Self doubt regarding complete lattice related to https://gateoverflow.in/27341/tifr2014-b-16

Question

https://gateoverflow.in/27341/tifr2014-b-16z
In this question, why ($\mathbb{N},∣)$ is not a complete lattice?

For $any \ finite$ subset of $\mathbb{N}$, $LCM$ of its elements will be $lub$ and $HCF$ will be $glb$ and these $LCM$ and $HCF$ will also be the elements of $\mathbb{N}$.
Even for $any \  infinite$ subset, $HCF$ will be $glb$ and $0$ will be $lub$.

Then why it's not complete lattice?
The only reason I could come up with is they might not considering $0 \ \epsilon \ \mathbb{N}$.
Is there any other reason?

Comments

what is y here? it is multiple of x (As it is given x.z=y)

Now, see the what is definition given for complete lattice

" It is called a complete lattice if every subset has a least upper bound and greatest lower bound"

Now take a subset of natural number

$\left \{ 1,2,3,8,9,12 \right \}$

Here 2,3 has no LUB

rt?

So, every subset of natural number cannot be a lattice

Hence, it is not a complete lattice for every subset

got it?

---

No. Srestha
LUB will be defined over ${1,2,3,8,9,12}$ not just on 2,3.
Moreover, it's LUB will be LCM of {1,2,3,8,9,12} i.e and 72 ϵ N.
LUB need not be an element of that subset but It must belong to the given set.

---

@Soumya

try to understand

The given definition is we need to check every subset of the set

And for that set may not 72 be present

Then how it will be complete?

rt?