1.3. Meets and Joins#

See also Join and meet.

Exercise 1.80.

  1. Because all elements in the set are greater than zero.

  2. Because among the set of all lower bounds, it is the greatest.

Exercise 1.85.

For 1., we have that for all \(a \in A\) (only one element) we have \(p \leq a\), satisfying the first requirement. The second requirement is that for all \(q\) such that \(q \leq a\) for all \(a \in A\), we have that \(q \leq p\). This requirement allows for other \(q\) that are equivalent to \(p\).

In 2. we remove the possibility of equivalent sets of elements by making the preorder skeletal.

The answer to 3. is yes.

Exercise 1.90.

The meet is the greatest common divisor and the join is the least common multiple.

Exercise 1.94.

There are only two possible worlds to consider; that \(a \leq b\) or that \(b \leq a\). If the first is true, then by the definition of \(f\) as a monotone map we have that \(a \leq_P b\) implies \(f(a) \leq_Q f(b)\). This implies the expression \(f(a) \vee f(b)\) is equivalent to \(f(b)\), which is \(\leq\) the expression \(f(a \vee b) = f(b)\). If the second case (\(b \leq a\)) is true, then again by the definition of a monotone map we have that \(b \leq a\) implies \(f(b) \leq f(a)\), the left hand expression is \(f(a)\), and the right hand expression is \(f(a \vee b) = f(a)\).