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Exercise 60 - Chapter 1

edited June 2018 in Exercises

Show that a skeletal dagger preorder is just a discrete preorder, and hence just a set.

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A dagger preorder is a preorder obeying the symmetry axiom:

$$ x \le y \Leftrightarrow y \le x $$ Also recall the skeletal preorders (Remark 1.26) and discrete preorders (Example 1.27):

Remark 1.26 (Partial orders are skeletal preorders).

A preorder is a partial order if we additionally have that 3. \( x \cong y \) implies \( x = y \).

Example 1.27 (Discrete preorders).

Every set \(X\) can be made into a discrete preorder. This means that the only order relations on \(X\) are of the form \( x \le x \); if \( x \neq y \) then neither \( x \le y \) or \( y \le x \) hold.

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  • 1.
    edited April 2018

    Recall that a skeletal dagger preorder is a preorder such that:

    1. \(x \le y \wedge y \le x \Rightarrow x = y\), and
    2. \(x \le y \Leftrightarrow y \le x\).

    Combining these properties(*), we have that \(x \le y \Rightarrow x = y\), which is the defining property of a discrete preorder.

    (*) Since \(x \le y\) and \(y \le x\) are equivalent by (2), we can remove \(y \le x\) from the precondition on (1). Formally, this involves substitution followed by idempotence of \(\wedge\).

    Comment Source:Recall that a _skeletal dagger preorder_ is a preorder such that: 1. \\(x \le y \wedge y \le x \Rightarrow x = y\\), and 2. \\(x \le y \Leftrightarrow y \le x\\). Combining these properties(\*), we have that \\(x \le y \Rightarrow x = y\\), which is the defining property of a discrete preorder. (\*) Since \\(x \le y\\) and \\(y \le x\\) are equivalent by (2), we can remove \\(y \le x\\) from the precondition on (1). Formally, this involves substitution followed by idempotence of \\(\wedge\\).
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