Suppose we have two ordered sets, namely $\{1,2\}\times \mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$, where in both the order is dictionary.
How to prove that these ordered sets have different order types?
P.S. For example, when I was working with $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$ I have noticed that these sets have different order types since in $\mathbb{N}$ all elements have the finite number of elements which are less than this element but this property fails in $\mathbb{N}\times \mathbb{N}$.
But this reasoning cannot be applied to my initial example.
Would be very thankful for help!
In a linear order, $x$ is a successor iff $\{y:y<x\}$ has a largest member. There are exactly $2$ non-successors in $\{1,2\}\times \Bbb N.$ In $\Bbb N\times \Bbb N$ there are infinitely many non-successors. An order-isomorphic bijection from one linearly-ordered set to another must map non-successors to non-successors. So a bijection from $\Bbb N\times \Bbb N$ to $\{1,2\}\times \Bbb N$ cannot be an order-isomorphism.
For a different approach, ,the lex-orders on $\{1,2\} \times \Bbb N$ amd on $\Bbb N\times \Bbb N$ are well-orders, and $\{1,2\}\times \Bbb N$ is an initial segment of $\Bbb N \times \Bbb N.$ That is, $\{1,2\}\times \Bbb N=\{y\in \Bbb N\times \Bbb N: y<(3,1)\}.$
Let $<$ be a well-order on a set $A.$ Let $x\in A$ and let $pred_<(x)=\{y:y<A\}$ be an initial segment of $A$ ("pred" for "predecessors".) There is no order-isomorphism from $A$ to $pred_<(x).$
Proof. Suppose a bijection $f:A\to pred_<(x)$ exists. We have $f(x)<x$ so let $y_0$ be the $<$-least $y\in A$ such that $f(y)<y.$ Let $z_0=f(y_0).$ We have $z_0<y_0$ and we have $\forall y<y_0\;(f(y)\geq y).$ So $f(z_0)\geq z_0=f(y_0)$. So $z_0<y_0\land f(z_0)\geq f(y_0),$ so $f$ does not preserve order.