subset of R uncountable and not dense

[bluemath]

Hello,

I'm trying to find a subset of R uncountable, which is not dense and with no interval (it exist)

R/Q is not ok because it's dense then I think the solution is to take R/Q and remove some elements but which ?

Thanks

 

[Ishuda]

Hello,

I'm trying to find a subset of R uncountable, which is not dense and with no interval (it exist)

R/Q is not ok because it's dense then I think the solution is to take R/Q and remove some elements but which ?

Thanks

Let \(\displaystyle S\, \subset\, \mathbb{R}\). Now, by dense, do you mean not dense in S, i.e. \(\displaystyle \exists\, x;\, x \in\, S\, and\, \, \exists \) a neighborhood N of x such that there is not a point of S in N? If so, take as a hint that there are just "as many points" between 0 and 1 as there are between \(\displaystyle -\infty\, and\, \infty\). So, what would happen if you joined two such well separated intervals?

Or, do you mean not dense in \(\displaystyle \mathbb{R}\), i.e. ... not a point of \(\displaystyle \mathbb{R}\) in N? If so, that is a horse of a different colour and all I can do is wish you luck in finding such.

 

[bluemath]

Let \(\displaystyle S\, \subset\, \mathbb{R}\). Now, by dense, do you mean not dense in S, .

Exactly that actually.

 

[Ishuda]

Exactly that actually.

O.K. You might look at
https://en.wikipedia.org/wiki/Smith–Volterra–Cantor_set

 

[bluemath]

Thanks a lot !

This example is really nice, wonderful job.

 

[pka]

This example is really nice, wonderful job.

However, in at least two posts you indicated that S should not be dense-in-itself.
The Cantor Set is dense-in-itself.

Let \(\displaystyle S_0=(0,\pi)\cup(\pi,2\pi)\), \(\displaystyle S_1=\mathbb{R}\setminus S_0\), and \(\displaystyle S=S_1\setminus\mathbb{Q}\)

Is it clear that \(\displaystyle \pi\in S~?\). Is it clear that \(\displaystyle S\) is totally disconnected ?

Can \(\displaystyle S\) be dense-in-itself?