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Options

Let \((P, \le)\) be a **preorder** and \(p \in P\) an element.
Consider the set \(A = \{p\}\) with one element.

1) Show that \(\bigwedge A \cong p\).

2) Show that if \(P\) is in fact a poset then \(\bigwedge A = p\).

3) Are the analogous facts true with \(\bigwedge\) is replace by \(\bigvee\)?

Discussed in comments 1 - 6.

Let \((P, \le)\) be a **poset** and \(p \in P\) an element.
Consider the set \(A = \{p\}\) with one element.

Show that \(\bigvee A = p\).

Does \(\bigwedge A = p\)?

## Comments

Let \(a \in P\), then \(a = p\) by definition of \(A\). Therefore \(p \leq a, \forall a \in A\). Since \(A\) is a single element set, there are no other elements in \(q \in A\) such that \(a \leq a, \forall a \in A\), therefore \(\bigvee A = p\).

By the same reasoning, I believe \(\bigwedge A = p\), but both the wording of the questions and the fact that I'm a category theory newbie make me doubt myself a little. Anyone want to boost my confidence and tell me I'm right? :-)

`Let \\(a \in P\\), then \\(a = p\\) by definition of \\(A\\). Therefore \\(p \leq a, \forall a \in A\\). Since \\(A\\) is a single element set, there are no other elements in \\(q \in A\\) such that \\(a \leq a, \forall a \in A\\), therefore \\(\bigvee A = p\\). By the same reasoning, I believe \\(\bigwedge A = p\\), but both the wording of the questions and the fact that I'm a category theory newbie make me doubt myself a little. Anyone want to boost my confidence and tell me I'm right? :-)`

James - you're definitely on the right track, but your logic is a bit shaky at one point. You're trying to show that if \(A = \{p\}\) then the least upper bound of \(A\) is \(p\).

I think you showed that \(p\) is an upper bound of \(A\), although you didn't actually say that.

Then, to show \(p\) is the

leastupper bound of \(p\), you need to show that for all \(q \in P\), if \(q\) is an upper bound of \(A\) then \(p \le q\).You mentioned that there are no other elements \(a \in A\) other than \(a = p\). This is true. But the poset \(P\) may have lots of other elements. So, you need to consider any \(q \in P\) that's an upper bound of \(A\) and show \(p \le q\).

`James - you're definitely on the right track, but your logic is a bit shaky at one point. You're trying to show that if \\(A = \\{p\\}\\) then the least upper bound of \\(A\\) is \\(p\\). I think you showed that \\(p\\) is an upper bound of \\(A\\), although you didn't actually say that. Then, to show \\(p\\) is the _least_ upper bound of \\(p\\), you need to show that for all \\(q \in P\\), if \\(q\\) is an upper bound of \\(A\\) then \\(p \le q\\). You mentioned that there are no other elements \\(a \in A\\) other than \\(a = p\\). This is true. But the poset \\(P\\) may have lots of other elements. So, you need to consider any \\(q \in P\\) that's an upper bound of \\(A\\) and show \\(p \le q\\).`

I'd say \( p \le p \) is true, meaning it must be a meet of A. (too trivial?)

Example 1.62 makes it clear that \( p \lor p = p \land p = p \)

`I'd say \\( p \le p \\) is true, meaning it must be a meet of A. (too trivial?) Example 1.62 makes it clear that \\( p \lor p = p \land p = p \\)`

\(A = \{p\} \subseteq P\)

\(p \leq p\) by reflexivity

Suppose we have a \(q\) such that \(q \leq a, \forall a \in A\)

Then \(q \leq p\)

So \(\bigwedge A = p\)

Suppose we have a \(q\) such that \(a \leq q, \forall a \in A\)

Then \(p \leq q\)

So \(\bigvee A = p\)

`\\(A = \\{p\\} \subseteq P\\) \\(p \leq p\\) by reflexivity Suppose we have a \\(q\\) such that \\(q \leq a, \forall a \in A\\) Then \\(q \leq p\\) So \\(\bigwedge A = p\\) Suppose we have a \\(q\\) such that \\(a \leq q, \forall a \in A\\) Then \\(p \leq q\\) So \\(\bigvee A = p\\)`

In the latest draft, this exercise is now #66. The text has also changed, making \(P\) a preorder instead of a poset. But this seems to make the exercise false (unless we treat all meets as the same): consider the two-element poset \(P = \{a, b\}\) with \(a \le b\) and \(b \le a\). Then the set \(\{a\}\) has two possible meets, one of which is \(b\), and \(a \neq b\).

`In the latest draft, this exercise is now #66. The text has also changed, making \\(P\\) a preorder instead of a poset. But this seems to make the exercise false (unless we treat all meets as the same): consider the two-element poset \\(P = \\{a, b\\}\\) with \\(a \le b\\) and \\(b \le a\\). Then the set \\(\\{a\\}\\) has two possible meets, one of which is \\(b\\), and \\(a \neq b\\).`

The preorder version is here https://forum.azimuthproject.org/discussion/2064

`The preorder version is here https://forum.azimuthproject.org/discussion/2064`

I'll move the preorder version here along with all the comments, and then delete it.

This was another version of the same exercise, but with "preorder" replacing "poset". This causes some problems, which are worth understanding:

Exercise 66 (preorder version).Let \((P,\le)\) be a preorder and \(p \in P\) an element. Consider the set \(A = \{p\} \) with one element. Show that$$ \bigwedge A = p .$$ Is the same true when \( \bigwedge \) is replaced by \( \bigvee \)?

`I'll move the preorder version here along with all the comments, and then delete it. This was another version of the same exercise, but with "preorder" replacing "poset". This causes some problems, which are worth understanding: **Exercise 66 (preorder version).** Let \\((P,\le)\\) be a preorder and \\(p \in P\\) an element. Consider the set \\(A = \\{p\\} \\) with one element. Show that $$ \bigwedge A = p .$$ Is the same true when \\( \bigwedge \\) is replaced by \\( \bigvee \\)?`

Jonathan Castello responded as follows to the preorder version of Exercise 66:

He was right! In a preorder we can't talk about "the" meet of a subset; there may be several meets. If \(a\) and \(b\) are two meets of a subset, we have both \(a \le b\) and \(b \le a\), but we can't always conclude \(a = b\), unless we're in a poset.

`[Jonathan Castello](https://forum.azimuthproject.org/discussion/2034/introduction-jonathan-castello/p1) responded as follows to the preorder version of Exercise 66: > Is this technically true in arbitrary preorders? It would seem that the two-element preorder suggested by Exercise 65 is a counterexample. > Or is the equality relation here meant to be \\(\le \cap \ge\\)? He was right! In a preorder we can't talk about "the" meet of a subset; there may be several meets. If \\(a\\) and \\(b\\) are two meets of a subset, we have both \\(a \le b\\) and \\(b \le a\\), but we can't always conclude \\(a = b\\), unless we're in a poset.`

Michael Hong responded as follows:

`[Michael Hong](https://forum.azimuthproject.org/discussion/1855/introduction-michael-hong/p1) responded as follows: > Through reflexivity, you get the relation \\(p \leq p\\). Therefore \\(\wedge A = p=\vee A\\). > Another way to show this might be to use transitivity instead. So using reflexivity, we have \\(p \leq p\\) and using transivity, we also get \\(p \leq p \leq p\\) giving us a transitive identity triangle of \\(p\\). Then, taking joins and meets of this identity preorder, we get \\(p \wedge p = p\\) and \\(p \vee p = p\\). > ![exercise66](http://aether.co.kr/images/exercise66.svg)`