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# Exercise 58 - Chapter 1

edited June 2018

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.

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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
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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.