Challenge Micromass' big counterexample challenge

[WWGD]

That is basically what @Samy_A constructed, just with a much shorter argument.

Didn't he remove whole intervals? I am saying:
Take x irrational, then $\mathbb R$ -{$x$} is your desired open set .

EDIT: It is open as the complement of the closed set {x}, and clearly $\mathbb Q \subset G \subset \mathbb R$.

 

[mfb]

Didn't he remove whole intervals?

If you take the union of all his sets (to get the final set) you should get the same with both approaches.

 

[zwierz]

I hope some of these statements were surprising to some of you

o yes, especially for those who did not read Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis. :)