# Let $N$ be the set of positive integers and define $m \leq n$ to mean that $m$ divides $n$. Is $N$ a lattice?

EDT: Clarification: How could one determine if $$N$$, according to the question in the title, is a lattice?

I have made the following assumption, which I wonder whether it is correct or not:

The set $$N$$ can be defined as $$N:=\left(\mathbf{Z}_+, \mid\right)$$ where $$\mathbf{Z}_+ = \left\{1,\,2,\,3 \, ... \right\}$$.

The number $$1$$ divides every number, and any prime number (according to definition) can only be divided by $$1$$ or itself.

Thus, if we draw a fragment of the Hasse diagram (just focusing on any of the two primes $$p_i$$ and $$p_j$$) that is corresponding to the set $$N$$, it could look something like:

Considering Euclid's proof of infinite primes - also considering that in our case, there are no limitations of how great of a value a prime number can have - we can assume that e.g. $$p_i \rightarrow \infty$$.

This leads us to the conclusion that:

$$1 \cup p_i=\infty$$

$$p_i\cup p_j=p_ip_j=\infty$$ (since $$\forall \, k$$ $$p_k\geq 1$$ and $$p_i \rightarrow \infty$$)

$$\vdots$$

Since, for every pair of elements which contains $$p_i$$ as defined above, there exist no least upper bound (here is where I am very unsure though) and the set $$N$$ therefore is not fulfilling the definition of a lattice - which is that there for every pair of elements must exist a least upper bound as well as a greatest lower bound. Therefore, $$N$$ can't be a lattice.

Am I on right track?

• Every finite set of positive integers has a greatest common divisor (greatest lower bound) and a least common multiple (least upper bound). This is a lattice. – Robert Israel Feb 22 at 17:28
• You should use the body of the Question to give a self-contained statement of the problem you want help with. Here the title strongly hints at what your problem is, but the body of the post launches into an attempted solution (by contradiction?) and the thread of reason is quickly lost on Readers (because of the lack of a clear problem statement that would set up your argument). – hardmath Feb 22 at 17:32
• Okey, I've made a clarification. @hardmath – user615771 Feb 22 at 17:36
• @CarlSchmidt There is no problem that $\mathbb N$ is infinite. Only finite sets of elements are required to have an upper bound in a lattice. You are perhaps thinking of a complete lattice, which $(\mathbb N,|)$ is not. – Mike Earnest Feb 22 at 17:43
• The word "partial" in "partial order" does not refer to the fact that the order is not complete as a lattice. For example, the Boolean poset of subsets of $\mathbb N$ is partial, yet it is still a complete lattice. The fact that the question is not asking if $(\mathbb N,|)$ is a complete lattice is what tells you that infinite sets (like all $x\in \mathbb N$) do not need an upper bound. – Mike Earnest Feb 22 at 17:56