#poset #lattice

Question

Let A = {$2^n | n$ is a positive integer}. A relation R on A is defined by $a^Rb$ $\iff$ a is a divisor of b.

Then the set A with respect to R is_________.

(A) a poset but not a lattice

(B)  a lattice but not a distributive lattice

(C) a distributive lattice but not bounded lattice

(D) not a poset.

Comments

A is a poset:

• Reflexive  because $2^n$ is divisor of $2^n$.
• Antisymmetric because if $2^n$ is divisor of $2^m$, and $2^m$ is divisor of $2^n$, then definitely $2^n$ = $2^m$.
• Transitive because if $2^n$ is divisor of $2^m$ and $2^m$ is divisor of $2^l$, then $2^n$ is also divisor of $2^l$. Take any value for $n,m$ and $l$.

A is also a lattice:
Suppose $2^p$ and $2^q$ are two elements, then,
LUB =  $max(2^p, 2^q)$ and GLB = $min(2^p, 2^q)$

If you draw Hasse diagram of A, it forms a chain with no upper bound and every chain is distributive
Now, this lattice is definitely not a bounded lattice because greatest element is not bounded.
Proof: Assume that $2^k$ is greatest element. But then $2^{k+1}$ will be a greater element than $2^k$. Hence, this lattice is not bounded.

So I think option C is correct.
(Please correct me if I made a mistake.)

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Option C ???

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Yes.

Correct.