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Exercise 70 - Chapter 1

edited June 2018 in Exercises

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 \(\bigvee\)?

Original poset version.

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

  • 1.

    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? :-)

    Comment Source: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? :-)
  • 2.
    edited May 2018

    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 \(p \le q\).

    Comment Source: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\\).
  • 3.

    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 \( p \lor p = p \land p = p \)

    Comment Source: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 \\)
  • 4.

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

    \(p \leq p\) 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 \(\bigvee A = p\)

    Comment Source:\\(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\\)
  • 5.
    edited April 2018

    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 \(a \neq b\).

    Comment Source: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\\).
  • 6.
    Comment Source:The preorder version is here https://forum.azimuthproject.org/discussion/2064
  • 7.
    edited May 2018

    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 \( \bigvee \)?

    Comment Source: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 \\)?
  • 8.
    edited May 2018

    Jonathan Castello 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.

    Comment Source:[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.
  • 9.
    edited May 2018

    Michael Hong 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

    Comment Source:[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)
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