Order type of $\mathbb{Z_+} \times \mathbb{Z_+}$ and $\mathbb{Z_+}\times \mathbb{Z_+}\times\mathbb{Z_+}$ with dictionary order are not the same.

Question

In Topology, Munkres mentions that the sets $\mathbb{Z_+} \times \mathbb{Z_+}$ and $\mathbb{Z_+}\times \mathbb{Z_+}\times\mathbb{Z_+}$, both ordered with the dictionary order, are well ordered sets with different order types.

I was able to show that $\mathbb{Z_+}$ and $\mathbb{Z_+} \times \mathbb{Z_+}$ have different order types this way: If $f: \mathbb{Z_+} \rightarrow \mathbb{Z_+} \times \mathbb{Z_+}$ is a bijective map between the sets, let $f(a)=(1,1)$ and $f(b) = (2,1)$ where $a<b$. As there are only finitely many numbers between $a$ and $b$, but infinitely many between $(1,1)$ and $(2,1)$, this cannot be order preserving.

I tried to do something similar for the sets $\mathbb{Z_+} \times \mathbb{Z_+}$ and $\mathbb{Z_+}\times \mathbb{Z_+}\times\mathbb{Z_+}$, but I wasn't able to arrive at a contradiction.

Answer 1

Consider the increasing sequence $a_n=(1, n, 1) \in \Bbb{Z}_+^3$. It has the following properties:

1. It has an upper bound in $\Bbb{Z}_+^3$, namely $(2, 1, 1)$.
2. For any $n$, there are infinitely many elements of $\Bbb{Z}_+^3$ between $a_n$ and $a_{n+1}$.

These properties would both be preserved by any order-preserving bijection between $\Bbb{Z}_+^3$ and $\Bbb{Z}_+^2$. On the other hand, no increasing sequence in $\Bbb{Z}_+^2$ can have both of these properties (if a sequence has property 2., then its first coordinate must be unbounded, which means the sequence itself is unbounded). So there can be no such bijection.