# Immediate succesors in $[0,1)\times \mathbb{Z}_+$ and $\mathbb{Z}_+\times [0,1)$

I am working on one example from Munkres book "Topology"and I would like to clarify one question.

Example: Consider the set $[0,1)$ of real numbers and the set $\mathbb{Z}_+$ of positive integers, both in their usual orders; give $\mathbb{Z}_+\times [0,1)$ the dictionary order. This set has the same order type as the set of nonnegative reals; the function $$f(n\times t)=n+t-1$$ is the required bijective order-preserving correspondence. On the other hand, the set $[0,1)\times \mathbb{Z}_+$ in the dictionary order has quite a different order type; for example, every element of this ordered set has an immediate successor.

My questions:

1) I've checked that $[0,1)\times \mathbb{Z}_+$ has the same order type as the set of nonnegative reals, right?

2) Any element $(t,n)$ from $[0,1)\times \mathbb{Z}_+$ has immediate successor, namely $(t,n+1)$. Right?

3) But elements in $\mathbb{Z}_+\times [0,1)$ have not immediate successors, right?

1. No. You checked that $\mathbb{Z}_+\times[0,1)$ has the same order type as the set of nonnegative reals.
2. Right.
3. Right, for every element of that set.
• Thanks for answer! Regarding my first question: I guess that the set $[0,1)\times \mathbb{Z}_+$ has the same order type as the set of nonnegative reals because i can consider the bijective function $f((t,n))=n+t-1$. What's wrong? – ZFR Sep 2 '18 at 22:10
• This function dois not preserve the order. For instance, $(0,2)<\left(\frac12,1\right)$, but $f(0,2)>f\left(\frac12,1\right)$. – José Carlos Santos Sep 2 '18 at 22:18
• Ahhh! I did not check it.. oops. Thanks for clarifying! – ZFR Sep 2 '18 at 22:22

The order type of the dictionary order depends on which way round the two factors are, because one of the factors has a much greater effect on the ordering than the other. Going by the convention used in your example (so that $\mathbb{Z}_+\times [0,1)$ is isomorphic to the non-negative reals) we see that $[0,1) \times \mathbb{Z}_+$ is quite different because increasing the value of the component in $[0,1)$ a little bit now ensures that many copies of $\mathbb{Z}$ will be included between the two elements, while increasing the component in $\mathbb{Z}$ by one only steps up by one element rather than an entire copy of $[0,1)$.

• Could you clarify youar answer, please. It's a bit hard to understand. – ZFR Sep 2 '18 at 22:47