Order Type of $\mathbb Z_+ \times \{1,2\}$ and $\{1,2\} \times \mathbb Z_+$ I'm currently working on §10 of "Topology" by James R. Munkres. I've got a problem with task 3:

Both $\{1,2\} \times \mathbb Z_+$ and $\mathbb Z_+ \times \{1,2\}$ are well-ordered
  in the dictionary order. Do they have the same order type?

$A := \{1,2\} \times \mathbb Z_+$
$B := \mathbb Z_+ \times \{1,2\}$
They have the same order type if there is an order preserving bijection between them.
Since both have the same cardinality, I could construct a function
$f: A \to B$
$f(\min A) = \min B$
$f(\,\min\,(A-\{\min A\})\,) = \min\,(B - \{\min B\})$
and so forth.
This function preserves the order. Now my study partner disagrees with this, because f reaches every element of B whose second component is 1, but not the others. So the function would not be surjective. But f being injective would surely imply different cardinalities for A and B.
Can you tell us the correct solution?
 A: No. This is not the correct solution. Note that $\Bbb Q$ and $\Bbb Z$ have the same cardinality. Does that mean that they have the same order types too?
Let me give you a hint, one of these orders has an element which is not a minimum, nor it is a successor; the other one does not.
A: Because every element with first coordinate $1$ always lies before any element with first coordinate $2$, the set $A$ looks like two copies of $\mathbb{Z}^+$, one after the other:
$$(1,0),(1,1),(1,2),\ldots,(2,0),(2,1)(2,2),\ldots$$
While we can order $B$ as
$$(0,1),(0,2),(1,1),(1,2),(2,1),(2,2),\ldots$$
which looks just like $\mathbb{Z}^+$ (we just double all the points).
So intuitively we expect the order types to be different.
If $f: A \rightarrow B$ is an order preserving bijection, then suppose $f(2,0) = (n,i)$, for some $n \in \mathbb{Z}^+, i \in \{1,2\}$. But $(2,0)$ has infinitely many predecessors, but no element in $B$ has. Contradiction, as $f$ should be a bijection between the set of predecessors.
