# Math Help - poset is a lattice

1. ## poset is a lattice

Q)determine whether the poset ({1,2,4,8,16},|)
solution :
i made a hasse diagram of this question but i dont understand how this diagram becomes a lattice.i dont under about least upper bound and lesat lower bound ,,,can 1 is least lower bound and 16 is least upper bound

2. ## Re: poset is a lattice

its greates upper bound

3. ## Re: poset is a lattice

Originally Posted by annie12
Q)determine whether the poset ({1,2,4,8,16},|)
solution :
i made a hasse diagram of this question but i dont understand how this diagram becomes a lattice.i dont under about least upper bound and lesat lower bound ,,,can 1 is least lower bound and 16 is least upper bound
It is the least upper bound and greatest lower bound.

The Hasse diagram is a lattice.

4. ## Re: poset is a lattice

but can you explain me what is meant by least upper bound and gretest lower bound

5. ## Re: poset is a lattice

Let $A = \{1,2,4,8,16\}$. Let $B \subseteq A$ be a nonempty subset. The least upper bound and greatest lower bound of $B$ is found by a two step process. First, let $U = \{x \in A \mid \forall b \in B, b|x\}$ and $L = \{x \in A \mid \forall b \in B, x|b\}$. The set $U$ is the set of upper bounds of $B$ and the set $L$ is the set of lower bounds of $B$. To find the least upper bound and greatest lower bound, let $U_2 = \{x \in U \mid \forall u \in U, x|u\}$ and let $L_2 = \{x \in L \mid \forall l \in L, l|x\}$. If $U_2$ is nonempty, it must contain exactly one element. Same for $L_2$. If $U_2$ contains an element, that is the least upper bound. If $L_2$ contains an element, that is the greatest lower bound. In a lattice, both $U_2$ and $L_2$ will be nonempty for any choice of a nonempty subset $B \subseteq A$.

6. ## Re: poset is a lattice

its difficult ,can you explain me with the lattice i have given

7. ## Re: poset is a lattice

Suppose $B = \{2\}$. Then $U = \{2,4,8,16\}$ since $2|2, 2|4, 2|8, 2|16$ and $L = \{1,2\}$ since $1|2, 2|2$. Then $U_2 = \{2\}$ (so, 2 is the least element of $\{2,4,8,16\}$) since $2|2,2|4,2|8,2|16$, but 4 does not divide 2, 8 does not divide 2, and 16 does not divide 2. $L_2 = \{2\}$ (so, 2 is the greatest element of $\{1,2\}$) because $1|2, 2|2$, but 2 does not divide 1. This means the least upper bound and greatest lower bound of 2 are both 2.

Suppose $B = \{2,16\}$. Then $U = \{16\}$ since $2|16, 16|16$, but for any other element of $A$, 16 would not divide it. Hence, 16 is the only upper bound of $\{2,16\}$. Obviously, $U_2 = \{16\}$, so that is the least upper bound. Then $L = \{1,2\}$ and $L_2 = \{2\}$, so 2 is the greatest lower bound.

Suppose $B = \{1,4,8\}$. Then the least upper bound will be 8 and the greatest lower bound will be 1.

Etc.