# 1.3. Meets and Joins

## Contents

# 1.3. Meets and Joins#

See also Join and meet. Itâ€™s hard to remember these terms; for people to â€śmeetâ€ť simply means that they come together and for objects to â€śjoinâ€ť also means they simply come together. Like conjunction and disjunction, the words donâ€™t really indicate that one option is better or worse. The author tries to comment on terminology at the bottom of Example 1.88, which only helps a bit.

One way to remember is to say them in order of low to high, just like you would list numbers (0,1,2,3 â€¦). In that case youâ€™d want to say â€śmeet and joinâ€ť rather than â€śjoin and meetâ€ť as Wikipedia does. Unfortunately thatâ€™s not also alphabetical order (like conjunction and disjunction). Another reason to say â€śmeets and joinsâ€ť rather than â€śjoin and meetâ€ť is that limits correspond to meets and colimits to joins, and weâ€™ll almost always say â€ślimits and colimitsâ€ť rather than â€ścolimits and limitsâ€ť when weâ€™re on that topic. Also, MJ was a famous basketball player and this section is on page 23.

## 1.3.1 Definition and basic examples#

*Exercise* 1.80.#

Because all elements in the set are greater than zero.

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

Reveal answer

Search Meet and join (not plural); consider making a redirect for â€śmeets and joinsâ€ť as well, which makes sense to search given multiple (isomorphic) meets and joins may exist.

*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.

Reveal answer

*Exercise* 1.90.#

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

Reveal answer

## 1.3.2 Back to observations and generative effects#

*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 \(f(a \vee b) = f(a)\).

Reveal answer