Wikipedia defines Order isomorphic as:
Whenever two posets are order isomorphic, they can be considered to be "essentially the same" in the sense that one of the orders can be obtained from the other just by renaming of elements.
and in the example section it has this example:
Negation is an order isomorphism from $(\mathbb {R} ,\leq ) \; to \; (\mathbb {R} ,\geq ) (where \; {\mathbb {R} }$ is the set of real numbers and $\leq$ denotes the usual numerical comparison), since −x ≥ −y if and only if x ≤ y.
and with that in mind I developed a simple example and I would like you guys to see if that is isomorphic here its:
$$P = (\mathbb {Z}^+,\ge)$$
$$Q = (\mathbb {Z}^+,\le)$$
Is it right to say that $P$ and $Q$ are isomorphic to each other ?
Best Answer
No, $(\mathbb Z^+,\geq)$ and $(\mathbb Z^+,\leq)$ are not isomorphic. For notational convenience to avoid confusion about the orders, let us refer to the first as $(P,\leq_P)$ - that is, $P$ is the set of positive integers and $x\leq_P y$ if and only if $x\geq y$ in the usual order. Likewise, let $(Q,\leq_Q)$ be the second example where $x\leq_Q y$ if and only if $x\leq y$ in the usual order.
The idea of "isomorphism" means that every property that can be expressed purely in terms of the poset structure must be shared by both. In particular, we can see this is false: $Q$ has a minimal element (since $1\leq_Q x$ for all $x$), however, clearly $P$ does not have a minimal element (since $x\not\leq_P x+1$ for all $x$). Thus, the sets cannot be isomorphic.
To formalize this, you would want to consider the idea of an order isomorphism - which is just a bijective map preserving order. Suppose that $f:Q\rightarrow P$ was an order isomorphism. What could $f(1)$ possibly be? Since $1\leq_P x$ for all $x\in Q$, we would need $f(1)\leq_Q f(x)$ for all $x\in Q$ implying $f(1)\leq_Q y$ for all $y\in P$, which is clearly not true of any candidate for $f(1)$.