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Exercise 58 - Chapter 1

edited June 2018 in Exercises

Check the two claims made in Proposition 1.53.

Proposition 1.53. For any preorder \( (P, \le_P ) \), the identity function is monotone.

If \( (Q, \le_Q ) \) and \( (R, \le_R) \) are preorders and \( f : P \rightarrow Q \) and \( g : Q \rightarrow R \) are monotone, then \( ( f .g) : P \rightarrow R \) is also monotone.

Comments

  • 1.

    Let \(x_p\) and \(y_p\) be elements of \(P\) such that \(x_p\leq y_p\).

    Since \(f\) is monotone, it maps to elements \(x_q\) and \(y_q\), where \(x_q\leq y_q\)

    Monotone function \(g\) maps \(x_q\) and \(y_q\) to \(x_r\) and \(y_r\) such that \(x_r\leq y_r\)

    Thus, \( ( f .g) : P \rightarrow R \) maps \(x_p\) and \(y_p\) to \(x_r\) and \(y_r\), since \(x_p\leq y_p\) and \(x_r\leq y_r\), \( ( f .g) \) is monotone

    Comment Source:Let \\(x_p\\) and \\(y_p\\) be elements of \\(P\\) such that \\(x_p\leq y_p\\). Since \\(f\\) is monotone, it maps to elements \\(x_q\\) and \\(y_q\\), where \\(x_q\leq y_q\\) Monotone function \\(g\\) maps \\(x_q\\) and \\(y_q\\) to \\(x_r\\) and \\(y_r\\) such that \\(x_r\leq y_r\\) Thus, \\( ( f .g) : P \rightarrow R \\) maps \\(x_p\\) and \\(y_p\\) to \\(x_r\\) and \\(y_r\\), since \\(x_p\leq y_p\\) and \\(x_r\leq y_r\\), \\( ( f .g) \\) is monotone
  • 2.

    Looks good, Deepak! Like 99.5% of mathematicians, I usually use the notation \(g \circ f\) or just \(gf\) for the composite of functions \(f : P \to Q\) and \(g : Q \to R\), but Fong and Spivak use \( (f.g) \), which has certain advantages.

    Comment Source:Looks good, Deepak! Like 99.5% of mathematicians, I usually use the notation \\(g \circ f\\) or just \\(gf\\) for the composite of functions \\(f : P \to Q\\) and \\(g : Q \to R\\), but Fong and Spivak use \\( (f.g) \\), which has certain advantages.
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