1. ## Poset Hasse Diagram

Hi again, I am stuck on another question so if anyone can help me that would be greatly appreciated.

For part a I think the matrix is:

$\left[ \begin{array}{cccccc} 1 & 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right]$

because the ordered pairs are:

(a,a), (a,c), (a,d), (a,e), (b,b), (b,d), (b,e), (b,f), (c,c), (c,e), (c,f), (d,d), (d,f), (e,e), (f,f)

Is this correct?

And I can't seem to get anything for part b and c. Can anyone help me here?

Thanks

2. the ordered pairs are:

(a,a), (a,c), (a,d), (a,e), (b,b), (b,d), (b,e), (b,f), (c,c), (c,e), (c,f), (d,d), (d,f), (e,e), (f,f)
You missed (a,f), and also in the matrix.

For (b), try finding the supremum of a and b.

For (c), the dimension is at least 2 because otherwise the order would have been linear. It is indeed possible to find a realizer consisting of two total orders. This concept is new to me, and I don't know any theorems about it, so I used some trial-and-error. One of the orders I have is a < c < b < e < d < f.

3. Originally Posted by emakarov
You missed (a,f), and also in the matrix.

For (b), try finding the supremum of a and b.

For (c), the dimension is at least 2 because otherwise the order would have been linear. It is indeed possible to find a realizer consisting of two total orders. This concept is new to me, and I don't know any theorems about it, so I used some trial-and-error. One of the orders I have is a < c < b < e < d < f.
oh yes of course, (a,f). Thanks for that.

For b) I have tried to find the supremum of a and b but have had no luck. Do you know how to do it?

For c) I understand the order you gave, but I am still having trouble finding the second. My theory here is a little shaky. Maybe a<d<b<e<c<f ?

Thanks for any help.

4. For b) I have tried to find the supremum of a and b but have had no luck. Do you know how to do it?
There are three elements that are >= both a and b (i.e., upper bounds of a and b): d, e and f. The supremum of a and b (the least upper bound), if it exists, is the least of these three, i.e., it is <= d, e and f. However, the set {d, e, f} does not have the least element.

For c) I understand the order you gave, but I am still having trouble finding the second. My theory here is a little shaky. Maybe a<d<b<e<c<f ?
It can't be this because both in the first order that I suggested:

a < c < b < e < d < f (*)

and in your variant, a < b; therefore, in the intersection we also have a < b. I suggest writing five pairs of incomparable elements and make sure that in the second linear order these pairs occur in the reverse order compared to (*). For example, in the second order it must be the case that b < a and b < c.

5. It can't be this because both in the first order that I suggested:

a < c < b < e < d < f (*)

and in your variant, a < b; therefore, in the intersection we also have a < b. I suggest writing five pairs of incomparable elements and make sure that in the second linear order these pairs occur in the reverse order compared to (*). For example, in the second order it must be the case that b < a and b < c.
One should check the following. There are 15 unordered pairs {x, y} where $x,y\in\{a,b,c,d,e,f\}$ and $x\ne y$. Ten pairs of comparable elements must occur in the same order in both linear orders, i.e., if x < y or y < x, then (x, y) occur in the first linear order iff it occurs in the second linear order. Five pairs of incomparable elements must occur in reverse order in the two linear order, i.e., if $x\not and $y\not, then (x,y) occurs in the first linear order iff (y,x) occurs in the second linear order.