Natural Number, Integer Number Density in Real Number

How can I prove that: Z & N has no density in R?
Lets be clear on what you mean.
The concept of density is a topological idea. So in R\displaystyle \mathcal{R} the basic open sets are generated by the open intervals (a,b)\displaystyle (a,b).
Definition: A set A\displaystyle \mathcal{A} is dense in R\displaystyle \mathcal{R} provided each open set OR\displaystyle \mathcal{O}\subset\mathcal{R} contains a point of A\displaystyle \mathcal{A}.
The interval (0,1)\displaystyle (0,1) is open in R\displaystyle \mathcal{R} what about that open set with respect to N & Z ?\displaystyle \mathbb{N}~\&~\mathbb{Z}~?
 
But I cannot just say that there isn't any number between 0 to 1 that hasn't have in the set.
Can I rely on the linear axis of real number and not to prove it by calculations?
There is a way to more rigorous way with calculations to prove It?
 
But I cannot just say that there isn't any number between 0 to 1 that hasn't have in the set.
Can I rely on the linear axis of real number and not to prove it by calculations?
There is a way to more rigorous way with calculations to prove It?
What I gave you is a proof.
The negation of the statement "every open set contains a point of the set", is
some open set does not contain any point of the set.
In any topological space X\displaystyle X to prove that AX\displaystyle A\subset X is not a dense subset it is
then sufficient to show that there exists a basic open set OX\displaystyle \mathcal{O}\subset X such that OA=.\displaystyle \mathcal{O}\cap A=\emptyset .[/QUOTE]
 
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To prove that something is false when the definition involved specifies all requires only a counter-example.

Consider the open interval (1, 2). Is any member of Z in that interval?

No.

But for Z to be dense in R, the definition requires that one or more members of Z must fall into any open interval. So Z does not meet the definition. If I have screwed this up somehow, pka will, I hope, correct me.
 
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