Digital Line Topology: Show Odd Integers are Dense in \mathbb{Z}

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Homework Statement


Show that the set of odd integers is dense in
the digital line topology on \mathbb{Z}

The Attempt at a Solution


if m in Z is odd then it gets mapped to the set {m}=> open
.
So is the digital line topology just the integers.
If I was given any 2 integers I could find an odd one in between if there is an element in between.
If I was given to consecutive integers I wouldn't be able to find an odd one in between but there are no elements in between in this set. And I thinking about this question correctly.
 
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Can you spell out what the 'digital topology' is? {m} doesn't mean much to me.
 
I think I got it figured out. thanks for having me define my terms better.
 
FYI:

I Googled "digital line topology" and found the following:

From http://www.math.csusb.edu/faculty/gllosent/About_me_files/555-Chapter2.pdf:
Example 1.10.

For each n\in\mathbb{Z}\,, de fine:

\displaystyle \textit{B}(n) =\left\{\matrix{\{ n\},\ & \text{if }\ n \text{ is odd.} \\ \ \\ \{ n-1,\,n,\,n+1\}, & \text{if }\ n \text{ is even.}}\right.

Consider \displaystyle {\frak{B}}= \{B(n)|n\in\mathbb{Z}\}: a (basis of the*) digital line topology.​

* added by me, SammyS.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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