# Dictionary order relations of $\Bbb{Z}_+\times [0,1)$ and $[0,1)\times \Bbb{Z}_+$

Actually, I am reading this context from the book “Topology” by James Munkres. First of all, I should recall the definition of Dictionary Order Relation, which goes as follows-

Suppose, $A, B$ are two sets with simple(or linear) order relation $<_A$ and $<_B$ respectively. Define an order relation $<$ on $A\times B$ by defining $$a_1\times b_1 <a_2\times b_2$$ if $a_1<_A a_2$, or if $a_1=a_2$ and $b_1<_B b_2$. It is called the the dictionary order relation on $A\times B$.

Now, then the author gives us examples of dictionary order relations on the sets $\Bbb{Z}_+\times [0,1)$ and $[0,1)\times \Bbb{Z}_+$. And gives us two figures to visualize these two dictionary order relations and the difference between that.

According to the book, the set $[0,1)\times \Bbb{Z}_+$ in the dictionary order has a quite different order type; for example, every element of this ordered set has an immediate successor.
But I cannot visualize these two dictionary order relations in detail. If $p$ is a point on $[0,1)\times \Bbb{Z}_+$, then $p$ is less than WHAT POINTS IN $[0,1)\times \Bbb{Z}_+$? I cannot understand the fact "every element of this ordered set($[0,1)\times \Bbb{Z}_+$) has an immediate successor".
Can anybody explain the dictionary order relation on $[0,1)\times \Bbb{Z}_+$ and the differnce between $\Bbb{Z}_+\times [0,1)$ and $[0,1)\times \Bbb{Z}_+$ in detail?

• Why do you need to visualize things? It's just a huge inconvenience having to visualize everything in mathematics. You have definitions. Use them. – Asaf Karagila Jul 23 '18 at 13:54

When you want to visualize the dictionary order (or as it is probably best known, the lexicographic order) one good trick is to think that the order is obtained by first "looking up2 and then to the right. So, in the order $\mathbb{N}\times \mathbb{Z}$, the elements greater than $(-1,0)$ will be $$(-1,1),(-1,2),\ldots,(0,0),(0,1),(0,2),\ldots,(1,0),(1,1),(1,2),\ldots.$$

In the structure, $(0,1]\times \mathbb{Z}_+$, the succesor of a given point $p=(r,n)$ will be obtained looking up, and it will be $(r,n+1)$. On the contrary, in the order $\mathbb{Z}_+\times (0,1]$, the successor of a point $p=(n,r)$ would be given by $(n,s)$ where $s$ is the next element after $r$ in the order $(0,1]$. Since $(0,1]$ is a dense linear order, such element $s$ cannot exists, and therefore $p=(n,r)$ does not have a successor in $\mathbb{Z}_+\times (0,1]$.

• Dario G, okay, you cleared my confusion. – Biswarup Saha Jul 23 '18 at 14:13
• I think the last $(0,1]\times \Bbb Z_+$ should be $\Bbb Z_+\times (0,1]$ instead – Alessandro Codenotti Jul 23 '18 at 16:39
• Corrected! Thanks – Darío G Jul 23 '18 at 16:42

order. N is the positive integers.

(0,1,2) × N looks like 1,2,3,... 1,2,3,... 1,2,3,...
N × (0,1,2) looks like 0,1,2, 0,1,2, 0,1,2, ...