so \(c \in T\)
• \(d\): \(f(d) = \varnothing\), \(d \notin f(d)\), so \(d \in T\)
• \(e\): \(f(e) = \{c,d,e\}\), \(e \in f(e)\), so \(e \notin T\)

\(T = \{b, c, d\}\)

Step 2: Verify T is not in the range of f

The range of \(f\) is: \(\{a,e\}, \{a,c,d\}, \{b,d\}, \varnothing, \{c,d,e\}\).
\(T = \{b,c,d\}\) does not appear in this list, so \(T \notin \text{rng}(f)\). \(\checkmark\)

Step 3: Show it is impossible to redefine f to include T

If we tried to define \(f(x_0) = T = \{b,c,d\}\) for some \(x_0 \in S\), we would need to reconstruct \(T\). But \(T\) is defined based on \(f\) itself. Changing \(f\) at \(x_0\) would change \(T\), leading to a contradiction (this is a finite version of Cantor's diagonal argument).

Key Concepts

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’s 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’s 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’s 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.