So I was thinking about the following: All real numbers are either rational or irrational. So if we take each rational number, $q$ as it's own set: $\{q\}$ and then consider the compliment of each of these sets: $\{q\}^{C}$, and intersect them: $\bigcap\limits_{q\in\mathbb{Q}}\{q\}^{C}$ Then the result will be the irrational numbers. If I am not mistaken this also means that $\{\mathbb{R} - \mathbb{Q}\}$ will be a Topology on $\mathbb{R}$.
I'm can't really wrap my head around the notation$ \cap_{q\in\mathbb{Q}} \{q\}^C$ . If we take let $q_1=\frac{1}{2}$ what is $\{1/2\}^C$?
Edit: Scratch the above, it totally makes sense.
@jmacdona I think that's exactly right!
If you got it, maybe you can explain it to me....
@jfuge $\{\frac{1}{2}\}^{C}$ = $(-\infty,\frac{1}{2})\cup(\frac{1}{2}, \infty) \subset \mathbb{R}$
Yeah it was that part that I understood.
I'm not sure that I understand how you conclude that $\{\mathbb{R}-\mathbb{Q}\}$ is a topology. Which definition do you use for topology?
@nklausen @jfuge A Topology is a collection of open sets. We have just shown that the Irrational numbers are are a collection of open sets.
I understand what A Topology is I just wasn't sure I understood ${ \frac{1}{2} }^c$
The set $\{ \frac{1}{2} \}^c = \{x\in \mathbb{R} \mid x\neq \frac{1}{2} \} = \mathbb{R} - \{ \frac{1}{2} \}$. @jfuge Does that make sense now?
Thanks for explaining this it really helped!