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Order Topology

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I've got the following quote from James Munkers book on general topology

The positive integers \(\displaystyle \mathbb{Z}^+\) form an ordered set with a smallest element .The order topology on \(\displaystyle \mathbb{Z}^+\) is the discrete topology , for every one-point set is open : if n>1 then the one-open set \(\displaystyle \{n\}= (n-1,n+1) \) is a basis element ; and if n=1 , the one-point set \(\displaystyle \{1\}=[1,2)\) is a basis element .
I can't understand the two marked phrases :confused:
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
I've got the following quote from James Munkers book on general topology

The positive integers $\mathbb{Z}^+$ form an ordered set with a smallest element. The order topology on $\mathbb{Z}^+$ is the discrete topology, for every one-point set is open : if $n>1$ then the one-open set $\{n\}=(n−1,n+1)$ is a basis element ; and if $n=1$ , the one-point set $\{1\}=[1,2)$ is a basis element.
I can't understand the two marked phrases :confused:
Well if you have a topological space $X$, then the order topology on $X$ is given by $\tau=\{(a,b):a,b\in X,\,a<b\}$. If we let $X=\mathbb{Z}^+=\{1,2,\ldots\}$, then these open sets in the order topology would be intervals that contain positive integers. Furthermore, we see that we can rewrite each interval as sets of integers: i.e. $(1,5)=\{2,3,4\}$, $(1,2) = \emptyset$, $(10,12)=\{11\}$, $(6,9)=\{7,8\}$, etc. Hence, each interval is a collection of integers, and is furthermore an element from $\mathcal{P}(\mathbb{Z}^+)$ (the power set of $\mathbb{Z}^+$). So it turns out that $\tau=\{(a,b):a,b\in \mathbb{Z}^+,\,a<b\} =\mathcal{P}(\mathbb{Z}^+)$, and hence the order topology on $\mathbb{Z}^+$ is the same as the discrete topology on $\mathbb{Z}^+$.

If $n>1$, it should be clear as to why $(n-1,n+1)$ are basis elements for the topology. But none of these particular basis elements can equal $\{1\}$! Since you restricted yourself to positive integers, we can't write $(0,2)=\{1\}$ since $0\notin\mathbb{Z}^+$. However, there must be a way to express $\{1\}$ as some combination of $B_i\in\mathcal{B}_{\tau}$ since $\{1\}\in\tau$. Thus, the only way to have $\{1\}\in\mathcal{B}_\tau$ (and hence in $\tau$) is to include 1 as an endpoint of some interval. Hence, $[1,2)=\{1\}$ does the job. Is this an open set though? Well, yes, since singleton sets are open by default.

I hope this clarifies somethings; I'm still not sure my explanation for the second part is good enough, but I'll think about it more in the next couple hours and update if necessary.
 
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