I want to prove that there is no isomorphism from the poset $(\mathbb{N}, \leq_{\mathbb{N}})$ onto the poset $R = (\mathbb{N} \times \{ 0,1\}, \leq_{R})$. Here $\leq_R$ is defined as $(x_1, y_1) \leq_R (x_2, y_2) \Leftrightarrow (y_1 <_{\{0,1\}} y_2) \lor(y_1 = y_2 \wedge x_1 \leq_{\mathbb{N}} x_2)$.
My proof idea goes as follows. $(0,1) \in R$ is a limit point of $R$. That is $[\forall x \in R](x <_R (0,1) \rightarrow [\exists y \in R](x <_R y <_R (0,1))) \wedge (0,0 <_R (0,1))$. In other words $(0,1)$ is not the $<_R$-successor of any point from $R$ and is distinct from the minimal element $(0,0)$. Otherwise let $x \in R$ with $(0,1) = min\{y \in R: x <_R y \}$. $x$ is an upper bound of $\mathbb{N} \times \{ 0\}$, for if there were some $z \in \mathbb{N} \times \{ 0\}$ with $x <_R z$ we would have that $(0,1) \leq_R z$ and so $z \in \mathbb{N} \times \{ 1\}$. Since $\mathbb{N} \times \{ 0\}$ has no upper bounds it follows that $(0, 1)$ is a limit point. However $(\mathbb{N}, \leq_{\mathbb{N}})$ has no limit points. So, there can be no isomorphism from this poset onto $R$.
Is my proof idea on the right track? Thanks in advance.
I think it's better to fomulate it in purely order theoretic terms: $(0,1)$ has infinitely many predecessors (all $(n,0), n \in \Bbb N$) but no element of $(\mathbb{N}, \le_{\Bbb N})$ has infinitely many predecessors ($n$ has $n$ many predecessors). So there can be no order isomorphism.
The "limit point" is true (working in the order topology) but why get topology into an order theory problem?