Natural Number, Integer Number Density in Real Number

shahar said:
How can I prove that: Z & N has no density in R?
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Lets be clear on what you mean.
The concept of density is a topological idea. So in \(\displaystyle \mathcal{R}\) the basic open sets are generated by the open intervals \(\displaystyle (a,b)\).
Definition: A set \(\displaystyle \mathcal{A}\) is dense in \(\displaystyle \mathcal{R}\) provided each open set \(\displaystyle \mathcal{O}\subset\mathcal{R}\) contains a point of \(\displaystyle \mathcal{A}\).
The interval \(\displaystyle (0,1)\) is open in \(\displaystyle \mathcal{R}\) what about that open set with respect to \(\displaystyle \mathbb{N}~\&~\mathbb{Z}~?\)
 
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?
 
shahar said:
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?
Click to expand...
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 \(\displaystyle X\) to prove that \(\displaystyle A\subset X\) is not a dense subset it is
then sufficient to show that there exists a basic open set \(\displaystyle \mathcal{O}\subset X\) such that \(\displaystyle \mathcal{O}\cap A=\emptyset .\)[/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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