How to show that $(S\cup\{0\},\ge)$ is order-isomorphic to $(S,\ge)$?

Question

Let $S$ denote the set containing all the natural numbers that are not divisible by $ 2 $.

And define the binary relation $ \ge $ on two natural number $ m , n $ , $m \ge n $ if $m = k n $ , for integer $k$, meaning that $ n $ divides $ m $.

How to show that $(S\cup\{0\},\ge)$ is order-isomorphic to $(S,\ge)$ ?

Answer 1

In that case, you can't prove it because they are not order-isomorphic.

Proof. Suppose that they are order-isomorphic. Then, there exists a bijection $f\colon S\cup\{0\}\to S$ such that $x\geq y\iff f(x)\geq f(y)~\forall~x,y\in S\cup\{0\}$.

Now, since $0$ is divisible by every integer (since $0=0\times a$ for all integers $a)$, we have $0\geq a~\forall~a\in S\cup\{0\}$. So, we have $f(0)\geq f(a)~\forall~a\in S\cup\{0\}$, i.e., $f(0)\geq k~\forall~k\in\textrm{Im}(f)=S$.

Since $f(0)\in S$, this means that there must exist some element in $S$ (which would be $f(0))$ which is divisible by every element in $S$. Since $0\in\textrm{Dom}(f)$, we know that $f(0)$ exists, say $f(0)=m$.

Now, since $f(0)=m\in S$, by definition of $S$, we get that $m+2\in S$ but note that $m$ is not divisible by $m+2$ (obvious), i.e., $m=f(0)\not\geq m+2$ which is a contradiction.

Hence, our assumption is wrong and $f$ doesn't exist.

Hence, we conclude that $(S\cup\{0\},\geq)$ is not order-isomorphic to $(S,\geq)$

An informal argument that one could use to show the same fact is that an order-isomorphism preserves maximum properties and since $0$ is a maximum for $(S\cup\{0\},\geq)$ whereas there is no maximum element in $(S,\geq)$, they cannot be order-isomorphic.

In fact, there's a more general theorem which gives some sufficient conditions for two posets to not be order-isomorphic. You can check it out here.

Answer 2

These poset are not order-isomorphic for the order considered.
It is enough to note that in $S \cup \{0\}$, we have $0 \geq s$, for all other $s$ since $s\cdot 0 = 0$, but in $S$ there exists no element in those conditions.

So $S \cup \{0\}$ has a minimal element—$0$—whereas $S$ doesn't.
By the way, both have infinitely many maximal elements—the prime numbers different from $2$.