A partial order $≤$ is defined on the set $S=\left \{ x, a_1, a_2, \ldots, a_n, y \right \}$ as $x$ $\leq _{i}$ $a_{i}$ for all $i$ and $a_{i}\leq y$ for all $i$, where $n ≥ 1$. The number of total orders on the set S which contain the partial order $≤$ is
In question one typing mistake is that in place of "x <=i ai" , "x <= ai" should be written.
To make this partial order a total order, we need the relation to hold for every two element of the partial order. Currently between any $a_i$ and $a_j,$ there is no relation. So, for every $a_i, a_j,$ we have to add either $(a_i, a_j)$ or $(a_j, a_i)$ in total order. So, this translates to giving an ordering for $n$ elements between $x$ and $y,$ which can be done in $n!$ ways. So, answer is (a).
The bottom figure is for a total order. We can permute the $a_i$ from $i = 1$ to $n,$ and each permutation will also be a total order containing the given partial order.
let A={0,1,2,3,∞}
here x=0, y=∞, a1=1, a2=2, a3=3.
∞ ∞
| |
1 and 3
2 1
3 2
0 0
is a Toset is also a Toset
bcz as mentiond in question , ≤ is applicable only for (x,a1) and (a3,y), however a1, a2, a3 (1,2,3) can be permutate in any order (3!=6) as long as x has leat value and y has greatest value in the set. same thing applicable for your query, i think.
In question it is given that
x ≤ ai for all i and ai ≤ y for all i,
@Rajesh, Its mentioned in question about the partial order ≤ & analysis has to be done wrt that only. Dont change the Poset. Hope it clears your doubt.
We are doing n!, because in the question it is given that
The number of total orders on the set S which contain the partial order ≤ is
Cond 1: it should be TOSET:-
So, as per the definiition of the toset it should contain all the elements of the set,
Cond 2: it should satisfy the relation x <= ai <= y.
That's why from cond 1 we have to take all n elements with x and y, and from cond 2 we can arrange in any order i.e. n!.
Hence, it is n! is the answer.
A partial order relation is total order , when it satisfies comparibility, means $a\leqslant b$ or $b\leqslant a$ . Then how poset looking like $a_{1},a_{2},.......$ has no relation between them? example https://math.stackexchange.com/questions/599086/total-order-relation
The given POSET is drawn as shown in the best answer here. To make this into a TOSET, we have to make it a chain. So, connect all the a's together, and link one end of the "a chain" with x, and the other end with y.
The whole "a chain" can be made in $n!$ ways, and x and y are fixed.
So, possible Total orders = $n!$
Option A