# Intuition for order type

Here is a definition from Munkres' book "Topology":

For any two sets $A$ and $B$ with order relations $<_A$ and $<_B$, respectively, we say that $A$ and $B$ have the same order type if there exists a bijective function $f:A \to B$ such that $$a_1 <_A a_2 \implies f(a_1) <_B f(a_2)$$

My intuitive understanding of this definition is that a "larger" element in one set can stay "larger" in another set under some bijective map $f$, but then I feel that I am not really understanding what it means by a "type" of the order as I am having hard time understanding the following excerpt :

Let $Z_{+}$ denote the set of positive integers. Then, it can be easily checked that: $$Z_+ \tag 1$$ $$\{1,...n\} \times Z_+ \tag 2$$ $$Z_+ \times Z_+ \tag 3$$ $$Z_+ \times (Z_+ \times Z_+) \tag 4$$

How do I check for instance that (1) and (2) have different order types, and what is the idea for different types?

• What ordering do you use for $\{1,2,\dots,n\}\times Z_+$? – TomGrubb May 20 '15 at 0:31
• I think Dictionary ordering – user74261 May 20 '15 at 0:34

Lets say we give $\{1,2,\dots,n\}\times Z_+$ the ordering $$(a,b)<(c,d)$$ if $a<c$ or if $a=c$ and $b<d$. Also we must assume $n>1$.
Assume we had an order preserving bijection $\phi$ from $Z_+$ to $\{1,2,\dots,n\}\times Z_+$. As $(1,1)$ is the minimal element of $\{1,2,\dots,n\}\times Z_+$, it follows that $\phi(1)=(1,1)$. Now if $2$ did not get sent to $(1,2)$, then since $\phi$ is bijective and since $\phi(1)=(1,1)$, then $\phi(a)=(1,2)$ for some $a>2$. But now $\phi(2)>\phi(a)$, so indeed it must be that $\phi(2)=(1,2)$ for $\phi$ to be order preserving. Continuing in this fashion, we can see that the only way for $\phi$ to be a bijection and be order preserving is to have $\phi(a)=(1,a)$ for all $a$ in $Z_+$. This is clearly not a bijection (as $n>1$), so such a $\phi$ cannot exist. Thus the sets have different order types.